*Clarification 1:* Some of you seem to be confusing weight with _apparent_ weight. This is understandable since apparent weight is often referred to as just "weight," especially in engineering. Astronauts in space are not weightless. Objects in free fall are not weightless. They just don't have _apparent_ weight. *Clarification 2:* The two sections of the Earth at 7:14 cancel each other's effects because of a balance between mass and distance. The bottom section might be more massive, but it's _also_ farther away from my clone. The top section is less massive, but closer to my clone. It's just the right amount to make the forces the same no matter how deep the hole is. Kind of cool, huh?
So, those two clones in space... if they were completely without motion, is there any movement they could make to start moving? Basically if you were still in space is there any arm or leg movement that could get you moving?
@@jamesweeks9583 No. They can twist and contort their bodies, but the best they'll ever be able to do is turn. The only way their _center of mass_ will ever move is from an external force (a force from outside them)... like the gravity exerted by the other clone. Without that external force, they're stuck in place forever. Other examples of external forces would be some kind of rocket pack or simply throwing something opposite the direction you want to move.
Did you mean to pin this comment? Also, Einstein's General Relativity says that it's not possible to distinguish what you call "apparent weight" from just "weight".
@@Alexagrigorieff I did. TH-cam has been randomly _unpinning_ my comments. I don't know why. It's happened on several other videos over the last couple days.
@@johnclark8359 I think you're asking if the trip time is still the same. I only know one very blunt approach to that question, and it's not a very good one, so hopefully a more mathy person can do the real work. That doesn't mean we can't have fun. According to one video I just looked up, the arc length of a cycloid is equal to eight times its radius. So to make things more fun, let's say we're not even limited to a cycloid that is entirely underground. So we increase its radius so much that its path takes us away from the surface of Earth and then hopefully back. I guess at that point our tunnel is more of a pipe. Two facts arise from this: 1. If the tunnel leads away from the surface of the planet, gravity will not compel us to move in the correct direction from our starting point. 2. If there is no limit to the radius, there is no limit to the length of the tunnel, since it's just 8 times the radius. So that's the distance, right? Or maybe two of those are the distance, since we have to get back to our starting point. What happens if we take "distance equals rate multiplied by time", plug a really really high number in for distance, and hold time at 40 minutes? Eventually, your average speed would be greater than the speed of light. But even before that, your peak speed reaches the speed of light. Since that can't happen, it must not be true that my really long tunnel you weren't asking about anyway could get you to the other side in just 40 minutes. At least not if you're being observed by someone who is standing on the surface of Earth watching this happen. But perhaps you, the traveler, would observe the trip to only take 40 minutes. Relativity and stuff. Maybe that's the 40 minute figure. It's for the traveler. Unlikely, though, as my really long pipe takes you super far from Earth for a while, and it's a highly eccentric orbit, meaning at some point you're out really far and barely moving at all. It would be hard to make up for that.
@Phoenix I think it's reasonable to assume that air resistance is not part of this theoretical exercise. Because the other issue is, unless it's a diameter tunnel (straight line through the core), gravity is going to pull the falling body against the wall of the tunnel, and we'd have to factor in the friction from that. Plus, he's using the same model for the Earth tunnel as for bodies in the vacuum of space.
@Phoenix maybe if you're tossed imperfectly down a tunnel that goes straight though the core. But if the tunnel is off-center, you'd roll instead. It would feel like the tunnel is at an incline, because from a flatworld perspective, it is. That means your acceleration due to gravity would be attenuated by your moment of inertia.
@Phoenix I can see the argument for that, but at the same time, does anyone really ever want to have a discussion on semantics? It seems like one of those things that a conversation _accidentally_ turns into, with no one actually intending it to happen. All we'd have to say is that the tunnel is frictionless and evacuated of all air, and other gasses and perfectly sealed off from any other materials that may turn into gasses at low enough pressure, and it's a physics problem again.
I also use computers to do my calculations. Elsie, Sally, Gladis and Mavis have been doing it for many years! They also make a good cup of tea and bake nice cakes.
It's a really easy one to solve, actually. You can just write d²r/dt² as v•dv/dr (where v = dr/dt) and then solve for v wrt r. Once you do that, you have a 1st order differential equation for r wrt t, and it's easily solvable by simply seperating the variables.
@@TheReligiousAtheists the thing is that is just a trick. You have to solve a similar d.e. before to know to use it(unless you are a genius). Computers don't need to do the same.
@@noether9447 It's actually a pretty standard trick in solving higher order DEs. The harmonic oscillator equation comes by solving a similar DE using the same trick, so it should be pretty well-known to a professional physicist like Nick
"In Physics, sometimes it's helpful to imagine one problem as a completely different problem." And you say _mathematicians_ are the ones who like to generalize things to death? 😁
Hey Nick, 1) Thank you for still giving us content during the rest of the insanity of the outside world 2) At each step in my path as a physics student, I keep learning news ways of "we can't solve this, so we approximate it or use a nifty trick". It was nice to see this one.
9:35 I appreciate that idea the most from studying physics. That there are so many phenomena in physics that mathematically are described in exactly the same way. Like voltage and pressure. Flow and current. Etc. Really helps me to understand physics.
@@noether9447 I think it's the other way around the physical phenomenon of current on a macro scale is the same as flow on a macro scale and because of that described by the same mathematics. Supper zoom 😉 to microscopic level though and the behavior of an electron in a wire is very different than the behavior of a molecule in a pipe. For example electrons don't experience turbulence, so the analogy flow and current only works on macro level.
dependant on mathematics but has systems that are mathematically impossible! We do ok at describing things and making predictions but we cant explain nothing and have no clue whats going on! :D
Your idea of rethinking a problem as a different, easier problem really resonates with me. I’ve written computer software for many years, and taught many people how to do it. I tell everyone that one sure sign of a great software developer is judicious laziness. Thanks for your excellent videos 🙏
Hey, Can you please do videos on the following topics 1. Temperatures below absolute zero 2. Gravitational waves property. If they travel at light speed, do they have other similar properties like reflection, refraction, diffraction, doppler shift polarization. What is their wavelength range? does special relativity apply to it? 3. Collapsing an air bubble with sound underneath a liquid surface 4. Doppler Shift on a single photon
I'm an out of the box thinker and sometimes I completely forget about the box, this might be a problem. Great video! This is an amazing way to visualize the falling through earth problem.
I am a highschool science student, here we are asked these kind of questions very frequently and the spring system helps reimagine countless situations which would otherwise be unsolvable, maybe you can make a video about simple harmonic motion and its practical applications!
the more i watch your videos, the more i get into phyiscs, this is my first time visualizing a 2 particle attraction as a ellipitical orbit problem and later using the earlier one as the special case of 2nd...that was awesome, love your videos, great work sir
That orbital one is really a great idea I have generally used the SHM method for periodic motion under gravity Great video as always Please continue making more
Man, I find your show amazing! The presentation style, the depth and the imagination are all really well balanced. Thanks for making us smarter ya crazy. And lot's of luck on your journey through space.
I have thought of this falling through the Earth problem abt 10 years ago as a kind of simple harmonic motion, I'm glad today 10yrs later someone confirmed it.
I had to solve this tunnel problems for preparing my competitive exams ! It was in second year of "classe préparatoire aux grandes écoles", two difficult quite general years (I wanted to enter vet school and had to do physics, maths, chemistry, litterature etc...). We had oral mock exams that were called "colles" (or "khôlles" as the litterary classes prépa jokingly wrote to seem more greek), which were kind of conversations with the teacher. For this one, we spent about an hour and a half solving it, and I was very surprised when I found out that only the Earth's density mattered, not the actual distance to travel ! It was four years ago and I'm almost a vet now, I'm happy you brought this back to my memories ;)
I've always called that the fundamental trick of Mathematics. If you have a problem that you don't know the solution to, just make it look like a problem that you do know how to solve :)
This question appeared in my physics prospectus twice, once while I was a second-year undergraduate in a mathematical physics course, and I was impressed at the derivation and homework set about. I remembered the drawing about how gravity decreased 1/r2 in space but 1/r in earth to the centre, a line. And in graduate school years later on the comprehensive exam (e.g., comps). I really like what you said in 9:05 about the Coriolis force and not drilling into the center; because during my MS exam the committee drilled the hole at one side, away from the poles, and I mathematically showed it would NOT descend. It would immediately hit the side (a friend of mind questioned me interrogatively) After I answered, I rewrote the problem, drilling the hole through the center and proceeded to derive, from first principles, and delivered a closed form solution, modeling it as a pendulum using classical dynamics. Frequency of small oscillations. My friend, mentioned above, said it would be interpreted as me being disrespectful at the committee the way I answered the question. I think they put the problem in like that on purpose! Nevertheless, I scored well on the exam and I know I answered that problem correct, and I got my vindication. Never looked back since.
I'm an out-of-the-box thinker. The diameter or outer boundary of the gravity inside an atom that holds subatomic particles together ... can grow larger than the atom, for larger elements with larger nuclei. If you can tap into this other type of gravity, you can warp the fabric of spacetime, and travel faster-than-light. = D
@@lionel4685 ... creating a warp bubble and falling on a downslope of gravity in the direction you are traveling. Light has the fastest speed within spacetime, but what if you bent spacetime itself? Hmm. What is your out-of-the-box thinking? That was the question at the end of this Science Asylum video.
Great explnations, that's why I like you videos so much. In many cases, I know the physics beforehand but still get a lot by the way you explain and give the physics better context. Thanks man! 👌
Thanks for confirming what happens when you fall through the earth, (I was guessing that gravity is effectively zero at the centre). I guess once you consider losses due to air resistance, then you wouldn't actually make it out the other side and would oscillate up and down until eventually stopping at the centre. Don't know computational / numerical methods, they are the only solution for those who can't remember all the formulas and rules in maths that solve stuff elegantly.
8:48 I think you should have been giving us the explanation for that, its not too complicated. For anyone reading this comment that wants the explanation: If you examine the force acting on you inside the earth you get a force that's proportional to the distance from the center. Then, if you chose a linear path and look at the component of that force along the path, you always get a force that is also proportional (with the same ratio) to some distance, this time its the distance to the center of the path. Therefore, regardless of the path, the system is always going to act like the same spring. Since the period of the motion of the spring only depends on it's stiffness the time will always stay the same.
I just realised I was smiling like a idiot while watching all the details and funny animations in this video.... They are great Nick. Can't even imagine the efforts you put into them.
Well, I haven't gone thru the Earth, but in 1996 I got a yachting friend to bring his fancy new GPS device to my house, added 180 degrees to both co-ordinates and got somewhere in Southern Spain. This was because I was planning a trip to Europe, in which I subsequently drove as near to that point as I could without intruding on private property. Thus, in a cabinet next to my dining table is a rock I can point to and tell a guest it is literally from "the other side of the world". I later learned of the "Earth Sandwich" concept while Blogging the 20th anniversary of the trip:- In 2006, the Internet comedian Ze Frank called for the preparation of an “Earth sandwich”-the simultaneous dropping of two pieces of bread on the ground at a pair of land antipodes, making the Earth into sandwich filling for the first time in history. Two of Frank’s fans responded to the call, one laying down bread in Spain, and the other in Auckland, New Zealand. I could have been there, dammit!
Nice "outside the box" solution. Usually solving a more general case is more difficult than solving a special case and thus not an effective type of outside the box thinking , but this is an example where it's easier and thus effective. The type of outside the box thinking that works most often for me is to begin by trying to identify unstated assumptions, so they can be scrutinized too, and possibly discarded or relaxed.
1. Weight is a function of gravitationally derived mass. 2. Mass is a derivative not an intrinsic value i.e. gravitational and inertial mass (both due to acceleration re the equivalence principle). . . You’ve confused mass and matter. 3. In free fall mass drops to zero because a body in free fall is at rest and resting bodies have no inertia aka acceleration (@ terminal velocity). But it still has matter. 4. Gravity doesn’t attract, it ______ matter. 5. The Earth isn’t a rigid sphere it’s a mollusk vide General Relativity C. 28. 6. If there were theoretically a hole through the theoretically spherical Earth, a body would fix at the center of the Earth, since that would be the center of _______ gravity. NB: I’m sorry about the blanks, but if you knew what I redacted you’d understand why.
It was bugging me too. But I guess it works if you ignore any friction when you bump into the tunnel walls & slide along them? The counterintuitive thing is that if you dig a straight tunnel parallel to a tangent to Earth's circumference, it looks as if every point was "level" - but "down" is always pointing to center of Earth - so the "straight" tunnel is actually "sloping inward", like a valley (in terms of gravity). You'd still fall/roll/slide through.
Elegant. Has been appreciated, the concept of the ISS in a state of continual free fall. See marbles transfer momentum when struck, extend this to free space where mass is fixed in by quark-antimatter-graviton into an absolute location. All photons and graviton interact by transfer of moment.
Another thing to consider would be the location of the moon during your free fall through the earth. Just as it effects water/tides it will effect your fall depending on its orientation to the axis of your tunnel and will change as it’s relative location changes during the fall. Interestingly the presence and location of the moon also means free falling objects near earth don’t all fall at 9.8 or straight down. The same considerations and more have to be made in orbital mechanics, which this problem is a type of as discussed in the beginning of the video. Great video! I think I just found a lesson plan.
Excellent lesson. About the thinking out of the box... For the tunnel through the center of the earth, the brute force method quickly gives you an equation that looks just like a spring problem. We'd probably all come up with that idea. But the orbit idea IS out of the box thinking. I've worked that problem several times and not once did I consider changing it to a special case orbiting problem. I like!!
@08:24 "And 42 minutes would still have been our answer." Are you kidding ? Everybody knows that "42 IS THE MEANING OF LIFE, THE UNIVERSE AND EVERYTHING". ! So long and thanks for all the fish !
That's just a coincidence of what our planet's properties happen to be, and how we define the minute. Try this calculation with data for a different planet, and you'll get a completely different number. This isn't like the fine structure constant, where it is a very specific rational number that shows up in a surprising place, in a way that is independent from any physical object we inherited, or any manmade unit definition.
If we ask, for example WolframAlpha r^2 * d^2r/ dt^2 = GM General analytical solution is ((r(t) sqrt(c_1 - (2 G M)/r))/c_1 + (G M ln(sqrt(c_1) r sqrt(c_1 - (2 G M)/r) + c_1 r - G M))/c_1^(3/2))^2 = (c_2 + t)^2 If we choose c_1=1 and c_2=0, we get (r sqrt(1 - (2 G M)/r) + G M ln(r sqrt(1 - (2 G M)/r) + r - G M))^2 = t^2
You don’t get to just choose those constants. They are determined by the initial conditions of the system. If you do it properly though, it might work.
Good video, and nice idea to see the problem as an orbit. But I think there are some slight errors in some of the calculations 5:10 “if we make their ellipses really skinny…the sum of the semi-major axes is half the distance between them” - I don’t understand why this would be the case. The clones are still in orbit around their centre of mass. They don’t fall to the middle, stop there magically, and then move back. Their orbits reach all the way to the other side. So the sum of the semi-major axes should be the full maximum distance between them. If I plug in the numbers, then the full orbit is approximately 32 hours. The point where they reach each other is 1/4 of that, so 8 hours. 5:43 “how long does it take for the moon to fall into the earth?…the sum of the semi-major axis is half the distance to the moon” - This seems to be the same mistake. The moon does not come to rest when reaching the Earth; if it would narrowly miss it, it would reach a point on the other side equally distant from the Earth. So the sum of the semi-major axis should be the FULL distance to the moon, not the half of it. If I do the calculation with that in mind, I get to the correct answer: 27.3 days for the full orbit (unsurprisingly the same as the moons normal orbit). And therefore 1/4 of that (6.8 days) to fall to the Earth. The final calculation for the fall through the earth correctly takes the full R as the semi-major axis, though.
Great animation, and lots of laughs. When nerd clone showed up and mentioned the earth's rotation I was busting a gut at *sigh* "he's right... "🤣 Cuz dat shit is a lot more work! Great to see your thoroughness in the coverage. Most channels would gloss over that point but no sir not here! ^_^
@@ScienceAsylum How do you explain human levitation and the observed levitation of ufos where there appears to be no prop wash? There are some good, believable youtube videos on human levitation. Then there's Crookes et al 19th century account of human levitation.
@@peterfred445 Given that they traveled thru space to get here, one would assume they got technology we don't, so anything we say would be just speculation. Maybe they're using the same kind of space warping that could be used for faster-than-light interstellar travel; maybe we're not seeing them, but a side-effect of their cloaking technology; maybe they were not really there in the first place and our memories and recording devices have been tampered with; maybe it's just rockets but using dark matter as propellant etc. There is no way to know without actually studying an alien craft directly.
This channel is such a gem, it is like a running commentary of melvin the mad's diary. This might pose a risk to the dangers of combining the batsuit with a white coat. Daily disclaimer: melvin the mage did not self levitate.
While solving a problem in which two similar situations are presented and knowing the data and result for the first situation, I have to calculate the result for the second situation for which new data is given, I think of the data of the second situation as some error has occured while measuring it and use the concept of calculation of experimental error in an result. The math still works! It's really all about our thinking. Kind of also verifies that math is made by "us"(and not the universe).
I really liked this video. I've been using it as an example for my algebra student of why its useful to learn how to write mathematical expressions in lots of different forms and recognize their equivalence -- depending on what you want to know, one form might make you spend _hours_ to get a messy answer (like your ODE) while form (similar to how you reframed the problem in this video) might let you solve it in minutes. The difference isn't usually _that_ stark, at least not at her level, but it's a good example. It also helps that she loves your channel, so anything I mention is related to one of your videos tends to be a good motivator.
"I spent like four hours trying to solve that thing symbolically before I gave up" ... Spoken by me about every undergraduate physics homework assignment ever!
I studied this in College Physics. The hole can be between any 2 points on earth. You could use it for travel. No matter the start or end, it takes 42 minutes to get there. You start at 0 mph and end at 0 mph. Perfect travel.
The final solution to the differential equation : T = (r_0^(1.5)*tan^-1(sqrt((r_0/R_e)-1))+sqrt(R_e*r_0^2-R_e^2*r_0))*(2GM)^-0.5 R_e = mass of the earth. r_0 = distance between moon and earth at T = 0.
I am definitely an out-of-the-box thinker, since as soon as I get a new box, my cat annexes it and moves in, so I have no choice but to think outside of the box.
I think it would have been beneficial to explain G= 6.67408 × 10-11 m3 / (kg s2) versus g=9.81 m / (s2). Most people watching are probably used to using "little g" to insert into equations involving gravity. Universal gravitational constant (G) is a proportional constant. It is the same in all cases. Local gravitational constant (g) is only at the surface of the earth. The surface of the moon or any other celestial body would have a different (g) at its surface unless it was the same mass as the earth.
Telling you man, the Nerd clone is my spirit animal :P Awesome video just like always! I loved the part at 7:48 were the falling clone comes from the other side of the Earth, but inhabitands of that part will see him coming upside down or... he will see them upside down... you know what I mean! That would be pretty cool lol!
For practical tunnels, you want a) to pick two points that are not going to pass into the mantle - i.e., you can draw a line that only passes through the crust - and b) you'll want the two points far enough apart that the slope of the tunnel "down" at each end is steep enough to overcome friction. Even then, obviously, you'd need some extra oomph in your propulsion to overcome the energy lost to friction, but this seems still like it'd be a very nice way to set up a rapid transit. What are the practical limits for how close together and far apart the points in question could be? Do we have a less-than-one-hour tunnel from LA to NYC possible?
5:53 ...so the ultimate question of Life, The Universe, and Everything is "How long would it take to fall through a hole through the center of the Earth?"
Good one ! ThE most interesting thing about this imo is not just that the time is the same for ALL straight line tunnels but rather that the size of the planet doesn't even matter ! Of course there are several real word issues that you mentioned but all rocky planets being about the same density is a pretty fair assumption for an ideal case. Brachistochrone tunnels are a very interesting extension to this imo.
Brilliant again (without the ™) the second part of the video also touches on why the principle of equivalence isn’t quite equivalent, which ought to be a nice stimulus to someone figuring out the implications for GR (as “an exercise for the reader”). Such excellent physics teaching here! Thank you!
@The Science Asylum thanks this works for Electromagnetic attraction too & force carriers btn 2 charged particles basically explains the Feynman Diagrams
Imagine falling for 40 minutes and when you slow down to a stop, you fall back for another 40 minutes. Also, this is like a pendulum where potential energy is converted into kinetic energy, which is then re converted into potential energy.
Hi Nick ! As per Einstein can i say that at centre of earth there is no space curvature/ "space pressure" /flat space and at surface on earth there is curved space /"space presure" ??
If you're in freefall, falling toward the center of the earth, the earth is pulling you in with an equal amount of force as your body is pulling the earth up towards you. It's the same effect that causes space probes to sap delta velocity from whatever gives them a gravity assist. If you slingshot around Jupiter you actually slow Jupiter down in order to speed up going around it.
The Science Asylum do you think you could make a video on the orbit of the Moon? It's both surprisingly complicated and interesting, I think. The whole orbit precesses around like an Euler disc every ~18.6 years (nodal precession), the actual orientation of the ellipse rotates around every ~8.85 years (apsidal precession). The eccentricity has multiple overlapping periods of oscillation, the short one being only ~6 months (see "evection"). Due to influence from the Sun, the Moon speeds up in its orbit when near new and full moon phases (see "variation"). As for the Moon itself, it appears to librate due to several mechanisms. Its rotational axis precesses much more rapidly than the Earth, also taking about 18.6 years (see Cassini's laws). There's the mechanism of tidal locking to talk about, too. All of these precise predictions of the Moon's position in the sky were essential in the 1700s and 1800s due to the lunar distance method of determining longitude, before the advent of accurate and cheap marine chronometers. Three-body problems are crazy!
That being the case, and it being filled only with air, I imagine it would probaby be so dense as to be liquid at the centre. So, yeah, that would certainly have to be factored in.
Crystal clear explanation. thank you! I take it that the same physics explains the orbits of the various planets and our Earth, around the Sun, right? Further, with orbiting systems, each body must know the position of the other body at all times, right? Thus my question: What is the velocity of the gravity communication vector; ie. how quickly does the body know of the changed position of the other body? What happens, for example, if it were to take 8 minutes for the Sun to determine that the Earth has moved its position? Would the Earth, or any other orbiting body, be able to achieve a stable orbit under these circumstances? Thanks for your time with this question. I may have a follow-up question after this one.
What I find fascinating about this is that the sum of the gravitational force on you from the various parts of the Earth farther from the center than you are balance. It's a pretty calc problem. But then, that the force generated by a sphere is the same as that by a point of the same mass at the center is a pretty calc problem. Remember, always assume a physical physicist of uniform density...
After 33 years and a week before my 34th time around the sun I finally get gravity. Omg mate thankyou soo much I have been trying to understand it for soo long you are brilliant
Its a simple harmonic motion problem. So you can find the period by using eom or considering it as a circular motions problem. Just need to find the angular frequency.
And our Earth has weight comparing with something. But it doesnt have weight comparing to itself. Formula says F=mg, m=density×volume, means F=density×volume×g. But density depends on mass of particles again. That is the loop
You are correct that weight comes from outside something. For example, our weight comes (mostly) from the Earth. The Earth's weight comes (mostly) from the Sun. It's a force. Forces are always exerted _between two objects._ If there was only one elementary particle in the universe, it wouldn't have weight.
This reminded me of a similar and more general trick I sometimes use when trying to figure out a "what will happen" physics question. First assume linearity (ie. nothing whacky happening somewhere in the middle of the thought process) and then take it to its limit which is often obvious. That gives you a trend which points to the answer. It's a bit like differential calculus which also looks at where things are headed as you approach infinity.
I Don't wanna be rude, but your solution it's more hard then solving eq at 2:06 , to use your method you need to know the period time equation of the bodies orbit, but to find this you have to solve the more complicated 2 body problem in the most general case, in 2D. But your method it's beautiful, I'll admit. The differential equation in 2:06 it's not difficult to solve, you just need to use the chain rule, divide both sides by the velocity, then your acceleration will become a dv/dr, with this you just move variables from one side to another and integrate both sides to get v as a function of r (use the initial conditions to determine the constant). Then you just have to isolate the velocity and integrate again, this last integral it's more difficult, but not too much. Plug in the right initial conditions and you're done! If it's not totally clear, the chain rule I'm referring to is: dv/dt = dv/dr * dr/dt, só dv/dr= a/v Alternatively, you can use conservation of energy and just integrate 1 time, just isolate velocity and integrate (you can see that the first step from the first method will lead you to the same equation you would get if just use energy conservation) Obs: I'm not exactly sure, but I think you forgot a minus sign in front of the GM, because acceleration will be directed towards the earth, opposite to the vector connecting earth to the body Another observation: the problem with the two bodies can move (starting from rest) it's not impossible to solve! It's basically the same as the one I talk about up here, you will just have to make a little pre-math to transform the 2 variables into just one, like x2-x1=R where x2 and x1 are the bodies position along a line connecting the two bodies One more correction: Actually, the period equation used in the video can be more or less easily obtained by Kepler's third law and newton gravitacional force law. So it's not to hard to use the method in the video
You can theoretically dig a tunnel between to points on Earth (excepting antipodes) and roll from one end to the other in somewhat less than 42 minutes. The quickest tunnel would be a hypocycloid.
I can get light to arrive instantaneously from the stars. What slows down light *from our perspective* is the gravity in the vicinity that slows down the *rate* of time. (It’s not that light slows down, it’s the rate of time that slows down.) So what happens where there is no gravity is that the rate of time speeds up relative to our rate of time so that starlight travels the same (or greater) distance at a faster rate of time. It’s still traveling the *same relativistic frame speed* but an entire second passes by from our perspective at a much faster rate, in a fraction of a second *where there is no gravity* to slow down time. So for most of the time, starlight is traveling through space at a faster rate of time relative to our rate of time. That means starlight arrives much more instantaneously because the rate of time is free from the effects of gravity for most of the way. With no mass of its own, starlight arrives instantaneously as it experiences no time of its own traveling at the speed of light. We see things in slow motion *where time is slowed down* by gravity. This relativistic effect allows for a young earth since it doesn’t take billions of lightyears for starlight to arrive through the *void* of space where there is no matter to slow down *time.*
*A few important/interesting notes:* When he says "weight", he means gravitational acceleration. If you stood on a scale in freefall, you wouldn't register any weight, no matter how far you are from Earth. Just to prevent any confusion. If Earth did not rotate, and were perfectly uniform, any hole you drill through the center of the Earth and jumped through, if you put another person in a coplanar circular orbit near Earth's surface, they would be at the *same linear position* as you along the entire orbit as you fall back and forth over and over. After all, it is called simple harmonic motion. It's circular. If I'm not mistaken, the principle he's talking about here 7:02 is called *Gauss's Law,* and it works in electromagnetism as well.
_"If you stood on a scale in free fall, you wouldn't register any weight."_ That's because scales don't actually measure weight. Also, yes, I'm talking about Gauss's law.
@@ScienceAsylum I guess it depends on your definition of weight. The way I learned it, weight is the magnitude of the normal force pushing you up against gravity. I think this definition is more useful because it explains why you weigh slightly less on an airplane flying east (which I've actually measured), even though your gravitational acceleration isn't that much lower. It goes more with people's intuitive concept of weight. Anyway, I just wanted to make sure no one would be confused. I wasn't trying to say you were wrong.
I eventually hit my kid's curiosity while re-watching this video. It was worth the effort of translating it live. Now he knows what an orbit is and why the moon does not fall onto us. I will pick another video tomorrow. Bedtime stories for nerds. Thank you!
Thanks Nick! I live watching your videos over & over again. I never get tired of them! So, I am imagining a person falling all the way through the center of the Earth & basically just oscillating from one side, to the other side & back to the other side, etc... But I am also imagining a satellite in space orbit around the Earth. & its maintaining its orbit, so its not losing nor gaining altitude. Am I correct in thinking that if the satellite is directly over the person's head when they arrive on one side of the Earth, that the satellite should, again be directly over their head when they arrive at the other side of the Earth?... Because if so, then shouldn't Pi be a factor in solving the maths to calculate the period of the person & the satellite? Because the person falling through the center of the Earth must cover approx. 3.14 times less the distance over the same time; as the satellite. So the velocity of the person falling through the center of the Earth must be less that the velocity of the satellite, if they both are to be synced up with each other with each period. Oh but wait! The satellite only goes half the circumference of the Earth as the person falls completely the diameter of the Earth ....& this is just half a period. ...oh never mind because (typing as I am thinking) I just figured out why its the "half of the halfs" thing that you talked about. So it wouldn't be Pi because twice the diameter of the Earth would equal the circumference of the Earth. & that's also why its about 42 minutes for both the satellite & the person. Ok. I answered my own question. But still thanks for all the cool videos!!!
*Clarification 1:* Some of you seem to be confusing weight with _apparent_ weight. This is understandable since apparent weight is often referred to as just "weight," especially in engineering. Astronauts in space are not weightless. Objects in free fall are not weightless. They just don't have _apparent_ weight.
*Clarification 2:* The two sections of the Earth at 7:14 cancel each other's effects because of a balance between mass and distance. The bottom section might be more massive, but it's _also_ farther away from my clone. The top section is less massive, but closer to my clone. It's just the right amount to make the forces the same no matter how deep the hole is. Kind of cool, huh?
So, those two clones in space... if they were completely without motion, is there any movement they could make to start moving? Basically if you were still in space is there any arm or leg movement that could get you moving?
@@jamesweeks9583 No. They can twist and contort their bodies, but the best they'll ever be able to do is turn. The only way their _center of mass_ will ever move is from an external force (a force from outside them)... like the gravity exerted by the other clone. Without that external force, they're stuck in place forever. Other examples of external forces would be some kind of rocket pack or simply throwing something opposite the direction you want to move.
Did you mean to pin this comment?
Also, Einstein's General Relativity says that it's not possible to distinguish what you call "apparent weight" from just "weight".
@@jamesweeks9583 They could throw their clothes away, or maybe spit (or other forms of ejection of bodily waste)
@@Alexagrigorieff I did. TH-cam has been randomly _unpinning_ my comments. I don't know why. It's happened on several other videos over the last couple days.
"There is an art, it says, or rather, a knack to flying. The knack lies in learning how to throw yourself at the ground and miss."
42.
What if the tunnel isn't straight but is a cycloid?
@@johnclark8359 then you have to say "Weeee!" the whole time. State law.
@@johnclark8359 I think you're asking if the trip time is still the same. I only know one very blunt approach to that question, and it's not a very good one, so hopefully a more mathy person can do the real work. That doesn't mean we can't have fun.
According to one video I just looked up, the arc length of a cycloid is equal to eight times its radius. So to make things more fun, let's say we're not even limited to a cycloid that is entirely underground. So we increase its radius so much that its path takes us away from the surface of Earth and then hopefully back. I guess at that point our tunnel is more of a pipe. Two facts arise from this:
1. If the tunnel leads away from the surface of the planet, gravity will not compel us to move in the correct direction from our starting point.
2. If there is no limit to the radius, there is no limit to the length of the tunnel, since it's just 8 times the radius. So that's the distance, right? Or maybe two of those are the distance, since we have to get back to our starting point. What happens if we take "distance equals rate multiplied by time", plug a really really high number in for distance, and hold time at 40 minutes? Eventually, your average speed would be greater than the speed of light. But even before that, your peak speed reaches the speed of light. Since that can't happen, it must not be true that my really long tunnel you weren't asking about anyway could get you to the other side in just 40 minutes.
At least not if you're being observed by someone who is standing on the surface of Earth watching this happen. But perhaps you, the traveler, would observe the trip to only take 40 minutes. Relativity and stuff. Maybe that's the 40 minute figure. It's for the traveler. Unlikely, though, as my really long pipe takes you super far from Earth for a while, and it's a highly eccentric orbit, meaning at some point you're out really far and barely moving at all. It would be hard to make up for that.
The image of the Earth red shifted while morphing into a tiny black hole... crazy attention to details 👏
You have the crazy “attention” to details… Nick has the crazy details.
I laughed at the clone repeatedly catching fire while falling because I am a terrible person.
I laughed as he kept throwing clones in holes.
6371 km in 20 minutes of acceleration means a top speed of 40250 km per hour, or 25000 mph
You can bet he's on fire
The clone in the hole burnt to ash. So I guess that was the ash hole clone.
@@AlleyKatt I know that guy sucks
You are a wonderful person. Anton Petrov says so.
"It's the ability to imagine one type of problem as a completely different problem."
Oh cool I'm really good at that!
"An easier problem."
Aww nuts.
😂
@Phoenix I think it's reasonable to assume that air resistance is not part of this theoretical exercise. Because the other issue is, unless it's a diameter tunnel (straight line through the core), gravity is going to pull the falling body against the wall of the tunnel, and we'd have to factor in the friction from that.
Plus, he's using the same model for the Earth tunnel as for bodies in the vacuum of space.
@Phoenix maybe if you're tossed imperfectly down a tunnel that goes straight though the core. But if the tunnel is off-center, you'd roll instead. It would feel like the tunnel is at an incline, because from a flatworld perspective, it is. That means your acceleration due to gravity would be attenuated by your moment of inertia.
@Phoenix another point for "no air resistance". It only holds us back.
@Phoenix I can see the argument for that, but at the same time, does anyone really ever want to have a discussion on semantics? It seems like one of those things that a conversation _accidentally_ turns into, with no one actually intending it to happen.
All we'd have to say is that the tunnel is frictionless and evacuated of all air, and other gasses and perfectly sealed off from any other materials that may turn into gasses at low enough pressure, and it's a physics problem again.
2:00 kudos to science asylum for trying to solve the equation symbolically for 4 hours 👏
I also use computers to do my calculations. Elsie, Sally, Gladis and Mavis have been doing it for many years! They also make a good cup of tea and bake nice cakes.
@@crouchingtigerhiddenadam1352 Who are they?
It's a really easy one to solve, actually. You can just write d²r/dt² as v•dv/dr (where v = dr/dt) and then solve for v wrt r. Once you do that, you have a 1st order differential equation for r wrt t, and it's easily solvable by simply seperating the variables.
@@TheReligiousAtheists the thing is that is just a trick. You have to solve a similar d.e. before to know to use it(unless you are a genius). Computers don't need to do the same.
@@noether9447 It's actually a pretty standard trick in solving higher order DEs. The harmonic oscillator equation comes by solving a similar DE using the same trick, so it should be pretty well-known to a professional physicist like Nick
"In Physics, sometimes it's helpful to imagine one problem as a completely different problem."
And you say _mathematicians_ are the ones who like to generalize things to death? 😁
So the answer is 42? Where have I heard that before?😉
CG Account so deepthought was right!
It should be 42 on all the planets.
Hitchhikers guide to Galaxy
@@hit3894 Finally! Thanks for all the fish!
Pangalatic gargleblaster anyone?
Hey Nick,
1) Thank you for still giving us content during the rest of the insanity of the outside world
2) At each step in my path as a physics student, I keep learning news ways of "we can't solve this, so we approximate it or use a nifty trick". It was nice to see this one.
Happy to help 😊
9:35 I appreciate that idea the most from studying physics. That there are so many phenomena in physics that mathematically are described in exactly the same way. Like voltage and pressure. Flow and current. Etc. Really helps me to understand physics.
Flow and current have such similar behaviour because they are described by the same piece of mathematics. Statistics.
@@noether9447 I think it's the other way around the physical phenomenon of current on a macro scale is the same as flow on a macro scale and because of that described by the same mathematics. Supper zoom 😉 to microscopic level though and the behavior of an electron in a wire is very different than the behavior of a molecule in a pipe. For example electrons don't experience turbulence, so the analogy flow and current only works on macro level.
@@noether9447 Don't forget Traffic :-)
@@noether9447 Flow and current have such similar behavior because they both amount to the same underlying phenomenon: electrostatic repulsion.
dependant on mathematics but has systems that are mathematically impossible! We do ok at describing things and making predictions but we cant explain nothing and have no clue whats going on! :D
LOl. Clone bursting into flames every time he passes through the core. Nice touch!
Your idea of rethinking a problem as a different, easier problem really resonates with me. I’ve written computer software for many years, and taught many people how to do it. I tell everyone that one sure sign of a great software developer is judicious laziness. Thanks for your excellent videos 🙏
I've often said 'you want to find the easiest way to do something? Get a lazy person'
@@adreanmarantz2103
"I choose a lazy person to do a hard job. Because a lazy person will find an easy way to do it." - Bill Gates
Hey,
Can you please do videos on the following topics
1. Temperatures below absolute zero
2. Gravitational waves property. If they travel at light speed, do they have other similar properties like reflection, refraction, diffraction, doppler shift polarization. What is their wavelength range?
does special relativity apply to it?
3. Collapsing an air bubble with sound underneath a liquid surface
4. Doppler Shift on a single photon
I knew that
But I wrote it because I have seen it on the Internet
Don't believe me, Google It
Seriously Nick😒
Why didn't you heart my comment?
I think you should make a hearty clone for that one
a great explanation for a tough problem
If a hole is drilled through the earth which on its own is impossible ,, objects or mass cannot pass through it period dumb guy .
Yup😊
@@8600esemusa speak English buddy
8:16 I loved that: some constant stuff multiplied by... Physics teachers should use that sentence all the time.
I'm an out of the box thinker and sometimes I completely forget about the box, this might be a problem.
Great video! This is an amazing way to visualize the falling through earth problem.
I am a highschool science student, here we are asked these kind of questions very frequently and the spring system helps reimagine countless situations which would otherwise be unsolvable, maybe you can make a video about simple harmonic motion and its practical applications!
Eddie Van Halen got famous by using harmonic motion.
Simple harmonic oskalator. Often studied as a free-body.
This is the best TH-cam channel that explains physics from very basic level.
the more i watch your videos, the more i get into phyiscs, this is my first time visualizing a 2 particle attraction as a ellipitical orbit problem and later using the earlier one as the special case of 2nd...that was awesome, love your videos, great work sir
That orbital one is really a great idea
I have generally used the SHM method for periodic motion under gravity
Great video as always
Please continue making more
During my EM lectures our lecturer told us to not think outside the box. That's too hard to solve.
Think outside the sphere!
😆
Man, I find your show amazing! The presentation style, the depth and the imagination are all really well balanced.
Thanks for making us smarter ya crazy. And lot's of luck on your journey through space.
I have thought of this falling through the Earth problem abt 10 years ago as a kind of simple harmonic motion, I'm glad today 10yrs later someone confirmed it.
I had to solve this tunnel problems for preparing my competitive exams ! It was in second year of "classe préparatoire aux grandes écoles", two difficult quite general years (I wanted to enter vet school and had to do physics, maths, chemistry, litterature etc...). We had oral mock exams that were called "colles" (or "khôlles" as the litterary classes prépa jokingly wrote to seem more greek), which were kind of conversations with the teacher. For this one, we spent about an hour and a half solving it, and I was very surprised when I found out that only the Earth's density mattered, not the actual distance to travel !
It was four years ago and I'm almost a vet now, I'm happy you brought this back to my memories ;)
Clone catching fire near centre of earth, Earth turning into a black hole when it is imagined a point mass, there little details are so cool!
I've always called that the fundamental trick of Mathematics. If you have a problem that you don't know the solution to, just make it look like a problem that you do know how to solve :)
This question appeared in my physics prospectus twice, once while I was a second-year undergraduate in a mathematical physics course, and I was impressed at the derivation and homework set about. I remembered the drawing about how gravity decreased 1/r2 in space but 1/r in earth to the centre, a line. And in graduate school years later on the comprehensive exam (e.g., comps). I really like what you said in 9:05 about the Coriolis force and not drilling into the center; because during my MS exam the committee drilled the hole at one side, away from the poles, and I mathematically showed it would NOT descend. It would immediately hit the side (a friend of mind questioned me interrogatively) After I answered, I rewrote the problem, drilling the hole through the center and proceeded to derive, from first principles, and delivered a closed form solution, modeling it as a pendulum using classical dynamics. Frequency of small oscillations. My friend, mentioned above, said it would be interpreted as me being disrespectful at the committee the way I answered the question. I think they put the problem in like that on purpose! Nevertheless, I scored well on the exam and I know I answered that problem correct, and I got my vindication. Never looked back since.
I'm an out-of-the-box thinker.
The diameter or outer boundary of the gravity inside an atom that holds subatomic particles together ... can grow larger than the atom, for larger elements with larger nuclei.
If you can tap into this other type of gravity, you can warp the fabric of spacetime, and travel faster-than-light.
= D
@@lionel4685
... creating a warp bubble and falling on a downslope of gravity in the direction you are traveling.
Light has the fastest speed within spacetime, but what if you bent spacetime itself? Hmm.
What is your out-of-the-box thinking?
That was the question at the end of this Science Asylum video.
Disclaimer : No clone has been harmed for the making of this video.
How do you know though? They're clones after all :P
Great explnations, that's why I like you videos so much. In many cases, I know the physics beforehand but still get a lot by the way you explain and give the physics better context.
Thanks man! 👌
Thanks for confirming what happens when you fall through the earth, (I was guessing that gravity is effectively zero at the centre). I guess once you consider losses due to air resistance, then you wouldn't actually make it out the other side and would oscillate up and down until eventually stopping at the centre.
Don't know computational / numerical methods, they are the only solution for those who can't remember all the formulas and rules in maths that solve stuff elegantly.
8:48
I think you should have been giving us the explanation for that, its not too complicated.
For anyone reading this comment that wants the explanation:
If you examine the force acting on you inside the earth you get a force that's proportional to the distance from the center.
Then, if you chose a linear path and look at the component of that force along the path, you always get a force that is also proportional (with the same ratio) to some distance, this time its the distance to the center of the path.
Therefore, regardless of the path, the system is always going to act like the same spring.
Since the period of the motion of the spring only depends on it's stiffness the time will always stay the same.
Thx
I just realised I was smiling like a idiot while watching all the details and funny animations in this video....
They are great Nick. Can't even imagine the efforts you put into them.
Well, I haven't gone thru the Earth, but in 1996 I got a yachting friend to bring his fancy new GPS device to my house, added 180 degrees to both co-ordinates and got somewhere in Southern Spain. This was because I was planning a trip to Europe, in which I subsequently drove as near to that point as I could without intruding on private property.
Thus, in a cabinet next to my dining table is a rock I can point to and tell a guest it is literally from "the other side of the world".
I later learned of the "Earth Sandwich" concept while Blogging the 20th anniversary of the trip:-
In 2006, the Internet comedian Ze Frank called for the preparation of an “Earth sandwich”-the simultaneous dropping of two pieces of bread on the ground at a pair of land antipodes, making the Earth into sandwich filling for the first time in history. Two of Frank’s fans responded to the call, one laying down bread in Spain, and the other in Auckland, New Zealand.
I could have been there, dammit!
Nice "outside the box" solution. Usually solving a more general case is more difficult than solving a special case and thus not an effective type of outside the box thinking , but this is an example where it's easier and thus effective. The type of outside the box thinking that works most often for me is to begin by trying to identify unstated assumptions, so they can be scrutinized too, and possibly discarded or relaxed.
1. Weight is a function of gravitationally derived mass.
2. Mass is a derivative not an intrinsic value i.e. gravitational and inertial mass (both due to acceleration re the equivalence principle). . . You’ve confused mass and matter.
3. In free fall mass drops to zero because a body in free fall is at rest and resting bodies have no inertia aka acceleration (@ terminal velocity). But it still has matter.
4. Gravity doesn’t attract, it ______ matter.
5. The Earth isn’t a rigid sphere it’s a mollusk vide General Relativity C. 28.
6. If there were theoretically a hole through the theoretically spherical Earth, a body would fix at the center of the Earth, since that would be the center of _______ gravity.
NB: I’m sorry about the blanks, but if you knew what I redacted you’d understand why.
Gotta admit, that's really genius.
Newton was the man!
Am I the only one who has a problem with "falling through the Earth" sideways? (42 minutes or not...)
It was bugging me too. But I guess it works if you ignore any friction when you bump into the tunnel walls & slide along them?
The counterintuitive thing is that if you dig a straight tunnel parallel to a tangent to Earth's circumference, it looks as if every point was "level" - but "down" is always pointing to center of Earth - so the "straight" tunnel is actually "sloping inward", like a valley (in terms of gravity). You'd still fall/roll/slide through.
I didn't get your question
Yes. We're ignoring friction in the shorter tunnels.
Imagine a train on a frictionless track. In fact, people have proposed designs for just such a method of travel. Look up "gravity train" on Wikipedia.
Just bring some rollerblades
And this is exactly why I love this channel, all about change of perspective in physics!
Elegant. Has been appreciated, the concept of the ISS in a state of continual free fall.
See marbles transfer momentum when struck, extend this to free space where mass is fixed in by quark-antimatter-graviton into an absolute location. All photons and graviton interact by transfer of moment.
I'm still inside the box. I'm not sure if I'm dead or alive. Probably both.
-"Reimagining it is the ultimate physics trick!"
yes!
This video went way more into depth of gravity from your comment on my question in your last video comments. Greatly appreciated!!!
11:25 The most feasible tunnels are thus Madrid-Auckland, Beijing-Buenos Aires and Jakarta-Bogotá. Cool fact for the nerd in us.
saw the notification and clicked with lightning speed. still watching and marvelled already
NEXT PLEASE
"Unfortunately we can't make a whole through the poles. There is no land on both sides."
The 5000K-hot Earth's Core: Am I joke to you?
Another thing to consider would be the location of the moon during your free fall through the earth. Just as it effects water/tides it will effect your fall depending on its orientation to the axis of your tunnel and will change as it’s relative location changes during the fall. Interestingly the presence and location of the moon also means free falling objects near earth don’t all fall at 9.8 or straight down. The same considerations and more have to be made in orbital mechanics, which this problem is a type of as discussed in the beginning of the video. Great video! I think I just found a lesson plan.
Excellent lesson. About the thinking out of the box... For the tunnel through the center of the earth, the brute force method quickly gives you an equation that looks just like a spring problem. We'd probably all come up with that idea. But the orbit idea IS out of the box thinking. I've worked that problem several times and not once did I consider changing it to a special case orbiting problem. I like!!
I watched this video after 3 people, after completing it. I got curious then I refreshed and there is 1157 views!!!!
@08:24 "And 42 minutes would still have been our answer." Are you kidding ? Everybody knows that "42 IS THE MEANING OF LIFE, THE UNIVERSE AND EVERYTHING". ! So long and thanks for all the fish !
That's just a coincidence of what our planet's properties happen to be, and how we define the minute. Try this calculation with data for a different planet, and you'll get a completely different number.
This isn't like the fine structure constant, where it is a very specific rational number that shows up in a surprising place, in a way that is independent from any physical object we inherited, or any manmade unit definition.
@@carultch Please watch (or read) H2G2 : "The HitchHiker Guide to the Galaxy."
If we ask, for example WolframAlpha
r^2 * d^2r/ dt^2 = GM
General analytical solution is
((r(t) sqrt(c_1 - (2 G M)/r))/c_1 + (G M ln(sqrt(c_1) r sqrt(c_1 - (2 G M)/r) + c_1 r - G M))/c_1^(3/2))^2 = (c_2 + t)^2
If we choose c_1=1 and c_2=0, we get
(r sqrt(1 - (2 G M)/r) + G M ln(r sqrt(1 - (2 G M)/r) + r - G M))^2 = t^2
You don’t get to just choose those constants. They are determined by the initial conditions of the system. If you do it properly though, it might work.
Good video, and nice idea to see the problem as an orbit.
But I think there are some slight errors in some of the calculations
5:10 “if we make their ellipses really skinny…the sum of the semi-major axes is half the distance between them” - I don’t understand why this would be the case. The clones are still in orbit around their centre of mass. They don’t fall to the middle, stop there magically, and then move back. Their orbits reach all the way to the other side. So the sum of the semi-major axes should be the full maximum distance between them. If I plug in the numbers, then the full orbit is approximately 32 hours. The point where they reach each other is 1/4 of that, so 8 hours.
5:43 “how long does it take for the moon to fall into the earth?…the sum of the semi-major axis is half the distance to the moon” - This seems to be the same mistake. The moon does not come to rest when reaching the Earth; if it would narrowly miss it, it would reach a point on the other side equally distant from the Earth. So the sum of the semi-major axis should be the FULL distance to the moon, not the half of it. If I do the calculation with that in mind, I get to the correct answer: 27.3 days for the full orbit (unsurprisingly the same as the moons normal orbit). And therefore 1/4 of that (6.8 days) to fall to the Earth.
The final calculation for the fall through the earth correctly takes the full R as the semi-major axis, though.
Flying = falling and missing. -Douglas Adams
Uncontrolled crash = ablative lithobraking -Kerbal Space Program
Great animation, and lots of laughs. When nerd clone showed up and mentioned the earth's rotation I was busting a gut at *sigh* "he's right... "🤣
Cuz dat shit is a lot more work! Great to see your thoroughness in the coverage. Most channels would gloss over that point but no sir not here! ^_^
I worked for 12 hours straight yesterday trying to get this video finished by today. Thanks for appreciating the work.
@@ScienceAsylum How do you explain human levitation and the observed levitation of ufos where there appears to be no prop wash? There are some good, believable youtube videos on human levitation. Then there's Crookes et al 19th century account of human levitation.
@@peterfred445 I can levitate for a brief moment, but I required a lot of beans to be consumed.
@@peterfred445 Given that they traveled thru space to get here, one would assume they got technology we don't, so anything we say would be just speculation. Maybe they're using the same kind of space warping that could be used for faster-than-light interstellar travel; maybe we're not seeing them, but a side-effect of their cloaking technology; maybe they were not really there in the first place and our memories and recording devices have been tampered with; maybe it's just rockets but using dark matter as propellant etc. There is no way to know without actually studying an alien craft directly.
Glad I found this channel! I'm wandering through the old material, and really enjoying it!
Out of the box thinking is what differentiates peoples that understand physics(maths or whatever) from people that knows it by heart
This channel is such a gem, it is like a running commentary of melvin the mad's diary.
This might pose a risk to the dangers of combining the batsuit with a white coat.
Daily disclaimer: melvin the mage did not self levitate.
I think I'm a "lost in the box" thinker...!
While solving a problem in which two similar situations are presented and knowing the data and result for the first situation, I have to calculate the result for the second situation for which new data is given, I think of the data of the second situation as some error has occured while measuring it and use the concept of calculation of experimental error in an result.
The math still works! It's really all about our thinking. Kind of also verifies that math is made by "us"(and not the universe).
I really liked this video. I've been using it as an example for my algebra student of why its useful to learn how to write mathematical expressions in lots of different forms and recognize their equivalence -- depending on what you want to know, one form might make you spend _hours_ to get a messy answer (like your ODE) while form (similar to how you reframed the problem in this video) might let you solve it in minutes. The difference isn't usually _that_ stark, at least not at her level, but it's a good example. It also helps that she loves your channel, so anything I mention is related to one of your videos tends to be a good motivator.
"I spent like four hours trying to solve that thing symbolically before I gave up"
... Spoken by me about every undergraduate physics homework assignment ever!
Can't imagine how you filmed the clone template...
That redshift at 6:27 was a nice touch
I've been watching these videos for a couple of years, now; and, this is my favorite, so far.
I studied this in College Physics. The hole can be between any 2 points on earth. You could use it for travel. No matter the start or end, it takes 42 minutes to get there. You start at 0 mph and end at 0 mph.
Perfect travel.
The final solution to the differential equation :
T = (r_0^(1.5)*tan^-1(sqrt((r_0/R_e)-1))+sqrt(R_e*r_0^2-R_e^2*r_0))*(2GM)^-0.5
R_e = mass of the earth.
r_0 = distance between moon and earth at T = 0.
I am definitely an out-of-the-box thinker, since as soon as I get a new box, my cat annexes it and moves in, so I have no choice but to think outside of the box.
Dang it cat!
I think it would have been beneficial to explain G= 6.67408 × 10-11 m3 / (kg s2) versus g=9.81 m / (s2).
Most people watching are probably used to using "little g" to insert into equations involving gravity.
Universal gravitational constant (G) is a proportional constant. It is the same in all cases.
Local gravitational constant (g) is only at the surface of the earth. The surface of the moon or any other celestial body would have a different (g) at its surface unless it was the same mass as the earth.
Telling you man, the Nerd clone is my spirit animal :P
Awesome video just like always!
I loved the part at 7:48 were the falling clone comes from the other side of the Earth, but inhabitands of that part will see him coming upside down or... he will see them upside down... you know what I mean! That would be pretty cool lol!
Congratulations man, this is your best video so far, very professional stuff. I start to believe you it's really OK to be a little crazy.
For practical tunnels, you want a) to pick two points that are not going to pass into the mantle - i.e., you can draw a line that only passes through the crust - and b) you'll want the two points far enough apart that the slope of the tunnel "down" at each end is steep enough to overcome friction. Even then, obviously, you'd need some extra oomph in your propulsion to overcome the energy lost to friction, but this seems still like it'd be a very nice way to set up a rapid transit. What are the practical limits for how close together and far apart the points in question could be? Do we have a less-than-one-hour tunnel from LA to NYC possible?
Excellent video Nick. You produce outstanding videos.
5:53 ...so the ultimate question of Life, The Universe, and Everything is "How long would it take to fall through a hole through the center of the Earth?"
Good one ! ThE most interesting thing about this imo is not just that the time is the same for ALL straight line tunnels but rather that the size of the planet doesn't even matter ! Of course there are several real word issues that you mentioned but all rocky planets being about the same density is a pretty fair assumption for an ideal case. Brachistochrone tunnels are a very interesting extension to this imo.
Brilliant again (without the ™) the second part of the video also touches on why the principle of equivalence isn’t quite equivalent, which ought to be a nice stimulus to someone figuring out the implications for GR (as “an exercise for the reader”). Such excellent physics teaching here! Thank you!
@The Science Asylum thanks this works for Electromagnetic attraction too & force carriers btn 2 charged particles basically explains the Feynman Diagrams
Imagine falling for 40 minutes and when you slow down to a stop, you fall back for another 40 minutes.
Also, this is like a pendulum where potential energy is converted into kinetic energy, which is then re converted into potential energy.
Hi Nick ! As per Einstein can i say that at centre of earth there is no space curvature/ "space pressure" /flat space and at surface on earth there is curved space /"space presure" ??
If you're in freefall, falling toward the center of the earth, the earth is pulling you in with an equal amount of force as your body is pulling the earth up towards you. It's the same effect that causes space probes to sap delta velocity from whatever gives them a gravity assist. If you slingshot around Jupiter you actually slow Jupiter down in order to speed up going around it.
The Science Asylum do you think you could make a video on the orbit of the Moon? It's both surprisingly complicated and interesting, I think. The whole orbit precesses around like an Euler disc every ~18.6 years (nodal precession), the actual orientation of the ellipse rotates around every ~8.85 years (apsidal precession). The eccentricity has multiple overlapping periods of oscillation, the short one being only ~6 months (see "evection"). Due to influence from the Sun, the Moon speeds up in its orbit when near new and full moon phases (see "variation"). As for the Moon itself, it appears to librate due to several mechanisms. Its rotational axis precesses much more rapidly than the Earth, also taking about 18.6 years (see Cassini's laws). There's the mechanism of tidal locking to talk about, too.
All of these precise predictions of the Moon's position in the sky were essential in the 1700s and 1800s due to the lunar distance method of determining longitude, before the advent of accurate and cheap marine chronometers.
Three-body problems are crazy!
Poincare also found problems with the 3 body problem with a view that it was physically impossible according to the current doctrine.
Great video ! You explained it very well with the graphics Well done. Take a medal out the bucket !
Assuming that the hole thru the earth isn't a perfect vacuum, wouldn't the air pressure also have to factored in?
That being the case, and it being filled only with air, I imagine it would probaby be so dense as to be liquid at the centre. So, yeah, that would certainly have to be factored in.
Most of the time I can’t even find the box, but that’s why I’m here. You’re showing me the box and then how it works. Thanks.
All your videos are excellent, but this one is the best I've seen so far.
Crystal clear explanation. thank you!
I take it that the same physics explains the orbits of the various planets and our Earth, around the Sun, right?
Further, with orbiting systems, each body must know the position of the other body at all times, right?
Thus my question: What is the velocity of the gravity communication vector; ie. how quickly does the body know of the changed position of the other body?
What happens, for example, if it were to take 8 minutes for the Sun to determine that the Earth has moved its position? Would the Earth, or any other orbiting body, be able to achieve a stable orbit under these circumstances?
Thanks for your time with this question. I may have a follow-up question after this one.
What I find fascinating about this is that the sum of the gravitational force on you from the various parts of the Earth farther from the center than you are balance.
It's a pretty calc problem. But then, that the force generated by a sphere is the same as that by a point of the same mass at the center is a pretty calc problem.
Remember, always assume a physical physicist of uniform density...
After 33 years and a week before my 34th time around the sun I finally get gravity. Omg mate thankyou soo much I have been trying to understand it for soo long you are brilliant
I have started using brilliant a week ago and I have done nearly 1k problems xD it is so much fun, I just do it for hours without pause
This is unbelievably interesting and educational.
Its a simple harmonic motion problem. So you can find the period by using eom or considering it as a circular motions problem. Just need to find the angular frequency.
And our Earth has weight comparing with something. But it doesnt have weight comparing to itself. Formula says F=mg, m=density×volume, means F=density×volume×g. But density depends on mass of particles again. That is the loop
You are correct that weight comes from outside something. For example, our weight comes (mostly) from the Earth. The Earth's weight comes (mostly) from the Sun. It's a force. Forces are always exerted _between two objects._ If there was only one elementary particle in the universe, it wouldn't have weight.
This reminded me of a similar and more general trick I sometimes use when trying to figure out a "what will happen" physics question.
First assume linearity (ie. nothing whacky happening somewhere in the middle of the thought process) and then take it to its limit which is often obvious. That gives you a trend which points to the answer. It's a bit like differential calculus which also looks at where things are headed as you approach infinity.
I Don't wanna be rude, but your solution it's more hard then solving eq at 2:06 , to use your method you need to know the period time equation of the bodies orbit, but to find this you have to solve the more complicated 2 body problem in the most general case, in 2D. But your method it's beautiful, I'll admit.
The differential equation in 2:06 it's not difficult to solve, you just need to use the chain rule, divide both sides by the velocity, then your acceleration will become a dv/dr, with this you just move variables from one side to another and integrate both sides to get v as a function of r (use the initial conditions to determine the constant). Then you just have to isolate the velocity and integrate again, this last integral it's more difficult, but not too much. Plug in the right initial conditions and you're done!
If it's not totally clear, the chain rule I'm referring to is:
dv/dt = dv/dr * dr/dt, só dv/dr= a/v
Alternatively, you can use conservation of energy and just integrate 1 time, just isolate velocity and integrate (you can see that the first step from the first method will lead you to the same equation you would get if just use energy conservation)
Obs: I'm not exactly sure, but I think you forgot a minus sign in front of the GM, because acceleration will be directed towards the earth, opposite to the vector connecting earth to the body
Another observation: the problem with the two bodies can move (starting from rest) it's not impossible to solve! It's basically the same as the one I talk about up here, you will just have to make a little pre-math to transform the 2 variables into just one, like x2-x1=R where x2 and x1 are the bodies position along a line connecting the two bodies
One more correction: Actually, the period equation used in the video can be more or less easily obtained by Kepler's third law and newton gravitacional force law. So it's not to hard to use the method in the video
To each their own 🤷♂️
I hate how simple the answers are versus how complicated I think they are.
That ellipsis analogy blew my mind!!!
You can theoretically dig a tunnel between to points on Earth (excepting antipodes) and roll from one end to the other in somewhat less than 42 minutes. The quickest tunnel would be a hypocycloid.
I can get light to arrive instantaneously from the stars. What slows down light *from our perspective* is the gravity in the vicinity that slows down the *rate* of time. (It’s not that light slows down, it’s the rate of time that slows down.)
So what happens where there is no gravity is that the rate of time speeds up relative to our rate of time so that starlight travels the same (or greater) distance at a faster rate of time. It’s still traveling the *same relativistic frame speed* but an entire second passes by from our perspective at a much faster rate, in a fraction of a second *where there is no gravity* to slow down time. So for most of the time, starlight is traveling through space at a faster rate of time relative to our rate of time.
That means starlight arrives much more instantaneously because the rate of time is free from the effects of gravity for most of the way. With no mass of its own, starlight arrives instantaneously as it experiences no time of its own traveling at the speed of light. We see things in slow motion *where time is slowed down* by gravity.
This relativistic effect allows for a young earth since it doesn’t take billions of lightyears for starlight to arrive through the *void* of space where there is no matter to slow down *time.*
42 minutes is SHOCKINGLY fast, and really gives a sense of just how fast those velocities get!
*A few important/interesting notes:* When he says "weight", he means gravitational acceleration. If you stood on a scale in freefall, you wouldn't register any weight, no matter how far you are from Earth. Just to prevent any confusion.
If Earth did not rotate, and were perfectly uniform, any hole you drill through the center of the Earth and jumped through, if you put another person in a coplanar circular orbit near Earth's surface, they would be at the *same linear position* as you along the entire orbit as you fall back and forth over and over. After all, it is called simple harmonic motion. It's circular.
If I'm not mistaken, the principle he's talking about here 7:02 is called *Gauss's Law,* and it works in electromagnetism as well.
_"If you stood on a scale in free fall, you wouldn't register any weight."_
That's because scales don't actually measure weight.
Also, yes, I'm talking about Gauss's law.
@@ScienceAsylum I guess it depends on your definition of weight. The way I learned it, weight is the magnitude of the normal force pushing you up against gravity. I think this definition is more useful because it explains why you weigh slightly less on an airplane flying east (which I've actually measured), even though your gravitational acceleration isn't that much lower. It goes more with people's intuitive concept of weight. Anyway, I just wanted to make sure no one would be confused. I wasn't trying to say you were wrong.
Yeah, I think it's the difference between a physicist's definition and an engineer's definition.
I eventually hit my kid's curiosity while re-watching this video. It was worth the effort of translating it live. Now he knows what an orbit is and why the moon does not fall onto us. I will pick another video tomorrow. Bedtime stories for nerds. Thank you!
That’s wonderful!
Thanks Nick! I live watching your videos over & over again. I never get tired of them!
So, I am imagining a person falling all the way through the center of the Earth & basically just oscillating from one side, to the other side & back to the other side, etc...
But I am also imagining a satellite in space orbit around the Earth. & its maintaining its orbit, so its not losing nor gaining altitude.
Am I correct in thinking that if the satellite is directly over the person's head when they arrive on one side of the Earth, that the satellite should, again be directly over their head when they arrive at the other side of the Earth?...
Because if so, then shouldn't Pi be a factor in solving the maths to calculate the period of the person & the satellite?
Because the person falling through the center of the Earth must cover approx. 3.14 times less the distance over the same time; as the satellite.
So the velocity of the person falling through the center of the Earth must be less that the velocity of the satellite, if they both are to be synced up with each other with each period.
Oh but wait!
The satellite only goes half the circumference of the Earth as the person falls completely the diameter of the Earth ....& this is just half a period.
...oh never mind because (typing as I am thinking) I just figured out why its the "half of the halfs" thing that you talked about. So it wouldn't be Pi because twice the diameter of the Earth would equal the circumference of the Earth.
& that's also why its about 42 minutes for both the satellite & the person.
Ok. I answered my own question.
But still thanks for all the cool videos!!!