Hello Matt. Very glad to find your great channel here! This is something I have been wondering for years and I did guess correctly that the gravity of a planet or star must be zero at the center of it's mass. But I have a question; Why is it then the fussion of a star occurs at it's center where the gravity is at it's weakest? Shouldn't it happen else where? I would very much appreciate of your feedbacks.,,🤗
@@TheGolfdaily Whilst gravity is being cancelled out or balanced at the center of the star, this doesn't mean that nuclear fusion cannot occur. It is not gravity that causes fusion but rather the pressure from the weight of all the matter from the surface down to the center of the star. Trillions upon trillions of tonnes matter all squeezing the center of the star, forcing hydrogen atoms together to fuse into larger helium atoms.
I love this video. More videos should be like this. Audience interaction that is easy to hear, a real board that the teacher draws on which is clearly visible. It's practically perfect.
I have been pondering this in the bathroom the last few days since Christmas and concluded that gravity balances out in the centre of earth due to equal pull of the earth all around you. I was glad to see you have posted this specific point to explain the same. Thanks! I also thought there would be slight unequal pulls due to the irregular shape of the earth and the pull of the sun/moon etc. but am glad to know the principle I thought is right.
I thought Guass's law was for enclosed surfaces (haven't taken it yet still in charge distributions but we should get to it in a week, can't wait for the double integrals which the law introduces a replacement for). But I guess since both Coloumbs law and universal gravitational law are inversly propotional with r^(2) guass's law should work perfectly well for both.
Fully agree. The graph in question is NOT agreed upon between al scientists! The Russians did test gravity while digging the deepest hole, they found first a bigger value 1 km under the ground!!! +je
+Matt Anderson What does it look like for live members of your audience to watch you teach this way? I imagine it might feel like "the man behind the curtain" scene in the Wizard of Oz. Pay no attention to the man behind the big glass board. You'll just get lost trying to read backwards. Watch the big display on the monitor at the front of the classroom.
Tunneling deeper into a sphere of uniform mass, would the matter above your head cancel out and such you would only need to count the mass in the sphere beneath your feet. For example if you get halfway to the centre of the earth you have a sphere of 1/2 the radius so it would be 1/8 of the mass so GM/r^2 * (0.125/0.5 squared) which is half the gravity.
Tim, Well done. Absolutely correct (provided, of course, we assume the earth is uniform density). With a mass of M=ρ(4/3)πr^3 (where ρ is density), you get F = GMm/r^2 = Gρ(4/3)πmr. So it goes linearly with radius r. Half the radius equals half the gravity. Cheers, Dr. A
This is a really cool video. I don’t have any physics knowledge and my math is kind of meh, but I’ve always been interested in gravity. This taught me a new look to gravity. Is gravity a universal constant? Or in other words, does gravity work the same everywhere? Is gravity directly related to the mass it is attached to? I.E. the bigger the planet or mass the stronger the gravity pull or is it related to the type of mass that is near it? If I took a large amount of mass from the moon and moved it to earth, would earths total gravity increase? (I assume I’d have to disperse the new mass perfectly even across the surface of earth to get an equal increase in gravity.
Hello Matt. Very glad to find your great channel here! This is something I have been wondering for years and I did guess correctly that the gravity of a planet or star must be zero at the center of it's mass. But I have a question; Why is it then the fussion of a star occurs at it's center where the gravity is at it's weakest? Shouldn't it happen else where? I would very much appreciate of your feedbacks.,,🤗
It's because the pressure is at its highest. There is a huge column of star matter going out radially that is being pulled toward the center. That spot in the center needs to support this column, hence the pressure is enormously high. Cheers, Dr. A
@@yoprofmatt OK, so I think I have mixed up between pressure and gravity. Fusion is triggered by the pile up of mass around it and not by the high gravity, I can see that.. But I wonder, would gravity alone be able to create fusion of atoms providing the gravity is strong enough?
Here's a thought: Fusion happens between two atoms. They get squeezed together (hydrogen) to form a new atom (helium). There is, of course, gravity between these two hydrogen atoms, but it is incredibly weak. There has to be other atoms nearby to "push" these two atoms together. So if you pile up a whole bunch of other atoms behind these two target atoms, they can get pushed together hard enough to create fusion. This is what the star does. So in some sense, it is gravity that is ultimately responsible here, because it is pulling on each big group of atoms. But the atoms need to be able to push on each other as well, in order to force the two initial atoms together. This is, essentially, what we call pressure. The atoms are packed together very tightly, and each other around. Cheers, Dr. A
@@yoprofmatt Fantastic, thank you so much for taking the time to answer my questions. They were in my mind for years unanswered. I get the pictures now! Again, Thanks a lot Dr.A
eXtremeDR, Great question, which I don't know the answer to off hand. But there are two contributions: gravity (general relativity), and speed (special relativity). Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Incredible video! I have one question though. If we imagine that the earth is a perfect sphere with the perfect mass and density distribution. Does this mean that we dig a hole so that the radius will be of that of the moon the gravitational force would be similar to the moon?
Great video! but how is the person being pulled by different chunks of the earth in all directions when gravity pulls the person or any other object towards its CENTRE only?
If you were stood in the centre of earth (in a hole) then the centre of mass would be somewhere between the centre of earth and the outer surface, so gravity would still be pulling towards the centre of mass... the centre of an object isn’t always the centre of mass. I think if you were in the centre of earth then centrifugal force would be amplified by the smaller radius and faster rotation and the person would get stuck to the walls like a washing machine, though I’m not a physicist so I’m probably very wrong lol
Hello and great respect to You, professor! I apologize for my english in advance, but hope it will be good enough to ask a question. I'm far away from physic so my question might sound stupid or been answered allready. Does the Universe have mass and if yes does it mean it has its gravity as well which sholud be somewat a total gravity of everything. My point is what if by getting further away from the Earth its gravity doesn't weaken but getting more and more neutralized by the Universe one instead. And Eart's orbit is just the neutral zone between bought? If so and if gravity has a centerwise direction then we could think of finding the center of the Universe. And if something has a center it most probably has an edge as well... Again, sorry if it sounds stupid. I'm just a casual guy interested in physic forces.
Dear Matt, If I were standing on a rocky core (say 2400 km in diameter) at the centre of the gas giant Jupiter - would the force of gravity be just the same as if I were standing on Pluto ( a planet of the same diameter)?
Also wondering... but I think it would not be the same. Because, anywhere you choose to stand on the rocky core inside Jupiter, you would have a little more mass of Jupiter (mass of the rest of Jupiter) placed in the opposite direction from yourself. That mass difference is different from just empty space all around you (the Pluto case).
Hello sir u really explain well but I hv doubt..n I hv to ask you this.. So aas u said the resultant of garvity pull from every direction on the centre of earth is zero but the earth is not uniformly massed I mean if you cut out planet in quadrant then u will notice that the material inside the earth is not same.. there might be some iron ore in one side but the opposite side contain petroleum then the collective mass may vary .. please reply m eagerly waiting for your response 🙏
Wait, how does that white board work? If you're writing on it how you normally writing on a paper and someone was looking at it from the other side the letters should be inverted right? So is there like a double glass thingy or how is that working or am I just super dumb?
I heard another explanation. The mass (of the Earth) affects space, space-time to be precise. In the center of the earth, that space is maximally curved. A 10x10x10 m hole doesn't change that much. So if you get into that hole in the center of the earth, you will still be squeezed together by the curvature of space.
Please, I have a surprising question now: How can there be an acceleration due to gravity on a body at rest on the ground? We know that the acceleration of stationary bodies is zero. Can you answer this question? I apologize if there are errors in my speech because I use translation. I am an Arab researcher.
I suppose we could compare the Earth's internal forces of gravity with the pressure gradient of a spinning sphere filled with water. The centrifugal force causes the water in the outer layers to have more pressure than the more inward layers and with the centre having no pressure. Videos of Jupiter show a vertical stack of 'discs of gas cloud' - all moving at different rates.
Andrew D Mackay if everything is pulling away from the center of the Earth then they Center would naturally form a hollow void and what you are describing is a LaGrange point I call it the LaGrange 0
sir Anderson if there is no gravity at the centre of the earth then what pulls us towards the centre and if gravitational force = 0 then how the lava and all molten materials stay there without any movement. Pls answer me
It's not gravity which pulls you, it's mass which pulls you and that effect is called gravity (gravitational force). As long as you had more mass in some direction, you would be pulled to the center. If the earth didn't move around the sun (it actually doesn't to be fully correct, it moves around the mass center of the whole solar system, including Jupiter's mass etc), and stood still in relation to the sun (hypothetically) you would experience the gravitation of the sun at the center of the earth.
Nice question. It is the mass that attracts us as we approach the center of earth, the mass of upper side balances the mass of down side and as we are at the center , the resultant force from all the sides becomes zero so gravity at center only is zero
so if there is a mesurement for a gravity pull towards the center of the earth,,,where does it stop.? wat is at the ficrim of the power of gravity. is it like the question. how long is a piece of string? or could there be an avent horizon, shoo i have so many question
Hi Rolf, It's important to remember that every bit of mass exerts a gravitational pull on every other mass. The only reason gravity becomes zero at the center of the earth is because all the mass of the earth is pulling you equally in all directions, thus completely canceling out. Cheers, Dr. A
Matt Anderson what if I m going towards the center. Will the gravity decrease or increase . If it will decrease according to the graph ...then how . Because the formula of gravitational force states that ..(G*m1*m2)/r^2. Then why it is decreasing rather then increasing. It should decrease when we go down . Cauz the radius is getting smaller. ?
Condition 1: the location in question is infinitely far away from all other objects Condition 2: the location is located in a specific position where the gravitational field of all nearby objects adds up to zero.
I really wanted to hear the answer of last question of the student, Imagine jumping through a large antipode hole of the earth if heat is not a problem, what will happen when we reach the core ? Will we slow down and if we do slow down then do we have enough momentum to make it to the other end or will we keep juggling and will be stuck in the core eventually? :D Really curious
According to Newton, gravity is a force because it causes an acceleration. If you think of a man who falls from a skyscraper, at the beginning of the fall his speed is less than the speed ad the end (acceleration is how much speed changes during time), according to F=ma. Gravity field of a planet (for example) is an infinite spherical volume that reaches every part of the universe and if you go far from the planet' surface, the force will be less as further you go, but it is never 0 (imagine an hyperbole as grafic) According to Einstein and his general relativity, gravity is not a force, but deformation of space-time because of the presence of mass. Imagine a big bed sheet in air. If you put a ball in its centre, the central part of the sheet will go down, and if you put a more little ball in another point of the sheet near the first ball, the second ball will fall into the first ball, because near the first ball the space-time is deformed by the mass of the first ball.
This may be true, there may be no force there pulling you, but at that point you will have collided with a black hole smaller than you and it would not be seemingly very comfortable. In fact, there is no way of telling if you would be alive to process the feeling. The aspect you missed here is that all of Earth's chunks are not just being forced inward toward the center of their own accord, but because they are all interacting with the gravity of a tiny black hole at the center of our Earth.
I think that the gravity_outward_from_the_center increases curly not linear because the material density changes between the layers. The earth is likely diving in an ocean of gravity (coming from all directions). It is considered same as a fishing net with so many holes that can allow gravity go through it. If the earth is more condense, the holes will become smaller and the net can block or catch the gravity better. A guy asked me "Hey, what is the gravity?" = A fish asked another fish "hey, what is the water?". I can answer the question from the fish, but I need to learn fish language.
I didn't like when the teacher would ask us questions, we're all forming question in our heads while it's happening, and it's much better when they didn't break up a lecture and just explain, then talk about it.
where i see a discrepancy in the theorem is how can gravity zero out in the center if its relative to the center of the mass. does this not mean that the center of a mass must be the strongest point of gravity within that mass. if there is a connection between mass and the ability to compress that mass and energy to that of no magnitude center must be the strongest point of gravity. for gravity to cancel out suggests that gravity is not relative to the center of a mass but really the center of mass. let me repeat the center of mass not the center of a mass. this means gravity must be the strongest at the mid point of the radius and decreases as you get closer to the center. this also would suggest that there is a void or pocket of space in the very center where gravity is zero and the shape of this pocket would be relative to the density of mass or counter gravity from each angle. so its likely not a perfect circle and its hard to say exactly how big this void would be and or its temperature if there was any at all.
Assuming that the tunnel passes through the axis of rotation. He would plummet right towards the center of the earth, after crossing that point he would decelerate until he reaches the other side of the tunnel. He would then fall right back to the center and this cycle would continue
Maulik Srivastava hm, do you think that if we take out air resistance aswell, would this cycle continue indefinitely? Or would you still eventually start slowing down bit by bit until you stopped at the point were the center would be?
I believe that the inner core is centrifugally forcing outwardly against the outer core creating intensive pressure between inner and outer core and I think it's strongest towards the equator and weaker at the poles. It's only my theory that I know of and I wonder how right or wrong I am
I don't think the earth is spinning fast enough to worry about this. If you compare the centripetal acceleration at the surface of the earth to the gravitational acceleration, you'll see that the gravitational acceleration is much bigger. Cheers, Dr. A
Was this at a college? I never did attend college but this seems like very basic science and I assumed that at a college the science being taught would be more advanced.
+u2zero2u This was for my introductory physics course at San Diego State University. This particular discussion was a bit of a digression from our usual topics. Check some of my other videos for more advanced material. And stay tuned as we continue to add more topics. Thanks for watching. Cheers, Dr. A
This is the very example that shows that we do not understand the forces behind Gravity. Just wish that we really can dig a hole up to the center of Earth (probably easyer on the Moon) to measure the real gravity pull.
What if one hemisphere was bigger and heavier than the other hemisphere that the forse will be unequal and wouldnt be cancelled out so what would happen then?
manan Chauhan, Great question. You'd have a nonzero (but still small) force. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Learn CBSE, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
Gravity at the center of the earth is 0 but a very large chunk of our earths radius is made of lighter material (crust). If the earths density was balanced the surface of earth would be where gravity is the strongest but the earth has a very dense core, this means that the outer core is where gravity would be the strongest. The gravity on the surface of the earth is 9.81m/s2 where as the gravity on the outer core is 10.57m/s2. When you go past the threshold between the outer core and the lower mantle the gravity starts to drop until you get to the center of the earth where the gravity is 0.
This means gravitational constant "G" is also zero at the centre of the earth. Technically it's not wrt to the other external source of gravitational mass. BTW i like your Board
+EEK Not necessarily. It just means that the gravity due to your surroundings add up to zero as vectors, due to the spherical symmetry. G is a universal constant (as far as we know), and is not a function of the mass distribution of your surroundings. There is a concept in theoretical cosmology called MOND, or Modified Newtonian Dynamics, where proponents of this concept believe that G might not be a universal constant, and could be different in different parts of the galaxy and universe to explain the inconsistencies in galactic rotation patterns that gave rise to the concept of dark matter. Dark matter is material thought to exist out of ignorance, because all of the mass that we can account for in galaxies (stars, planets, moons, stellar remnants, black holes, nebulas, gas, dust, interstellar medium, etcetera) to the rotational patterns is simply not enough to match observational results. Therefore, people who believe dark matter is the explanation, believe that there exists material which only is evidenced by its gravity, and never will show evidence. MOND is the alternative explanation to dark matter. The dark matter explanation is more commonly accepted.
+carultch The following links are just FYI if interested en.wikipedia.org/wiki/Talk:Newton%27s_law_of_universal_gravitation#Does_.22g_.3DGM.2Fd.5E2.22_conform_with_.22F_.3D_GMm.2Fd.5E2.22_mathematically.3F en.wikipedia.org/wiki/Talk:Newton%27s_law_of_universal_gravitation#Since_this_is_technically_incorrect.2C_is_Gauss.27s_law_also_incorrect.3F
EEK I don't understand the first link. The answer is actually YES, if you think of g as a value measured in Newtons/kilogram instead of an acceleration. That is to say, it is force per unit mass. It is true that the larger body will also accelerate toward the smaller body, albeit negligible. However, it is still true that the force on the smaller body, per unit mass of the smaller body, is g=G*M/r^2. Similarly, the force on the larger body, per unit mass of the larger body, is g=G*m/r^2. The confusion in that discussion comes from thinking of g as the "acceleration due to gravity", rather than what it really is, which is gravitational field strength (gravitational force per unit mass). Understanding the difference can help greatly when working with problems (like Atwood's machine) where other forces come in to play, reminding you that g is not necessarily the acceleration, except in the special case of free fall.
I'd love scientific feedback on this up I could: One of the commenters below mentioned that because the Earth would be pulling at you from the sides that one could not be "center enough" to experience weightlessness. Here is my response to his comment: "OGW I understand what you're saying and if you were to look at your comment, if you will, "through a microscope," you're correct. That said, because the GRADIENT would be SO Incredibly gradual, I imagine you could travel quite some several MILES within a big cavernous space at the core and you'd still be virtually weightless (back to microscopes, etc., if you had a microgravity weight scale and were even 10 to 20 MILES "up" from the very center, it would take a very sensitive scale to measure any weight at all! Would Matt or any other scientists like to chime in? I will even go out on a limb so far as to say that if you had a 100 mile in diameter "room" in which to have your laboratory, things would be pretty much weightless everywhere in that room - certainly enough for some great experiments, flying around etc. Would make a GREAT vacation-hotel (if a little claustrophobic, I suppose :-)
voicetube, I agree with you. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Wow. Such nice way of teaching. Teach some of that to Indian teachers as well. BTW I knew the answer but my reasoning was not right. I thought at center, the mass of earth contributing to gravitation is zero.
i knew that about gravity at the center but...wtf is going on with pressure?...should not pressure start to fall together with gravity? and then wtf if the gore is not solid but plasma
I always guess you'd be spun around from the multiple angles of gravity pushing you and I kept on thinking of it throughout time I finally just went you know what I'm going to figure this out for real
There is not universal agreement on what you call "weight". Is it just weight = mg? If you're in free-fall then, do you still have weight? I like to regard weight as simply "what is the force on you from the scale underneath you?" When you're standing on a scale on the earth, your weight is mg. When you're in free-fall (like an astronaut is), your weight = zero, because a scale underneath your feet is not pushing back up on you (it is also in free-fall). This is what we call "weightless." You still have mass, of course, and you are likely still in a region of gravitational acceleration g (or very close to it), but your body feels different. There is no force pushing up on your feet. Your stomach kind of lurches into your throat. You might feel a little nauseous. This is weightlessness. So yes, some textbooks refer to your "weight" as mg. And this is usually what people are referring to when they say, "What's your weight?" (Which you should never ask, of course.) But a more technically correct (and snarky) answer would be: "What's my weight? Am I standing on a scale? It depends on the net force and my acceleration, so can you please elucidate what forces are present and then I can tell you. Let's draw a free-body diagram and work it out. And also, none of your business." Cheers, Dr. A
@@captainjackson18 g is the sum total of all of G*M/r^2 for every M in the universe. Some values of M are more important than others, particularly the M of the astronomical body closest to you. If you dealt with an exotic shaped object that weren't a sphere, then you'd need something more sophisticated than g=G*M/r^2 to calculate its surface gravity. However, most objects large enough to have a significant gravity, are pretty much guaranteed to be spheres or close to it. The structural forces to maintain an irregular shape become impractically large for anything over a few hundred km in size.
Totally, what I thought, but I don't understand his comment about the high pressures at the center of the earth. I would expect it is a vacuum...or nearly so. Pressure should be highest at the center of mass (highest density), which we now know is NOT the center of the earth...it is actually a ring some x distance from the center of the earth. Hey, there's a good subject. Calculate where the this ring is and explain why the center is a vacuum. Unless I am wrong, and then I'd really love to hear the explanation.
This guy is blowing my mind right now. The info is impressive but no where near as impressive as him writing backwards and me not knowing till the video was almost over
Why is there a core if the center is weightless? If the earth has been turning on its axis since it was formed and the center of its mass has no gravity would it not be a hole created by the centrifugal forces? Or does the corresponding masses gravitational pull on eachother, outweigh the forces created by the rotation? If so, would a drop of water in a weightless environment be denser in the center? Excuse my grammar, this is my second language. I am just trying to understand how the center of mass can be the densest point as we see in the illustrations of modern theories about earth's structure.
Rino M, I'm not sure I have a very good answer for you. We might need to ask a geologist. But remember there is a tremendous amount of pressure at the center of the earth. Which we have completely ignored here. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
If you got stuck between 2 huge planets right in the middle, would their gravity tear you apart, or would these two gravitational fieds cancel each other out? If they do not tear you apart, then gravity is not a real force. But what is it then?
You need properties like mass, radius, orbital velocities, and angular momentum of the system. In _Newtonian_ mechanics, the *gravitational* *tidal* *acceleration* would determine what happens. I doubt the two masses would be aligned perfectly to cancel gravitational effects. In _General_ relativity, tidal forces are replaced with *spacetime* *curvature* and *gravitational* *waves.* There would be an extreme effect near black holes and neutron stars, but you will still need mass/energy density and radius.
With Einstein's general relativity, gravity simply becomes the warping of spacetime. That's our best understanding. For now. Who knows how that might change in the future. Cheers, Dr. A
I guess my other question would be, does all mass have a gravitational force linked to its total mass? I wonder if we will ever figure out how to manipulate gravity. In this video I understood that if I was at the center of the earth and all sides of me pulled at the same strength, I would float. But if we found a way to strengthen or weaken gravity in a single direction I would be pulled or pushed that same direction yes? I don’t think I would be as simple as adding or subtracting mass but maybe it is.
Travis, Great question. The model we are using here is Newton's Law of Gravitation. This says that every mass in the universe pulls on every other mass in the universe, and the force is proportional to the mass of each object and inversely proportional to the square of the distance between them. The only way (that we know of currently) to strengthen the force is to increase the mass (assuming you don't change the distance). However, Einstein's general theory of relativity effectively "killed" Newton's gravity, instead saying that mass effectively warps space-time (the fabric of the universe). One way to increase gravity in some region of space-time is not to increase the mass in that region, but to send a gravitational wave through that region. This rippling of space-time may be thought of as locally increasing (and decreasing) gravity in the Newtonian sense. These waves were predicted by Einstein in 1916 and found experimentally 100 years later (the LIGO experiment). We are, of course, still observers of these events, and have no known way to generate gravitational waves of any measurable size (it takes the collision of black holes to be big enough for us to see it). But never say never. Who knows what will be our next grand theory that replaces general relativity? And will it have some secret key for manipulating our universe in currently unseen ways? Stay tuned. Cheers, Dr. A
Dr. Anderson, wouldn't an object at the center of the Earth experience a minute whole-surface tidal force similar to the'spaghettification' of an individual falling into a black hole? I would expect that while a spherical object at the Earth's center would experience no net gravitational force, it would experience a nearly infinitesimal force across its entire surface that would pull uniformly outward at each surface point.
D Campbell, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
Sir it was very helpful but I have a query that the force exerted by the centre of earth on us on the surface is maximum and the force at the centre of earth is zero.ok good. But sir you show that when the object is at centre then the the forces acting by chunk of rocks from different directions cancel out each other. But what about the force acting by the centre of earth on the men? You tell about the forces acting on the object towards the centre of chunk of rocks but what about the force acting on the person from the exact centre of earth ?
+Dan M Yes. And in the case of a hypothetical uniform density, it would be simple harmonic motion. Force of gravity would linearly grow with radial position. In reality, because Earth is not uniform, it is a little more interesting. Here's a graph of what it really looks like. The PREM data shows the best information we have, regarding interior density distribution and gravity of Earth. It grows nearly linearly throughout the cores. Peak gravity is only slightly higher than surface gravity, and occurs at the bottom of the mantle. This slowly decreases in a tapered manner, until it reaches the surface gravity we know and love. upload.wikimedia.org/wikipedia/commons/thumb/5/50/EarthGravityPREM.svg/2000px-EarthGravityPREM.svg.png
I’m a bit confused..the graph seems to show that by the time you reach the center of the earth the gravitational force would be 0, in other words you would stop falling right there in the middle. Wouldn’t you need a separate force to act upon you in order to“ re accelerate” you through to come out on the other side of the planet?
@@kingslaphappy1533 There would sort of be a separate "force", but it really isn't a force. It is your inertia that would keep you in motion as you pass through the equilibrium position at the center of the Earth.
A few disclaimers, and unrealistic assumptions to keep it simple: 1. assume we add enough air to the planet's atmosphere from elsewhere in the solar system, in order to maintain the surface pressure we currently have, of 100 kPa. 2. assume the temperature in the hole is uniform at 300 K. 3. assume a uniform density of Earth, rather than a linearly increasing density, or a PREM model of density, that are more consistent with experimental results. This means gravity is a linear function of the radius, as can be derived from Gauss's law of gravitation: g(r) = g0 * r/R where g0 is the surface gravity at radius R. The relation between pressure and radial altitude is: dP(r)/dr = -rho(r)*g(r) Where rho(r) is the density at radial position r, and g(r) is the gravitational field strength. From the ideal gas law. Note that we are using K for the gas constant, because R is spoken-for. M is the effective molar mass of air, T is the temperature. rho(r) = P(r)*M/(K*T) Put it together: dP(r)/dr = -P(r) * M/(K*T) * g0/R *r Group all the constants to the left: dP(r)/dr = -M*g0/(K*T*R) * P(r)*r Call that cluster of constants C: C = M*g0/(K*T*R) And our differential equation becomes: dP(r)/dr = -C*P(r)*r Solving it, and implementing our boundary condition that P(R) = P0, we get: P(r) = P0*exp((M*g0)/(2*K*T*R)*(R^2 - r^2)) At r=0: P(0) = P0*exp((M*g0)/(2*K*T)*R) Data: P0 = 100 kPa M = 28.97 kg/kmol K = 8314 J/kmol-K T = 300 K R=6.371e6 m g0 = 9.8 N/kg With this data, the pressure of the air at the center of Earth would need to be 3*10^159 kPa, if we were to maintain atmospheric pressure at the surface of Earth, and maintain this hole's air temperature at a uniform 300 Kelvin.
I think the gravity is the result of attraction of an object and the x-particles(god particles? Might have played the important role in aging in everything in the universe, one of the factors causing chemical and biological reactions), just like proton and electron attract each other, and as the x-particles are moving toward the object, they impose a force on things in their way as the result of resistance when they go through the things, and they push the things to direction of the object they are heading to. X-particles can travel through everything but has to face resistance and that is why up to the surface on the surface of the earth the gravity is strongest, and as the x-particles travel deeper into the earth, they have to face more resistance from layers materials in the earth and the force of x-particles coming the opposite side of the earth, the speed of the X-particles getting slower, and the force moving toward the center of the earth becomes weaker, and at the center of the Earth where all forces of x-particles from all directions are equal and cancel one another out and become zero. I believe that X-particles don't stop moving at the center of the earth, but they continue moving on their course and their force become weaker and weaker(their speed decrease) as they are moving outward to the other side of the earth. Does earth pull objects inward? I think the Earth indirectly pulls things inward, not directly - they directly pull x-particles inward (the result of attraction law in physic) and as the particles constantly fly toward the Earth they push everything in their ways toward the earth and that's why we see objects with mass falling toward the Earth. Our weight is the result of the x-particles fly to the earth and has to travel through our body and face resistance and become a force pushing our body down to the Earth, the more mass the more resistance, the heavier.
I said zero at 0:35. How the hell does a physics class take another 3 minutes and 15 seconds to get that? Also gravity decreasing linearly is also pretty obvious seeing as the mass above you is pulling you up and the mass below you is pulling you down. The further you go down the more mass is above you pulling you up. I stupidly decided to study finance.
![รอจะบอกคำนั้น - KARN The Parkinson x DAOU PITTAYA [LIVE SESSION]](http://i.ytimg.com/vi/wACfLFATe2Y/mqdefault.jpg)
How is this demon writing backwards with ease?
No demon. Check it out here: www.learning.glass
Cheers,
Dr. A
VERY cool solution!
There was one guy who could do this for real. Leonardo Davinchi.
Hello Matt.
Very glad to find your great channel here!
This is something I have been wondering for years and I did guess correctly that the gravity of a planet or star must be zero at the center of it's mass.
But I have a question;
Why is it then the fussion of a star occurs at it's center where the gravity is at it's weakest? Shouldn't it happen else where?
I would very much appreciate of your feedbacks.,,🤗
@@TheGolfdaily Whilst gravity is being cancelled out or balanced at the center of the star, this doesn't mean that nuclear fusion cannot occur. It is not gravity that causes fusion but rather the pressure from the weight of all the matter from the surface down to the center of the star. Trillions upon trillions of tonnes matter all squeezing the center of the star, forcing hydrogen atoms together to fuse into larger helium atoms.
I love this video. More videos should be like this.
Audience interaction that is easy to hear, a real board that the teacher draws on which is clearly visible. It's practically perfect.
Unknow0059,
Thanks for the comment, that made my day.
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Great idea and thought behind these Video... 👍👍..... Thanks Prof Anderson and have a nice weekend...
I have been pondering this in the bathroom the last few days since Christmas and concluded that gravity balances out in the centre of earth due to equal pull of the earth all around you. I was glad to see you have posted this specific point to explain the same. Thanks! I also thought there would be slight unequal pulls due to the irregular shape of the earth and the pull of the sun/moon etc. but am glad to know the principle I thought is right.
Stu,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
I swear i felt like the smartest person in the universe when i guessed it would be 0 long before he said so
Wonderful lecturer that makes physics fun and understandable !!!
I thought Guass's law was for enclosed surfaces (haven't taken it yet still in charge distributions but we should get to it in a week, can't wait for the double integrals which the law introduces a replacement for). But I guess since both Coloumbs law and universal gravitational law are inversly propotional with r^(2) guass's law should work perfectly well for both.
Fully agree. The graph in question is NOT agreed upon between al scientists! The Russians did test gravity while digging the deepest hole, they found first a bigger value 1 km under the ground!!! +je
What I would like to know is, how did this professor learn to right backwards so we can see it correctly?
HeadingForTomorrow thats is what im thinking all the time,,
+HeadingForTomorrow The board is flipping the image before it gets put in the video.
+FutureChaosTV Excellent! Take a look here: th-cam.com/video/CWHMtSNKxYA/w-d-xo.html
More board info here: learningglasssolutions.com
Cheers, Dr. A
+Matt Anderson
What does it look like for live members of your audience to watch you teach this way? I imagine it might feel like "the man behind the curtain" scene in the Wizard of Oz.
Pay no attention to the man behind the big glass board. You'll just get lost trying to read backwards. Watch the big display on the monitor at the front of the classroom.
The live members see it backwards, of course. But right next to the glass I have HD monitors where they can see it flipped.
Cheers, Dr. A
Tunneling deeper into a sphere of uniform mass, would the matter above your head cancel out and such you would only need to count the mass in the sphere beneath your feet.
For example if you get halfway to the centre of the earth you have a sphere of 1/2 the radius so it would be 1/8 of the mass so GM/r^2 * (0.125/0.5 squared) which is half the gravity.
Tim,
Well done.
Absolutely correct (provided, of course, we assume the earth is uniform density). With a mass of M=ρ(4/3)πr^3 (where ρ is density), you get F = GMm/r^2 = Gρ(4/3)πmr. So it goes linearly with radius r. Half the radius equals half the gravity.
Cheers,
Dr. A
just realized you had to write backwords...
Not really. See the trick here: th-cam.com/video/CWHMtSNKxYA/w-d-xo.html
Cheers,
Dr. A
Could just be a reverse image
Great explanation, your students are very lucky to have you as their teacher
Lovely Video!! The concept of gravitational force at the earth was explained beautifully. TH-cam need tecahers like you.
Thank you so much 😀
Cheers,
Dr. A
Would that be correct also for black hole?
What is the textbook for this course
This is a really cool video. I don’t have any physics knowledge and my math is kind of meh, but I’ve always been interested in gravity. This taught me a new look to gravity. Is gravity a universal constant? Or in other words, does gravity work the same everywhere?
Is gravity directly related to the mass it is attached to? I.E. the bigger the planet or mass the stronger the gravity pull or is it related to the type of mass that is near it?
If I took a large amount of mass from the moon and moved it to earth, would earths total gravity increase? (I assume I’d have to disperse the new mass perfectly even across the surface of earth to get an equal increase in gravity.
Thanks Travis. I posted an answer on your other comment.
Cheers,
Dr. A
So if you built an earth mass death star type space station, would the structure be under more or less load in the center?
Does time dilation continue all the way to the center?
Hello Matt.
Very glad to find your great channel here!
This is something I have been wondering for years and I did guess correctly that the gravity of a planet or star must be zero at the center of it's mass.
But I have a question;
Why is it then the fussion of a star occurs at it's center where the gravity is at it's weakest? Shouldn't it happen else where?
I would very much appreciate of your feedbacks.,,🤗
It's because the pressure is at its highest. There is a huge column of star matter going out radially that is being pulled toward the center. That spot in the center needs to support this column, hence the pressure is enormously high.
Cheers,
Dr. A
@@yoprofmatt
OK, so I think I have mixed up between pressure and gravity. Fusion is triggered by the pile up of mass around it and not by the high gravity, I can see that..
But I wonder, would gravity alone be able to create fusion of atoms providing the gravity is strong enough?
Here's a thought: Fusion happens between two atoms. They get squeezed together (hydrogen) to form a new atom (helium). There is, of course, gravity between these two hydrogen atoms, but it is incredibly weak. There has to be other atoms nearby to "push" these two atoms together. So if you pile up a whole bunch of other atoms behind these two target atoms, they can get pushed together hard enough to create fusion. This is what the star does. So in some sense, it is gravity that is ultimately responsible here, because it is pulling on each big group of atoms. But the atoms need to be able to push on each other as well, in order to force the two initial atoms together. This is, essentially, what we call pressure. The atoms are packed together very tightly, and each other around.
Cheers,
Dr. A
@@yoprofmatt
Fantastic, thank you so much for taking the time to answer my questions. They were in my mind for years unanswered. I get the pictures now!
Again, Thanks a lot Dr.A
What is the time dilation effect between the earth surface and its core?
eXtremeDR,
Great question, which I don't know the answer to off hand. But there are two contributions: gravity (general relativity), and speed (special relativity).
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
How is the curvature of space affected at the center?
How deep do you go before you start losing gravity?
Incredible video! I have one question though. If we imagine that the earth is a perfect sphere with the perfect mass and density distribution. Does this mean that we dig a hole so that the radius will be of that of the moon the gravitational force would be similar to the moon?
Yes, correct.
Cheers,
Dr. A
Great video! but how is the person being pulled by different chunks of the earth in all directions when gravity pulls the person or any other object towards its CENTRE only?
If you were stood in the centre of earth (in a hole) then the centre of mass would be somewhere between the centre of earth and the outer surface, so gravity would still be pulling towards the centre of mass... the centre of an object isn’t always the centre of mass.
I think if you were in the centre of earth then centrifugal force would be amplified by the smaller radius and faster rotation and the person would get stuck to the walls like a washing machine, though I’m not a physicist so I’m probably very wrong lol
sir you are great what a manifestation
Hello and great respect to You, professor! I apologize for my english in advance, but hope it will be good enough to ask a question. I'm far away from physic so my question might sound stupid or been answered allready. Does the Universe have mass and if yes does it mean it has its gravity as well which sholud be somewat a total gravity of everything. My point is what if by getting further away from the Earth its gravity doesn't weaken but getting more and more neutralized by the Universe one instead. And Eart's orbit is just the neutral zone between bought? If so and if gravity has a centerwise direction then we could think of finding the center of the Universe. And if something has a center it most probably has an edge as well...
Again, sorry if it sounds stupid. I'm just a casual guy interested in physic forces.
yeah it has center and the center of univers or earth is kabba. Search more about kabba
Dear Matt, If I were standing on a rocky core (say 2400 km in diameter) at the centre of the gas giant Jupiter - would the force of gravity be just the same as if I were standing on Pluto ( a planet of the same diameter)?
Also wondering... but I think it would not be the same. Because, anywhere you choose to stand on the rocky core inside Jupiter, you would have a little more mass of Jupiter (mass of the rest of Jupiter) placed in the opposite direction from yourself. That mass difference is different from just empty space all around you (the Pluto case).
Yes.
Hello sir u really explain well but I hv doubt..n I hv to ask you this..
So aas u said the resultant of garvity pull from every direction on the centre of earth is zero but the earth is not uniformly massed I mean if you cut out planet in quadrant then u will notice that the material inside the earth is not same.. there might be some iron ore in one side but the opposite side contain petroleum then the collective mass may vary .. please reply m eagerly waiting for your response 🙏
Which class ???
Why the hell did you cut it off!!!!!!>?????
Wow i can listen to this giy for hours where is the whole lecture professor
Wait, how does that white board work? If you're writing on it how you normally writing on a paper and someone was looking at it from the other side the letters should be inverted right? So is there like a double glass thingy or how is that working or am I just super dumb?
Double H,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
@@yoprofmatt that's crazy haha ok thank you
I heard another explanation. The mass (of the Earth) affects space, space-time to be precise. In the center of the earth, that space is maximally curved. A 10x10x10 m hole doesn't change that much. So if you get into that hole in the center of the earth, you will still be squeezed together by the curvature of space.
Is it possible to go inside the center of earth as there is no gravitational acceleration?
Please, I have a surprising question now: How can there be an acceleration due to gravity on a body at rest on the ground? We know that the acceleration of stationary bodies is zero. Can you answer this question? I apologize if there are errors in my speech because I use translation. I am an Arab researcher.
I suppose we could compare the Earth's internal forces of gravity with the pressure gradient of a spinning sphere filled with water. The centrifugal force causes the water in the outer layers to have more pressure than the more inward layers and with the centre having no pressure. Videos of Jupiter show a vertical stack of 'discs of gas cloud' - all moving at different rates.
Andrew D Mackay if everything is pulling away from the center of the Earth then they Center would naturally form a hollow void and what you are describing is a LaGrange point I call it the LaGrange 0
sir Anderson if there is no gravity at the centre of the earth then what pulls us towards the centre and if gravitational force = 0 then how the lava and all molten materials stay there without any movement. Pls answer me
the earths core is solid, it's the layer in between the core and the mantle that is molten.
It's not gravity which pulls you, it's mass which pulls you and that effect is called gravity (gravitational force). As long as you had more mass in some direction, you would be pulled to the center. If the earth didn't move around the sun (it actually doesn't to be fully correct, it moves around the mass center of the whole solar system, including Jupiter's mass etc), and stood still in relation to the sun (hypothetically) you would experience the gravitation of the sun at the center of the earth.
Nice question. It is the mass that attracts us as we approach the center of earth, the mass of upper side balances the mass of down side and as we are at the center , the resultant force from all the sides becomes zero so gravity at center only is zero
so if there is a mesurement for a gravity pull towards the center of the earth,,,where does it stop.? wat is at the ficrim of the power of gravity. is it like the question. how long is a piece of string? or could there be an avent horizon, shoo i have so many question
Hi Rolf,
It's important to remember that every bit of mass exerts a gravitational pull on every other mass. The only reason gravity becomes zero at the center of the earth is because all the mass of the earth is pulling you equally in all directions, thus completely canceling out.
Cheers,
Dr. A
rmussel1:
Yes, it's excellent that you understand this. Well done, and keep up with the physics!
Cheers,
Dr. A
when you dont know how to reply
Matt Anderson what if I m going towards the center. Will the gravity decrease or increase . If it will decrease according to the graph ...then how . Because the formula of gravitational force states that ..(G*m1*m2)/r^2. Then why it is decreasing rather then increasing. It should decrease when we go down . Cauz the radius is getting smaller. ?
Thnks sir
gravity at the center of black hole = 0?
in the earth the weight of body is maximum ..?
So if you dug a hole deep enough, not to the center, you should feel less gravity?
Yes
It will just be less gravity, like on the moon
Sir pls tell the two conditions where value of g is zero ?
Condition 1: the location in question is infinitely far away from all other objects
Condition 2: the location is located in a specific position where the gravitational field of all nearby objects adds up to zero.
What about tidal forces? It probably doesn't feel the same as if you were in outer space.
I really wanted to hear the answer of last question of the student, Imagine jumping through a large antipode hole of the earth if heat is not a problem, what will happen when we reach the core ? Will we slow down and if we do slow down then do we have enough momentum to make it to the other end or will we keep juggling and will be stuck in the core eventually? :D Really curious
you are great sir. Thanks a lot!!
Udaya Ojha,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
I thought I was the only person that wondered this
I was surprised that I as bus driver knew it was 0 because it was only logical but some students said stupid things :(
Prof ....what is gravity actually is ....plz ans
According to Newton, gravity is a force because it causes an acceleration. If you think of a man who falls from a skyscraper, at the beginning of the fall his speed is less than the speed ad the end (acceleration is how much speed changes during time), according to F=ma. Gravity field of a planet (for example) is an infinite spherical volume that reaches every part of the universe and if you go far from the planet' surface, the force will be less as further you go, but it is never 0 (imagine an hyperbole as grafic)
According to Einstein and his general relativity, gravity is not a force, but deformation of space-time because of the presence of mass. Imagine a big bed sheet in air. If you put a ball in its centre, the central part of the sheet will go down, and if you put a more little ball in another point of the sheet near the first ball, the second ball will fall into the first ball, because near the first ball the space-time is deformed by the mass of the first ball.
This may be true, there may be no force there pulling you, but at that point you will have collided with a black hole smaller than you and it would not be seemingly very comfortable. In fact, there is no way of telling if you would be alive to process the feeling. The aspect you missed here is that all of Earth's chunks are not just being forced inward toward the center of their own accord, but because they are all interacting with the gravity of a tiny black hole at the center of our Earth.
I think that the gravity_outward_from_the_center increases curly not linear because the material density changes between the layers. The earth is likely diving in an ocean of gravity (coming from all directions). It is considered same as a fishing net with so many holes that can allow gravity go through it. If the earth is more condense, the holes will become smaller and the net can block or catch the gravity better. A guy asked me "Hey, what is the gravity?" = A fish asked another fish "hey, what is the water?". I can answer the question from the fish, but I need to learn fish language.
I didn't like when the teacher would ask us questions, we're all forming question in our heads while it's happening, and it's much better when they didn't break up a lecture and just explain, then talk about it.
Aaron ikr this is the most annoying thing
where i see a discrepancy in the theorem is how can gravity zero out in the center if its relative to the center of the mass. does this not mean that the center of a mass must be the strongest point of gravity within that mass. if there is a connection between mass and the ability to compress that mass and energy to that of no magnitude center must be the strongest point of gravity. for gravity to cancel out suggests that gravity is not relative to the center of a mass but really the center of mass. let me repeat the center of mass not the center of a mass. this means gravity must be the strongest at the mid point of the radius and decreases as you get closer to the center. this also would suggest that there is a void or pocket of space in the very center where gravity is zero and the shape of this pocket would be relative to the density of mass or counter gravity from each angle. so its likely not a perfect circle and its hard to say exactly how big this void would be and or its temperature if there was any at all.
Matt, do you have a video where you provide the answer to the question the young man posed at 6:30? Thanks
He made one
Assuming that the tunnel passes through the axis of rotation.
He would plummet right towards the center of the earth, after crossing that point he would decelerate until he reaches the other side of the tunnel. He would then fall right back to the center and this cycle would continue
Maulik Srivastava hm, do you think that if we take out air resistance aswell, would this cycle continue indefinitely? Or would you still eventually start slowing down bit by bit until you stopped at the point were the center would be?
This is a question I’ve been to to answer theoretically all along but the video is not complete 😢
Jesus if I had a profesor like him in school I would be 100% times more interested in physics! Great explanation! Thank you!
I believe that the inner core is centrifugally forcing outwardly against the outer core creating intensive pressure between inner and outer core and I think it's strongest towards the equator and weaker at the poles. It's only my theory that I know of and I wonder how right or wrong I am
I don't think the earth is spinning fast enough to worry about this. If you compare the centripetal acceleration at the surface of the earth to the gravitational acceleration, you'll see that the gravitational acceleration is much bigger.
Cheers,
Dr. A
Very interesting. Thanks.
But in which direction r u going to stand????🤔
I would likely be curled up in a ball screaming my head off.
Cheers,
Dr. A
still watching this video in 2023❤
Was this at a college? I never did attend college but this seems like very basic science and I assumed that at a college the science being taught would be more advanced.
+u2zero2u This was for my introductory physics course at San Diego State University. This particular discussion was a bit of a digression from our usual topics. Check some of my other videos for more advanced material. And stay tuned as we continue to add more topics. Thanks for watching.
Cheers, Dr. A
+Matt Anderson I defiantly will and thanks for uploading these great vids!
This is the very example that shows that we do not understand the forces behind Gravity. Just wish that we really can dig a hole up to the center of Earth (probably easyer on the Moon) to measure the real gravity pull.
What if one hemisphere was bigger and heavier than the other hemisphere that the forse will be unequal and wouldnt be cancelled out so what would happen then?
manan Chauhan,
Great question. You'd have a nonzero (but still small) force.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Respected Professor Pls teach me how to write inverted? Humble request! I too teach physics and Maths!
Learn CBSE,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
Professor humbled to get your response Stay blessed and keep showering great concepts onstudents
Yes 🥰 very nice
Gravity at the center of the earth is 0 but a very large chunk of our earths radius is made of lighter material (crust). If the earths density was balanced the surface of earth would be where gravity is the strongest but the earth has a very dense core, this means that the outer core is where gravity would be the strongest. The gravity on the surface of the earth is 9.81m/s2 where as the gravity on the outer core is 10.57m/s2. When you go past the threshold between the outer core and the lower mantle the gravity starts to drop until you get to the center of the earth where the gravity is 0.
This means gravitational constant "G" is also zero at the centre of the earth. Technically it's not wrt to the other external source of gravitational mass. BTW i like your Board
+EEK
Not necessarily. It just means that the gravity due to your surroundings add up to zero as vectors, due to the spherical symmetry.
G is a universal constant (as far as we know), and is not a function of the mass distribution of your surroundings.
There is a concept in theoretical cosmology called MOND, or Modified Newtonian Dynamics, where proponents of this concept believe that G might not be a universal constant, and could be different in different parts of the galaxy and universe to explain the inconsistencies in galactic rotation patterns that gave rise to the concept of dark matter.
Dark matter is material thought to exist out of ignorance, because all of the mass that we can account for in galaxies (stars, planets, moons, stellar remnants, black holes, nebulas, gas, dust, interstellar medium, etcetera) to the rotational patterns is simply not enough to match observational results. Therefore, people who believe dark matter is the explanation, believe that there exists material which only is evidenced by its gravity, and never will show evidence. MOND is the alternative explanation to dark matter. The dark matter explanation is more commonly accepted.
+carultch I'm really enjoying all the posts from you and the others. Some excellent debates happening here. Well done.
Cheers, Dr. A
+carultch
The following links are just FYI if interested
en.wikipedia.org/wiki/Talk:Newton%27s_law_of_universal_gravitation#Does_.22g_.3DGM.2Fd.5E2.22_conform_with_.22F_.3D_GMm.2Fd.5E2.22_mathematically.3F
en.wikipedia.org/wiki/Talk:Newton%27s_law_of_universal_gravitation#Since_this_is_technically_incorrect.2C_is_Gauss.27s_law_also_incorrect.3F
EEK
I don't understand the first link. The answer is actually YES, if you think of g as a value measured in Newtons/kilogram instead of an acceleration.
That is to say, it is force per unit mass. It is true that the larger body will also accelerate toward the smaller body, albeit negligible. However, it is still true that the force on the smaller body, per unit mass of the smaller body, is g=G*M/r^2. Similarly, the force on the larger body, per unit mass of the larger body, is g=G*m/r^2.
The confusion in that discussion comes from thinking of g as the "acceleration due to gravity", rather than what it really is, which is gravitational field strength (gravitational force per unit mass). Understanding the difference can help greatly when working with problems (like Atwood's machine) where other forces come in to play, reminding you that g is not necessarily the acceleration, except in the special case of free fall.
I'd love scientific feedback on this up I could:
One of the commenters below mentioned that because the Earth would be pulling at you from the sides that one could not be "center enough" to experience weightlessness. Here is my response to his comment:
"OGW I understand what you're saying and if you were to look at your comment, if you will, "through a microscope," you're correct. That said, because the GRADIENT would be SO Incredibly gradual, I imagine you could travel quite some several MILES within a big cavernous space at the core and you'd still be virtually weightless (back to microscopes, etc., if you had a microgravity weight scale and were even 10 to 20 MILES "up" from the very center, it would take a very sensitive scale to measure any weight at all! Would Matt or any other scientists like to chime in?
I will even go out on a limb so far as to say that if you had a 100 mile in diameter "room" in which to have your laboratory, things would be pretty much weightless everywhere in that room - certainly enough for some great experiments, flying around etc. Would make a GREAT vacation-hotel (if a little claustrophobic, I suppose :-)
voicetube,
I agree with you.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Wow. Such nice way of teaching. Teach some of that to Indian teachers as well. BTW I knew the answer but my reasoning was not right. I thought at center, the mass of earth contributing to gravitation is zero.
Thanks sir . ^_^
i knew that about gravity at the center but...wtf is going on with pressure?...should not pressure start to fall together with gravity? and then wtf if the gore is not solid but plasma
Interesting video!
Thanks for the comment.
Cheers,
Dr. A
I always guess you'd be spun around from the multiple angles of gravity pushing you and I kept on thinking of it throughout time I finally just went you know what I'm going to figure this out for real
I would think it would be the opposite. You would feel inward force from all directions
The force of gravity and the weight of an object is the same right?
There is not universal agreement on what you call "weight". Is it just weight = mg? If you're in free-fall then, do you still have weight? I like to regard weight as simply "what is the force on you from the scale underneath you?" When you're standing on a scale on the earth, your weight is mg. When you're in free-fall (like an astronaut is), your weight = zero, because a scale underneath your feet is not pushing back up on you (it is also in free-fall). This is what we call "weightless." You still have mass, of course, and you are likely still in a region of gravitational acceleration g (or very close to it), but your body feels different. There is no force pushing up on your feet. Your stomach kind of lurches into your throat. You might feel a little nauseous. This is weightlessness.
So yes, some textbooks refer to your "weight" as mg. And this is usually what people are referring to when they say, "What's your weight?" (Which you should never ask, of course.) But a more technically correct (and snarky) answer would be: "What's my weight? Am I standing on a scale? It depends on the net force and my acceleration, so can you please elucidate what forces are present and then I can tell you. Let's draw a free-body diagram and work it out. And also, none of your business."
Cheers,
Dr. A
@@yoprofmatt so, F = GMm/r^2 is not equal to mg in all cases?
@@captainjackson18 g is the sum total of all of G*M/r^2 for every M in the universe. Some values of M are more important than others, particularly the M of the astronomical body closest to you.
If you dealt with an exotic shaped object that weren't a sphere, then you'd need something more sophisticated than g=G*M/r^2 to calculate its surface gravity. However, most objects large enough to have a significant gravity, are pretty much guaranteed to be spheres or close to it. The structural forces to maintain an irregular shape become impractically large for anything over a few hundred km in size.
Totally, what I thought, but I don't understand his comment about the high pressures at the center of the earth. I would expect it is a vacuum...or nearly so. Pressure should be highest at the center of mass (highest density), which we now know is NOT the center of the earth...it is actually a ring some x distance from the center of the earth. Hey, there's a good subject. Calculate where the this ring is and explain why the center is a vacuum. Unless I am wrong, and then I'd really love to hear the explanation.
This guy is blowing my mind right now. The info is impressive but no where near as impressive as him writing backwards and me not knowing till the video was almost over
Alex Barbre,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
if the gravity at the center of the earth is 0 what is holding the particles inside the center together??
Pressure from the outer layers.
Wait, he writes on the glass like a mirror? Because he needs to flip everything so we can read it
I need that board!
Hmm.. I want this kind of lecturer please..
What’s the answer to the question about jumping through the earth?
th-cam.com/video/bVt8F7ZTrhI/w-d-xo.html
Why is there a core if the center is weightless? If the earth has been turning on its axis since it was formed and the center of its mass has no gravity would it not be a hole created by the centrifugal forces? Or does the corresponding masses gravitational pull on eachother, outweigh the forces created by the rotation? If so, would a drop of water in a weightless environment be denser in the center? Excuse my grammar, this is my second language. I am just trying to understand how the center of mass can be the densest point as we see in the illustrations of modern theories about earth's structure.
Rino M,
I'm not sure I have a very good answer for you. We might need to ask a geologist. But remember there is a tremendous amount of pressure at the center of the earth. Which we have completely ignored here.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
I so bad want to participate in his discussions!
Josiah,
You can. Join my new website: www.universityphysics.education
Cheers,
Dr. A
at first glance id say weightlessness.
If you got stuck between 2 huge planets right in the middle, would their gravity tear you apart, or would these two gravitational fieds cancel each other out? If they do not tear you apart, then gravity is not a real force. But what is it then?
You need properties like mass, radius, orbital velocities, and angular momentum of the system.
In _Newtonian_ mechanics, the *gravitational* *tidal* *acceleration* would determine what happens. I doubt the two masses would be aligned perfectly to cancel gravitational effects.
In _General_ relativity, tidal forces are replaced with *spacetime* *curvature* and *gravitational* *waves.* There would be an extreme effect near black holes and neutron stars, but you will still need mass/energy density and radius.
This is equivalent to cave paintings, scientists today have no idea what gravity is or what causes it.
With Einstein's general relativity, gravity simply becomes the warping of spacetime. That's our best understanding. For now. Who knows how that might change in the future.
Cheers,
Dr. A
I guess my other question would be, does all mass have a gravitational force linked to its total mass?
I wonder if we will ever figure out how to manipulate gravity. In this video I understood that if I was at the center of the earth and all sides of me pulled at the same strength, I would float. But if we found a way to strengthen or weaken gravity in a single direction I would be pulled or pushed that same direction yes?
I don’t think I would be as simple as adding or subtracting mass but maybe it is.
Travis,
Great question.
The model we are using here is Newton's Law of Gravitation. This says that every mass in the universe pulls on every other mass in the universe, and the force is proportional to the mass of each object and inversely proportional to the square of the distance between them. The only way (that we know of currently) to strengthen the force is to increase the mass (assuming you don't change the distance).
However, Einstein's general theory of relativity effectively "killed" Newton's gravity, instead saying that mass effectively warps space-time (the fabric of the universe). One way to increase gravity in some region of space-time is not to increase the mass in that region, but to send a gravitational wave through that region. This rippling of space-time may be thought of as locally increasing (and decreasing) gravity in the Newtonian sense. These waves were predicted by Einstein in 1916 and found experimentally 100 years later (the LIGO experiment). We are, of course, still observers of these events, and have no known way to generate gravitational waves of any measurable size (it takes the collision of black holes to be big enough for us to see it).
But never say never. Who knows what will be our next grand theory that replaces general relativity? And will it have some secret key for manipulating our universe in currently unseen ways?
Stay tuned.
Cheers,
Dr. A
Thank you
How big is that center of gravity? Are you saying that the earth is hollow at the center?
Thanks sir.
You are welcome.
Cheers,
Dr. A
Dr. Anderson, wouldn't an object at the center of the Earth experience a minute whole-surface tidal force similar to the'spaghettification' of an individual falling into a black hole? I would expect that while a spherical object at the Earth's center would experience no net gravitational force, it would experience a nearly infinitesimal force across its entire surface that would pull uniformly outward at each surface point.
There would be a tidal force, but it wouldn't be anywhere near the magnitude an object experiences while falling in to a black hole.
Whats freaking me out the most is how he writes so well backwards
He writes with left hand :>
D Campbell,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
Sir it was very helpful but I have a query that the force exerted by the centre of earth on us on the surface is maximum and the force at the centre of earth is zero.ok good.
But sir you show that when the object is at centre then the the forces acting by chunk of rocks from different directions cancel out each other. But what about the force acting by the centre of earth on the men? You tell about the forces acting on the object towards the centre of chunk of rocks but what about the force acting on the person from the exact centre of earth ?
So that last questioned asked.. You would jump in from the surface and reach zero velocity at the opposite surface correct? Just like a pendulum?
+Dan M
Yes. And in the case of a hypothetical uniform density, it would be simple harmonic motion. Force of gravity would linearly grow with radial position.
In reality, because Earth is not uniform, it is a little more interesting. Here's a graph of what it really looks like. The PREM data shows the best information we have, regarding interior density distribution and gravity of Earth. It grows nearly linearly throughout the cores. Peak gravity is only slightly higher than surface gravity, and occurs at the bottom of the mantle. This slowly decreases in a tapered manner, until it reaches the surface gravity we know and love.
upload.wikimedia.org/wikipedia/commons/thumb/5/50/EarthGravityPREM.svg/2000px-EarthGravityPREM.svg.png
I’m a bit confused..the graph seems to show that by the time you reach the center of the earth the gravitational force would be 0, in other words you would stop falling right there in the middle. Wouldn’t you need a separate force to act upon you in order to“ re accelerate” you through to come out on the other side of the planet?
@@kingslaphappy1533 There would sort of be a separate "force", but it really isn't a force. It is your inertia that would keep you in motion as you pass through the equilibrium position at the center of the Earth.
If we dug a tunnel through a diameter what would be the air pressure at the Center of the Earth?
A few disclaimers, and unrealistic assumptions to keep it simple:
1. assume we add enough air to the planet's atmosphere from elsewhere in the solar system, in order to maintain the surface pressure we currently have, of 100 kPa.
2. assume the temperature in the hole is uniform at 300 K.
3. assume a uniform density of Earth, rather than a linearly increasing density, or a PREM model of density, that are more consistent with experimental results.
This means gravity is a linear function of the radius, as can be derived from Gauss's law of gravitation:
g(r) = g0 * r/R
where g0 is the surface gravity at radius R.
The relation between pressure and radial altitude is:
dP(r)/dr = -rho(r)*g(r)
Where rho(r) is the density at radial position r, and g(r) is the gravitational field strength.
From the ideal gas law. Note that we are using K for the gas constant, because R is spoken-for. M is the effective molar mass of air, T is the temperature.
rho(r) = P(r)*M/(K*T)
Put it together:
dP(r)/dr = -P(r) * M/(K*T) * g0/R *r
Group all the constants to the left:
dP(r)/dr = -M*g0/(K*T*R) * P(r)*r
Call that cluster of constants C:
C = M*g0/(K*T*R)
And our differential equation becomes:
dP(r)/dr = -C*P(r)*r
Solving it, and implementing our boundary condition that P(R) = P0, we get:
P(r) = P0*exp((M*g0)/(2*K*T*R)*(R^2 - r^2))
At r=0:
P(0) = P0*exp((M*g0)/(2*K*T)*R)
Data:
P0 = 100 kPa
M = 28.97 kg/kmol
K = 8314 J/kmol-K
T = 300 K
R=6.371e6 m
g0 = 9.8 N/kg
With this data, the pressure of the air at the center of Earth would need to be 3*10^159 kPa, if we were to maintain atmospheric pressure at the surface of Earth, and maintain this hole's air temperature at a uniform 300 Kelvin.
I think the gravity is the result of attraction of an object and the x-particles(god particles? Might have played the important role in aging in everything in the universe, one of the factors causing chemical and biological reactions), just like proton and electron attract each other, and as the x-particles are moving toward the object, they impose a force on things in their way as the result of resistance when they go through the things, and they push the things to direction of the object they are heading to.
X-particles can travel through everything but has to face resistance and that is why up to the surface on the surface of the earth the gravity is strongest, and as the x-particles travel deeper into the earth, they have to face more resistance from layers materials in the earth and the force of x-particles coming the opposite side of the earth, the speed of the X-particles getting slower, and the force moving toward the center of the earth becomes weaker, and at the center of the Earth where all forces of x-particles from all directions are equal and cancel one another out and become zero. I believe that X-particles don't stop moving at the center of the earth, but they continue moving on their course and their force become weaker and weaker(their speed decrease) as they are moving outward to the other side of the earth. Does earth pull objects inward?
I think the Earth indirectly pulls things inward, not directly - they directly pull x-particles inward (the result of attraction law in physic) and as the particles constantly fly toward the Earth they push everything in their ways toward the earth and that's why we see objects with mass falling toward the Earth.
Our weight is the result of the x-particles fly to the earth and has to travel through our body and face resistance and become a force pushing our body down to the Earth, the more mass the more resistance, the heavier.
I dont even know why i searched up for this
perplexing concept
Agreed.
Cheers,
Dr. A
I said zero at 0:35. How the hell does a physics class take another 3 minutes and 15 seconds to get that? Also gravity decreasing linearly is also pretty obvious seeing as the mass above you is pulling you up and the mass below you is pulling you down. The further you go down the more mass is above you pulling you up. I stupidly decided to study finance.
My first guess was 0. Only thing that makes sense. I have never taken a class like this in my life. Wtf guys.
More amazing than gravity at center of earth is how you write backards without effort!
Wish that was true. A simple horizontal flip of the image is all it takes.
Cheers,
Dr. A
hi...he just flopped the video after finishing.....actually he wrote is with right hand then flipped the video just it...
So now the center of the earth is 0 g. So what abt the inner core, did it fload everywhere in that inner core like at the very center of the earth?
Yes, I believe so. (I wasn't actually there, of course).
Cheers,
Dr. A
How is he mirror writing so well?