That really is a beautiful solution, and a wonderful proof. Thank you for sharing it. And the satellite part! it all makes sense now as its "height" is a function of a sine function (SHM) and the horizontal position is also one, but on a different plane thus having no effect. Wow.
Circle is a special case of ellipse with foci F1 = F2. And in the case of the tunnel (which is a straight line) foci are equal R. So basically this tunnel is a very flat orbit :-) Beautiful solution indeed :-)
thank you, sir, for the amazing solution to the physics problem of last week. Even though I was able to solve the question earlier, your explanation gave it more clarity, your explanations always do!! :)
Every video you post makes me excited about physics! Love every video and how challenging the problems are for me, being a sophomore in high school. I still love them though!
This is amazing you are working this out now, they thought I was crazy when I spoke of engineering this as an equilibrium solution for the Earth. For certain, I need more studying and practice on my communicating of my ideas.
I must admit I was thrown by the lat & long and looked up the web on how to calculate the length of the tunnel, which was an interesting sin cos function. Well now I learnt 2 things!
Given: x"(t) + K*x(t) = 0 as the differential equation. K is a constant, not necessarily the spring stiffness constant. Just a constant in general that represents the ratio of restoring force to inertia. Our differential equation involves a second derivative plus the a coefficient times the function, equaling zero. We need a function whose second derivative is a negative version of itself. Which is sine and cosine. Recognize the cycle of derivatives that happens with sine and cosine. Derivative of sine = cosine. Derivative of cosine = -sine. Derivative of -sine = -cosine. Derivative of -cosine = +sine. This is only true when you use radians as the unit of the argument. When you have a coefficient of the variable in the argument of the trig function, such as cos(w*t), the chain rule of derivatives applies. That coefficient comes out in front multiplied with the derivative of the function. Both sine and cosine are plausible solutions of this equation, as well as linear combinations of the two. Define the radian frequency for each function to be w, such that the argument of each function is w*t. Put arbitrary coefficients in front of each sine and cosine term. x(t) = C1*cos(w*t) + C2*sin(w*t) Alternatively, you can recognize that you can combine both trig functions, into one trig function with an arbitrary amplitude (A) and phase shift (phi). x(t) = A*cos(w*t + phi) In either case, you'll generate the second derivative of this function, and substitute x"(t) and x(t) back in to the original equation. The goal being, to solve for w in terms of K. You will end up determining that w = sqrt(K). With initial conditions, you create variants of this equation that allow you to solve for either A and phi, or C1 and C2. These are two independent data points, with a time and either a position or velocity at that time.
I love your lectures, Sir. I'm not a physicist - to be honest i do not understand very much of your calculation :), but i love physics. And it is very interresting about the time of satelites to orbit the Earth. These days the ISS is visible from Drammen (not far from Oslo), so i will watch the ISS, and see if it reappears after ruoghly one and a half hour. If i*m lucky i can also get a QSO with one of the astronauts, since there is always a licenced radioamateur on board the ISS. Again, thank you for interresting lectures, and keep up the good work. Greetings from Norway.
Great stuff ! .....& even more amazingly, the time will be the same # on virtually ANY size planet, moon, asteroid/body anywhere .....just as long as its density is the same as earth's !
really very beautiful solution thank you so much for the video sir and it means sir that time period is independent of our and vehicle's mass, means If I and you are sliding through the same tunnel (obviously that will be the best moment of my life) at the same time in different vehicles then we will reach at the same time on the other end?
The tunnel problem is great. Can I ask you something about throwing a pebble in the air? It is said that the trajectory of a pebble thrown in the air is a parabola. I claim it must be an ellipse under ordinary circumstances. My reasons: if the mass of the earth were concentrated in the centre the pebble would no hit the ground and it would go all the way to the centre but missing it. It will overpass it but eventually it will come back along an ellipse. It will not escape the earth. For that one had to throw it with the escape velocity. Am I right?
Yes it's an ellipse. If you slice out of the ellipse a portion which is very small, so small that g is constant to a very high degree of accuracy then it's a parabola to a very high degree of accuracy. The difference between g at ground level and 100 m up is 0.000000002 m/s^2.
Thanks for sharing this! I followed a different path, using the coordinates, but I didn't account for the fact that only the mass of earth belonging to the current radius attracts the object, and I got arround 100min
Professor quick question: Was there any reason for choosing Honolulu to Oslo? & are you CERTAIN the cylindrical boundary of your tunnel is a straight line? I am very curious. many blessings -amv
I thought you said in an earlier video that a magazine story proposing to go from Honolulu - Oslo. Though my memory refuses to serve me very well therefore I'm probably incorrect.
Respected Sir. That is really Great. I wonder how you keep yourself energatic? You have the same energy level which we see in your MIT lectures. God Bless you, Sir.
Dear professor Lewin. I couldn’t sleep two nights. And thinking about trigonometric identities. And integrals of acceleration. I think “little R” it is not constant. If the tunnel goes more deeply then will be less constant which will be a more accurate solution when r is not constant?
sir, even I thought the same initially that the time period shall not depend on the length of the tunnel but later I imagined something like this, if I dig a tunnel that passes through the radius of the earth and many more tunnels parallel to it till the edge of the earth(In 2D like the one you have drawn) which will have shortest length. if I drop balls in all the tunnels at the same time, would the time period in all the tunnels be same? how about the last and the shortest tunnels on either side? The length will be very very short compared to the one at the middle, so I thought time period should also be very small, shouldn't it? Please sir, explain it to me in brief, I'll catch it.
Lectures by Walter Lewin. They will make you ♥ Physics. Will there be any change in time taken to cross the tunnel because of the increase in amplitude of SHM? ?
ofcoz there will be a change in the time; the higher the initial speed the shorter the time will be to make it from O to H. THat's rather obvious. Please watch my 8.01 lectures and you will learn some important physics.
That's surprising and very interesting. But if it took Newton 40 years, I don't feel too bad about not discovering it in a week :-) It makes sense that it's the same for a satellite. What it does is basically the same oscillation, it just does it in Y direction as well as X.
It's true for ANY point inside ANY shell (with outside radius R of Earth and inside radius R-D) as long as the mass density in any thin shell (with thickness much smaller than D) inside the shell with radii R and R-D is constant. However the density can vary a lot with r. Newton already addressed this in great detail. It's essentially Gauss' Law.
I don't know if I'm being a bit dumb here but I thought omega was root(k/m) and if MG/R^3 is the constant then therefor shouldn't the equation be omega=root(MG/mR^3)?
Lectures by Walter Lewin. They will make you ♥ Physics. Do you know anywhere where I can ask questions that I have on physics or math, because I am teaching myself physics and maths and often some things are hard to get my head around without asking questions or having a teacher. If you could tell me who I can ask my questions to online, then that would be great! Thanks!
Amazing solution ....... but if I remember right; if this a simple harmonic motion , then this time is the "periodic time" which is "Time needed to make one complete cycle" , and if that is the case then this is the time needed to go from Oslo to Honolulu and back again to Oslo . then we actually need only half of this time . I am not saying the solution is wrong .... it is brilliant , but I think the time is doubled . Am i right ??
Satellites "tracks" the orbits are centric, the tunnel looks more like a straight line. Thus, I am not sure if this calculation are true, roughly probably yes. They would be 100% percent true if the tunnel would go exactly through the earth core centre/ not really possible. Then the pendulum mathematics is correct, this what I think. I would just divide the tunnel to the two parts and in one part the object is accelerating and i the other is breaking. The question is with which acceleration as the g (9,81) is in direction to the earth centre an the tunnel is not in this direction, thus we need to use a ramp equations (inclined plane) to get acceleration , connected with the angle. the other point is that we do not know if the g (9,81) is also 9,81 100km under the ground surface, but since this symetrical for acceleration and breaking this will be compensated.
The solution which I have posted holds for ANY straight deep tunnel anywhere on Earth under the assumption that the mass density is the same everywhere.
only youtube ? so look .. i'm arab . Egyptian Person and i pleased for knowing someone like you .. I learned a little physics but I did not understand anything from the teacher. I love physics but I did not have the opportunity to learn the right and it will be difficult to learn on TH-cam besides strengthening my English language I hope to learn physics from you
i think the basic concept is wrong. the person who enters the hole on oneside will oscillate inside the hole for some time and finally will stuck in the middle of the hole. ( simple pendulum)
Watch the first part of the video carefully, he assumed there won't be friction or air resistance. In the absence of these damping forces, your visualization doesn't happen. But this is very much similar to a simple pendulum.
if this motion is simple harmonic, then by conservation of energy the car or any body will not be able to make it back to Honululu if it starts from Honululu as well(Thanks to your video in which the 15 kg pendulum never reached your chin) :)
Sir I got the correct answer on the very first day . i was really happy to get it right thank you sir and please keep making such problems and I really loved this problem .
These kind of videos are the real gems hidden in garbage filled TH-cam. What a wonderful explanation!!
14:05 "appart from the fact that you never survive that travel trough the tunnle - thats a detail"
I love that kind of humor xD
Mind blown! Beautiful formula and when you linked it to an orbiting satellite, it all made perfect sense.
:)
Admiration for your love of teaching
😊
We love you, professor !
Bravo! What a fascinating result. Thank you, kind sir, for continuing to educate the world.
:)
I was having trouble finding the solution for a tunnel not passing through the earth, until I saw this. Thank you, as always, professor.
Elegant explanation! Thank you!
I remember this problem from my freshman Physics, still amazes me. Thankyou Dr. Lewin ...
:)
That really is a beautiful solution, and a wonderful proof. Thank you for sharing it.
And the satellite part! it all makes sense now as its "height" is a function of a sine function (SHM) and the horizontal position is also one, but on a different plane thus having no effect.
Wow.
Circle is a special case of ellipse with foci F1 = F2. And in the case of the tunnel (which is a straight line) foci are equal R. So basically this tunnel is a very flat orbit :-) Beautiful solution indeed :-)
thank you, sir, for the amazing solution to the physics problem of last week. Even though I was able to solve the question earlier, your explanation gave it more clarity, your explanations always do!! :)
:)
.....FROM BRAZIL....YOU ARE THE BEST...
Every video you post makes me excited about physics! Love every video and how challenging the problems are for me, being a sophomore in high school. I still love them though!
:)
Great sir
Beatifully simple!!!
Excellent explanation! Thank you!
Also, nice chart papers. how much does it cost?
This is amazing you are working this out now, they thought I was crazy when I spoke of engineering this as an equilibrium solution for the Earth. For certain, I need more studying and practice on my communicating of my ideas.
:)
Thank you for the challenge sir!
I must admit I was thrown by the lat & long and looked up the web on how to calculate the length of the tunnel, which was an interesting sin cos function. Well now I learnt 2 things!
:)
Good 🙏😊🙏😊
how can i solve that differential equation(9:52). I haven't been taught calculus yet. is there another way to solve it?
:)
Vasilis R you need to know about simple harmonic motion which requires some calculus, you can learn calculus on TH-cam.
Given:
x"(t) + K*x(t) = 0 as the differential equation. K is a constant, not necessarily the spring stiffness constant. Just a constant in general that represents the ratio of restoring force to inertia.
Our differential equation involves a second derivative plus the a coefficient times the function, equaling zero. We need a function whose second derivative is a negative version of itself. Which is sine and cosine.
Recognize the cycle of derivatives that happens with sine and cosine. Derivative of sine = cosine. Derivative of cosine = -sine. Derivative of -sine = -cosine. Derivative of -cosine = +sine. This is only true when you use radians as the unit of the argument. When you have a coefficient of the variable in the argument of the trig function, such as cos(w*t), the chain rule of derivatives applies. That coefficient comes out in front multiplied with the derivative of the function.
Both sine and cosine are plausible solutions of this equation, as well as linear combinations of the two. Define the radian frequency for each function to be w, such that the argument of each function is w*t. Put arbitrary coefficients in front of each sine and cosine term.
x(t) = C1*cos(w*t) + C2*sin(w*t)
Alternatively, you can recognize that you can combine both trig functions, into one trig function with an arbitrary amplitude (A) and phase shift (phi).
x(t) = A*cos(w*t + phi)
In either case, you'll generate the second derivative of this function, and substitute x"(t) and x(t) back in to the original equation. The goal being, to solve for w in terms of K. You will end up determining that w = sqrt(K).
With initial conditions, you create variants of this equation that allow you to solve for either A and phi, or C1 and C2. These are two independent data points, with a time and either a position or velocity at that time.
You only neee C of calculus ... Not like full package of it.
Awesome sir🤔🤔🤔🤔🤔
Thanks! Any more description of how the gravitational force for the part of earth that is "behind" the body does not have any effect?
use google
GORGEOUS PROBLEM !
I love your lectures, Sir. I'm not a physicist - to be honest i do not understand very much of your calculation :), but i love physics. And it is very interresting about the time of satelites to orbit the Earth. These days the ISS is visible from Drammen (not far from Oslo), so i will watch the ISS, and see if it reappears after ruoghly one and a half hour. If i*m lucky i can also get a QSO with one of the astronauts, since there is always a licenced radioamateur on board the ISS. Again, thank you for interresting lectures, and keep up the good work. Greetings from Norway.
thank u !
Great stuff ! .....& even more amazingly, the time will be the same # on virtually ANY size planet, moon, asteroid/body anywhere .....just as long as its density is the same as earth's !
Yes, that's amazing. I will mention that in the video of my next Physics Problem. Well the asteroid would have to be spherical!
Nice question and explanation.
:)
really very beautiful solution thank you so much for the video sir
and it means sir that time period is independent of our and vehicle's mass, means If I and you are sliding through the same tunnel (obviously that will be the best moment of my life) at the same time in different vehicles then we will reach at the same time on the other end?
yes
The tunnel problem is great.
Can I ask you something about throwing a pebble in the air?
It is said that the trajectory of a pebble thrown in the air is a parabola.
I claim it must be an ellipse under ordinary circumstances.
My reasons: if the mass of the earth were concentrated in the centre the pebble would
no hit the ground and it would go all the way to the centre but missing it. It will overpass it
but eventually it will come back along an ellipse. It will not escape the earth. For that one had to throw it
with the escape velocity.
Am I right?
Yes it's an ellipse. If you slice out of the ellipse a portion which is very small, so small that g is constant to a very high degree of accuracy then it's a parabola to a very high degree of accuracy. The difference between g at ground level and 100 m up is 0.000000002 m/s^2.
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you.
Thanks for sharing this! I followed a different path, using the coordinates, but I didn't account for the fact that only the mass of earth belonging to the current radius attracts the object, and I got arround 100min
You rock walter!
:)
Beautiful problem.
:)
Professor quick question:
Was there any reason for choosing Honolulu to Oslo? & are you CERTAIN the cylindrical boundary of your tunnel is a straight line? I am very curious.
many blessings
-amv
yes there was a reason - strictly private
I thought you said in an earlier video that a magazine story proposing to go from Honolulu - Oslo. Though my memory refuses to serve me very well therefore I'm probably incorrect.
yes I did say that. But I assumed that all of you would realize that that was just a made-up story. Such a tunnel is completely impossible.
very educational sir
:)
Respected Sir. That is really Great.
I wonder how you keep yourself energatic? You have the same energy level which we see in your MIT lectures.
God Bless you, Sir.
thanks
Thank you sir for wonderful solution
Now I am feeling physics
God give you long life
Dear professor Lewin. I couldn’t sleep two nights. And thinking about trigonometric identities. And integrals of acceleration. I think “little R” it is not constant. If the tunnel goes more deeply then will be less constant which will be a more accurate solution when r is not constant?
*watch my solution and get educated*
Really sir I love soo muuuucchhhhh🥰😏😏😏😏😏🤩🤩
sir, even I thought the same initially that the time period shall not depend on the length of the tunnel but later I imagined something like this, if I dig a tunnel that passes through the radius of the earth and many more tunnels parallel to it till the edge of the earth(In 2D like the one you have drawn) which will have shortest length. if I drop balls in all the tunnels at the same time, would the time period in all the tunnels be same? how about the last and the shortest tunnels on either side?
The length will be very very short compared to the one at the middle, so I thought time period should also be very small, shouldn't it?
Please sir, explain it to me in brief, I'll catch it.
:)
wonderful sir!!
:)
If you are quite near the center of earth, isn't the mass above your head also attracting you?
all that mass results in net effect of zero
use google
sir do you think we can survive when at the middle of the tunnel due to the gravitational force ?
you will not survive. Temperatures will be too high.
Lectures by Walter Lewin. They will make you ♥ Physics. will the g force enough to kill us if not take temperature into account?
the maximum g along the direction of the tunnel is 7.4 m/s^2 which is LESS than 9.8 m/s^2. Thus that will NEVER kill you. Yes the heat may kill you.
Lectures by Walter Lewin. They will make you ♥ Physics. so the conclusion is impossible to make this tunnel and we can never cheat with the nature
this prob can solved by pure math with given position?
What if the object has some initial velocity at the mouth of the tunnel ??
it will then have that same speed as it emerges from the tunnel at H.
Lectures by Walter Lewin. They will make you ♥ Physics. Will there be any change in time taken to cross the tunnel because of the increase in amplitude of SHM? ?
ofcoz there will be a change in the time; the higher the initial speed the shorter the time will be to make it from O to H. THat's rather obvious. Please watch my 8.01 lectures and you will learn some important physics.
That's surprising and very interesting. But if it took Newton 40 years, I don't feel too bad about not discovering it in a week :-) It makes sense that it's the same for a satellite. What it does is basically the same oscillation, it just does it in Y direction as well as X.
:)
Great proof! It should be interesting to compare the solution for the tunnel and the satellite when the relativistic effect is taken into account.
relativistic effects are negligibly small. All speeds involved are muchhhhh smaller than c.
You absolutely right, but for the mathematics only, to see how the two situations give the same result...
Didn´t you ignore the gravitation force of the shell you have already traveled through? I think this would increase the time for the journey.
The grav force due to the shell I traveled through is ZERO! use google.
Is it, for any given point in the shell?
It's true for ANY point inside ANY shell. Newton already addressed this in great detail. It's essentially Gauss' Law.
It's true for ANY point inside ANY shell (with outside radius R of Earth and inside radius R-D) as long as the mass density in any thin shell (with thickness much smaller than D) inside the shell with radii R and R-D is constant. However the density can vary a lot with r. Newton already addressed this in great detail. It's essentially Gauss' Law.
Thank you for your answer. Amazing how easily intuition misleads us!
Hi Walter, I have been wanting to have a chat with you, is it possible to contact you via e-mail ? Thanks a lot.
Love u sir
I don't know if I'm being a bit dumb here but I thought omega was root(k/m) and if MG/R^3 is the constant then therefor shouldn't the equation be omega=root(MG/mR^3)?
you applied equations without understanding them.
Lectures by Walter Lewin. They will make you ♥ Physics. What did I misunderstand?
use google
Lectures by Walter Lewin. They will make you ♥ Physics. Do you know anywhere where I can ask questions that I have on physics or math, because I am teaching myself physics and maths and often some things are hard to get my head around without asking questions or having a teacher. If you could tell me who I can ask my questions to online, then that would be great! Thanks!
use google and quora
Amazing solution ....... but if I remember right; if this a simple harmonic motion , then this time is the "periodic time" which is "Time needed to make one complete cycle" , and if that is the case then this is the time needed to go from Oslo to Honolulu and back again to Oslo . then we actually need only half of this time .
I am not saying the solution is wrong .... it is brilliant , but I think the time is doubled .
Am i right ??
watch my video again!
Oh ..... You have said it .... I didn't notice , I focused on the written formula , Thanks for your concern
Yeah!!! Got it right
Thankyou.
The one who is going to answer me is it the real Professor Walter Lewin or the admin of this channel employed by Professor?
so, straight lines inside the earth are equivalent to circles above the earth regarding travel time? cool
:)
Satellites "tracks" the orbits are centric, the tunnel looks more like a straight line. Thus, I am not sure if this calculation are true, roughly probably yes. They would be 100% percent true if the tunnel would go exactly through the earth core centre/ not really possible. Then the pendulum mathematics is correct, this what I think. I would just divide the tunnel to the two parts and in one part the object is accelerating and i the other is breaking. The question is with which acceleration as the g (9,81) is in direction to the earth centre an the tunnel is not in this direction, thus we need to use a ramp equations (inclined plane) to get acceleration , connected with the angle. the other point is that we do not know if the g (9,81) is also 9,81 100km under the ground surface, but since this symetrical for acceleration and breaking this will be compensated.
The solution which I have posted holds for ANY straight deep tunnel anywhere on Earth under the assumption that the mass density is the same everywhere.
Well, I did not get the math right. But proud to say I got the concept of the irrelevance of the location of the tunnel.
:)
sir,,do you have page or private profile on facebook that makes me easy contact you ?
sorry about bad english :D
I am not on Facebook and not on Linkedin.
only youtube ?
so look .. i'm arab . Egyptian Person and i pleased for knowing someone like you .. I learned a little physics but I did not understand anything from the teacher. I love physics but I did not have the opportunity to learn the right and it will be difficult to learn on TH-cam besides strengthening my English language
I hope to learn physics from you
:)
الله يخرب بييييتك جرصتنا قدام الاجانب 😂😂😂 learn english bro its for ur own good believe me😉
First! ... had to do it
i think the basic concept is wrong. the person who enters the hole on oneside will oscillate inside the hole for some time and finally will stuck in the middle of the hole. ( simple pendulum)
Watch the first part of the video carefully, he assumed there won't be friction or air resistance. In the absence of these damping forces, your visualization doesn't happen. But this is very much similar to a simple pendulum.
didn't realize everything happen at the same plane
en ik maar wachten op Walter met schepje die aan het graven gaat..... hihi
je wachtte niet lang genoeg
if this motion is simple harmonic, then by conservation of energy the car or any body will not be able to make it back to Honululu if it starts from Honululu as well(Thanks to your video in which the 15 kg pendulum never reached your chin) :)
*you are mistaken* - do your homework on SHM
I think sir he is referring to damped oscillation.
I dont care if you care about it or not.
The straight distance between Honolulu and Oslo is about 8763km !
Thank you very very very much 🤍
Sir I got the correct answer on the very first day . i was really happy to get it right thank you sir and please keep making such problems and I really loved this problem .
:)
this problem is beautifult is in kleppner and kolenkow book ,do you like this book?
good book