What would happen if the center of the Earth were empty: would it be possible to walk on the inner walls of the spherical cavity? In fact, at the geometric center of the Earth gravity is 0, but on the walls of the cavity it should be "negative", ie I should be attracted to the surface of the earth by the intermediate mass. Or am I wrong something?
Newtons Shell theorem states that any mass inside a hollow sphere with constant density will not have any resultant force on it from the gravity of the sphere. Meaning if that were to happen you would be in 0G, and can float anywhere. The theorem was also breifly used in this video, thats why we could ignore the mass which we already passed when digging to the centre of the Earth.
Given that the gravity acceleration in the center is zero and the rotation of the earth imposes a centrifugal force to all partial masses, which is the force that is pushing all the masses towards the center? By intuition, one could assert that the "attractor" should be a shell surface somewhere between the external surface and an internal spherical void.
If the value of g decreases as we approach the center, where g=0,then the force at the center is also 0. But there is great pressure at the center that generate temperature that melts the center.Why? Does it mean the shell theory is incorrect?
You cannot relate force and pressure in this way. Pressure is a scalar so added pressure does not cancel, however added force, if in the opposite direction, does. So, at the centre of the Earth, you wont be dragged by a force, as you are pulled in opposite directions which cancel in any direction. However, millions of tonnes of rock are pulled towards you, causing equal (and immense) pressure all around you.
Newtons Shell Theorem states that if you are in a hollow sphere with constant density then it doesn't exert a force on you (because if you are close to one side, then any force from the small mass closer to you, is prefectly cancelled by the large mass on the other side of the sphere). Hence, there is no need to consider any forces from mass which is on the exterior of Earth, as any mass you pass can act as a hollow sphere.
hmm, but you didn't take into account the different densities of the layers inside Earth, which are also growing in jumps while approaching its center! Which I guess wasn't your goal :p
Hello ten year old comment. This was my frustration with the video. I have been staring at an image that seems to show the highest gravitational force at the core-mantle boundary (gutenberg discontinuity). en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation I will keep searching for a demonstration that includes the varying densities as depth increases.
Wilson's law....what goes down...must go up inside a hollow earth.. Wilson already proved earth is hollow.... Wilson proved it and Newton altered it to humans on surface. Certain winners want humans to believe they live on surface. Newton's laws is fine for surface .. Wilson's law is perfect for inner surface. He proved the polars are a way to surface of earth.
Whoa. :) For this problem, F=ma because the mass of the particle isn't changing. Only when the mass is changing is it that you need F=dp/dt. Hope this helps.
This was exactly what I needed to code gravity in a game I'm making
This just helped me _so much_ with my physics homework. Thank you!
same here
I looked for "Gravitation inside earth" and landed straight on your video. Exactly what I needed. Excellent crystallization. Thanks!
Reidar Wasenius Excellent
Awesome teacher. Thank you for the help.
Good lecture!
Great vid - will set this as a challenge to my students 👌
What would happen if the center of the Earth were empty: would it be possible to walk on the inner walls of the spherical cavity? In fact, at the geometric center of the Earth gravity is 0, but on the walls of the cavity it should be "negative", ie I should be attracted to the surface of the earth by the intermediate mass. Or am I wrong something?
Newtons Shell theorem states that any mass inside a hollow sphere with constant density will not have any resultant force on it from the gravity of the sphere. Meaning if that were to happen you would be in 0G, and can float anywhere. The theorem was also breifly used in this video, thats why we could ignore the mass which we already passed when digging to the centre of the Earth.
@@Triple_Trouble739 Thanks Morgan, my daughter, who is in her first year of Physics, explained it to me too! I apologize, I am a humanist ...
Amazing, have watched all your videos on shell theorem, makes it so very clear.
Excellent.
But what about the forces the earth parts above apply on the body?
This should clear things up. Let me know if you need more explanation.
th-cam.com/video/PtpkiS4it88/w-d-xo.html
Don't forget to indicate your units!
Given that the gravity acceleration in the center is zero and the rotation of the earth imposes a centrifugal force to all partial masses, which is the force that is pushing all the masses towards the center?
By intuition, one could assert that the "attractor" should be a shell surface somewhere between the external surface and an internal spherical void.
Why do you assune that outer ring does not contribute to the point at -d
I'm not sure what you're asking. Can you ask this a different way?
@saulremi thanks for responding.
Let me draw this and then post my question.
why is the radius (r) not decreased to (r-d) ie effective radius ?
Why the radius not decreased to r-x ?
you sound like jake gyllenhaal. good stuff, btw
Thanks so much- this really helped!
If the value of g decreases as we approach the center, where g=0,then the force at the center is also 0. But there is great pressure at the center that generate temperature that melts the center.Why? Does it mean the shell theory is incorrect?
You cannot relate force and pressure in this way. Pressure is a scalar so added pressure does not cancel, however added force, if in the opposite direction, does. So, at the centre of the Earth, you wont be dragged by a force, as you are pulled in opposite directions which cancel in any direction. However, millions of tonnes of rock are pulled towards you, causing equal (and immense) pressure all around you.
Thank you so much.
Why is the weight of a person higher, when he is in a deep hole? (mine)
I saw a test on that example.
Shouldnt it be less weight?
K_ Ralph Correct. It should be less weight. The "r" in the equation for 0
Thanks a lot
What about the material that you pass on the way down. I think you can't cancel the R in the denominator because it should be r-D (distance down).
Newtons Shell Theorem states that if you are in a hollow sphere with constant density then it doesn't exert a force on you (because if you are close to one side, then any force from the small mass closer to you, is prefectly cancelled by the large mass on the other side of the sphere). Hence, there is no need to consider any forces from mass which is on the exterior of Earth, as any mass you pass can act as a hollow sphere.
Wilson's law proves humans live inside earth...
hmm, but you didn't take into account the different densities of the layers inside Earth, which are also growing in jumps while approaching its center! Which I guess wasn't your goal :p
Hello ten year old comment. This was my frustration with the video. I have been staring at an image that seems to show the highest gravitational force at the core-mantle boundary (gutenberg discontinuity). en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation
I will keep searching for a demonstration that includes the varying densities as depth increases.
Wilson's law....what goes down...must go up inside a hollow earth..
Wilson already proved earth is hollow....
Wilson proved it and Newton altered it to humans on surface.
Certain winners want humans to believe they live on surface.
Newton's laws is fine for surface ..
Wilson's law is perfect for inner surface.
He proved the polars are a way to surface of earth.
At 00:33 Newton’s second law is NOT F=Ma it is Rate of change of momentum.. for Christ’s sake you are a goddamn Physics teacher
Whoa. :) For this problem, F=ma because the mass of the particle isn't changing. Only when the mass is changing is it that you need F=dp/dt. Hope this helps.
you have anger issues
What about the material that you pass on the way down. I think you can't cancel the R in the denominator because it should be r-D (distance down).
r in the numerator is not the radius of the earth, it's the distance from the centre so you can cancel.
meant denominator, although given that I'm replying after a yr it's probs not worth correcting myself lol
hope this helps
Wilson's law proves humans live inside earth...