# Centre Of Gravity.

## Main Question or Discussion Point

Hi all.
I was presented today with a question that I think I know the right answer to but I just want to confirm it.
Imagine that a cylinder is removed from the earth, the cylinder being inside the earth, like the height of the cylinder is the diameter of the earth. The cylinder goes right through the earth. (sorry for my crappy description).

Say if one was to jump into this, would you fall right to the centre of gravity and remain still and experience no other force, or when you arrive at the centre of gravity get torn apart by the gravitational effect of the rest of the earth. I think the first case is correct.
Thanks!

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You are talking about a sort of giant hole through the entire earth, right? You would, depending on how tightly packed together you are, fall completely to the other side(negating air resistance) and back again forever, or you would be ripped apart on your way down. The second is more likely. You wouldn't stop at the center because you've got a ton of kinetic energy, which happens to be exactly equal to the amount you need to get exactly to the bottom.

I am yes.
So you would continuously oscillate through the center of gravity until you would come to a stop and then be torn apart?

Doc Al
Mentor
Say if one was to jump into this, would you fall right to the centre of gravity and remain still and experience no other force, or when you arrive at the centre of gravity get torn apart by the gravitational effect of the rest of the earth. I think the first case is correct.
The first is not quite correct, but the second is totally off. Why would you "remain still"? You have just fallen from a great height to the center of the earth! You'd be moving fast--and what's going to stop you? (Ignore air resistance and other complications, of course.)

But you certainly won't be torn apart: When you are at the center of the earth, inside this hollow cylinder, the gravitational force on you is zero.

tiny-tim
Homework Helper
You are talking about a sort of giant hole through the entire earth, right? You would, depending on how tightly packed together you are, fall completely to the other side(negating air resistance) and back again forever, or you would be ripped apart on your way down. The second is more likely. You wouldn't stop at the center because you've got a ton of kinetic energy, which happens to be exactly equal to the amount you need to get exactly to the bottom.
Hi Ed!

Flexo is right … you'd be in orbit … a very thin orbit!

But I don't think there's any danger of being ripped apart … the gravity inside a hollow sphere (say, empty up to radius a, and solid for a < r < b) is zero, so the gravity at radius a inside a solid sphere is the same as on the surface of a solid sphere of radius a.

I am yes.
So you would continuously oscillate through the center of gravity until you would come to a stop and then be torn apart?
You would not come to a complete stop in a vacuum. In a more real scenario, with air, that is roughly correct.

Bah, I was wrong about the tearing.

Well without air resistance you would never stop. With air restistance the oscillation would be damped (with each new oscillation the maximum amplitude would be a little less). As for the torn apart I don't see why you would (and if you were it would be in no way related to whether you 'stopped' in the center or not). The graviational force will never be greater in the center then on the surface (in fact it will pretty much be zero assuming a spherical earth and that the cylinder is of sufficiently small radius). It would however be very very hot.

Hmm I see. I obviously don't quite get this, how come the gravitational force from the rest of the mass of the world around you will not effect you?

Well the 9.81 N/kg you experience on the surface of the planet is the result of the entire mass of the earth pulling on you (yes gravity is THAT weak) (Netwon's GMm/r^2). However this attraction is a vector quantity (i.e. you are pulled in the direction of the mass). So imagine you're one lightyear 'to the left' of our sun. Then you'd be pulled towards it. However, imagine there was an identical copy (same mass) of our sun 'to the right' as well. Then the force to the left and right would be the same and basically cancel producing no net force. Same in the interior, you would feel a force in all directions within the sphere which when you add them up would all cancel leaving a total zero gravity.

Well the 9.81 N/kg you experience on the surface of the planet is the result of the entire mass of the earth pulling on you (yes gravity is THAT weak) (Netwon's GMm/r^2). However this attraction is a vector quantity (i.e. you are pulled in the direction of the mass). So imagine you're one lightyear 'to the left' of our sun. Then you'd be pulled towards it. However, imagine there was an identical copy (same mass) of our sun 'to the right' as well. Then the force to the left and right would be the same and basically cancel producing no net force. Same in the interior, you would feel a force in all directions within the sphere which when you add them up would all cancel leaving a total zero gravity.
Using the not entirely spherical earth, would you actually stretch out a little bit towards the equator, or is it such that as long as any vector has an equal and opposite one, nothing will occur?

tiny-tim
Homework Helper
Hmm I see. I obviously don't quite get this, how come the gravitational force from the rest of the mass of the world around you will not effect you?
'cos it's more or less equal-and-opposite … so it cancels!

Newton's Shell Theorem is at the root of what tiny-tim says. It is also at the root of why we may approximate a spherical shell as a point source when observed from the outside.

I think you are very much over estimating the strength of gravity (Hollywood is probably to blame for this) the fact that the earth is an ellipsoid would only make the net gravity at the center like a pound or something. Which is hardly a 'tearing you apart' level of force. However, the fact that the earth is not a perfect sphere and the fact that there is air restistance (and that air pressure would be inhomogenous as you passed through various temperatures) means that the real world result would not be you oscillating until you ran out of energy but in fact your path would not be perfectly straight so you'd probably slam into the wall at some point.

"Newton's Shell Theorem is at the root of what tiny-tim says. It is also at the root of why we may approximate a spherical shell as a point source when observed from the outside."

That's actually incorrect. The fact that the force of gravity can be approximated as an inverse square law means that you would feel no force ANYWHERE within a mass shell. However, regardless of the form of the law (if it was 1/r^5 for example) you would still experience zero net force in the CENTRE of a mass shell because that is simply a result of vector cancelation. But obviously in the situation mentioned you would have to pass through other points in order to get to the center but the lack of force at the centre itself is not dependent on the inverse square nature.

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Thanks very much everyone, it all makes sense now.

OMG this is the EXACT question i've been wondering since i was a kid, I can't believe your description of the experiment was so similar to mine

thank you guys i've one doubt less in my life

tiny-tim
Homework Helper
Welcome to PF!

OMG this is the EXACT question i've been wondering since i was a kid, I can't believe your description of the experiment was so similar to mine

thank you guys i've one doubt less in my life
Hi valleyman ! Welcome to PF!

What a coincidence!

I wonder what you were actually looking for?

Thank you, this is the first time i visit this forum, but I'm not looking for anything in particular. I found this link somewhere and I thought to give a look, as I'm a physics student at the university of Rome. I enjoyed this first visit, I think i'll come here much more

Hmm I see. I obviously don't quite get this, how come the gravitational force from the rest of the mass of the world around you will not effect you?
As it's been mentioned already, it does effect you but when you are subject to more forces at once, the forces add as vectors, and at the centre of earth every piece of earth around pulls you towards it and therefore all forces cancel each other out.

Your gory image of being ripped apart is probably inspired by thinking "well, if 2 friends start pulling me from different sides, they don't cancel each other out, they rip me in half!" :)

But that's actually true only because each of your evil friends will be likely pulling one different arm. The gravity would be pulling each and every particles you're made of with extremely minor differences between your left and right sides. You wouldn't even notice a minor swelling.

I forget who predicted this, but you would reach the other side of the earth in 42 minutes and come to a stop. This is all theory whereas the earth is a "pefect" sphere and the tunnel is a perfect vacuum!

"I forget who predicted this, but you would reach the other side of the earth in 42 minutes and come to a stop. This is all theory whereas the earth is a "pefect" sphere and the tunnel is a perfect vacuum!"

I don't really think that's something that a certain "person" predicted. The estimate could be done by pretty much anyone who knows integral calculus.

so this was just general knowledge since the beginning of time? nobody was the first to derive this? haha

No but it's like saying "I forget who first determined how long it would take a train leaving from station A going a speed of 120km/h to reach a station B that is 300km away".

well im going to be the first to derive why everyone argues with me on this website even when a statement i say is correct... thanks for your input maverick