15
$\begingroup$

Lets say the Earth is hollow and you are in the center of it (Same mass except all of it is on the outside like a beach ball) If you move slightly to one side now your distance is closer to that side therefore a stronger gravitational force however at the same time you have more mass now on the other side. At what rate would you fall? Which direction?

Also, is there a scenario where depending on the radius of the sphere you would fall the other direction or towards the empty center?

$\endgroup$
29
$\begingroup$

If the mass/charge is symmetrically distributed on your sphere, there is no force acting on you, anywhere within the sphere. This is because every force originating from some part of the sphere will be canceled by another part.

Like you said, if you move towards on side, the gravitational pull of that side will become stronger, but then there will also be "more" mass that is pulling you in the other direction. These two components cancel each other exactly.

$\endgroup$
  • 8
    $\begingroup$ This is one of my favorite properties of gravity! It has great ramifications for future large spacecraft construction, and other such things. However, this ideal answer only applies to an ideal hollow sphere (i.e. as soon as the sphere has something in it, like a uniform atmosphere, or a giant spaceship, anything else, you will begin to experience a net gravitational attraction toward the center of mass.) $\endgroup$ – Wug May 22 '15 at 0:56
  • 1
    $\begingroup$ So if you are 10 meters inside the sphere, gravity will attract you equally from all directions? $\endgroup$ – CJ Dennis May 22 '15 at 7:26
  • 1
    $\begingroup$ @CJDennis Yes, that is the point of the answer :-) $\endgroup$ – Ant May 22 '15 at 16:28
  • 1
    $\begingroup$ Why is this answer more popular than the Newton's Shell Theorem answer? $\endgroup$ – Aron May 23 '15 at 7:26
25
$\begingroup$

The answer is that inside a spherically symmetric shell of matter (your hollow earth or massive beach ball) there is no gravitational force anywhere - you will not "fall" in any direction, whether you are at the centre or not, regardless of the radius of the sphere.

This is a classic result of both Newtonian Gravity, and Einstein's General Theory of Relativity. In both cases it is called the [ ] Shell Theorem. Although, except at the centre, you are closer to one side than the other you can imagine that there is "more" farther away than there is closer, and because of the nature of the 1/r^2 law of Newtonian gravity everything cancels out. It is of course more complicated in General Relativity.

For a simple introduction, see the Wikipedia article on the Shell Theorem

You may also be interested to know that according to General Relativity time passes more slowly inside a hollow massive sphere than it does outside.

Whilst there is no gravitational force inside the sphere, and therefore no gravitational field, there is a non-zero gravitational potential - it just happens to be the same everywhere. Since the force of gravity depends on potential difference (just like voltage in the electrostatic case), if the potential is constant there can be no force. However, the potential does effect the passage of time in General Relativity, and since it is not zero inside the sphere time passes differently. Inside an "ordinary" matter sphere time passes more slowly.

$\endgroup$
  • $\begingroup$ +1: great answer, especially for the second half, giving a simply grasped, but not often spoken about, consequence. $\endgroup$ – WetSavannaAnimal May 21 '15 at 23:36
  • $\begingroup$ @WetSavannaAnimalakaRodVance Thanks - esp. since I finally disentangled many confusions about "flat" inside (Ricci flat and Riemann flat) and at infinity (Ricci flat but not Riemann flat) etc. thanks to other answers here to more technical related questions. $\endgroup$ – Julian Moore May 22 '15 at 8:19
  • 1
    $\begingroup$ "inside a spherically symmetric shell [...] there is no gravitational force anywhere" Only if there are no significant masses outside that shell. For example, I still feel a strong gravitational force towards the floor, even though I'm surrounded by the shell of my house. $\endgroup$ – David Richerby May 22 '15 at 9:47
  • 3
    $\begingroup$ @Taemyr Don't be ridiculous -- I live on a hill! :-) $\endgroup$ – David Richerby May 22 '15 at 11:20
  • 3
    $\begingroup$ @DavidRicherby To be fair, you feel the gravitational force towards the floor because the shell of your house at that point includes the planet attached to the floor, so your house is not spherically symmetric. Now, if the shell of your house were in freefall, then you would feel no gravitational force relative to your house. Of course, you would eventually feel the floor, but at that point your house wouldn't be in freefall... it would be in pieces. $\endgroup$ – iamnotmaynard May 22 '15 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.