I’ve read that at the center of large celestial bodies there’s zero gravity (or close to). While confirmation would be nice, if true, I’m wondering how large that area can actually be and moreover, does it scale up with more mass and/or even size - that is, does the sun have a larger center area of low (zero?) gravity than the earth and so on with evermore mass. Or is that area the same regardless of mass’ size?
Thank you
There is no area or volume of zero gravity inside planets or stars. It exists as a point, but since it’s a point, it has zero size.Go in any direction from that point, no matter how little. Now more mass is behind you than in front of you; you feel gravity pulling you back.Edit: Seems I was wrong, sorry.
- https://physics.stackexchange.com/questions/150238/gravitational-field-intensity-inside-a-hollow-sphere
- https://en.wikipedia.org/wiki/Shell_theorem
“If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object’s location within the shell.”
So it’s not zero but low gravity and increases the more mass-I leave behind me as I move out from the center?
That’s exactly what I meant, yes.
I’m not sure if it was correct though, edited my previous comment. Though maybe you did not ask about hollow bodies.
It’s basic math. You can do the gravitational calculations yourself. Basically any sphere of uniform density is going to exert gravity uniformly. So if you’re in the center the pull from the mass on any direction will be counteracted by the pull in the opposite direction. It’s one of the basic introduction to physics calculus examples.
Example:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html
So to your question about what the zone of negligible gravity would be, you can define negligible gravity, and then figure out how large that zone would be based on the material on the outside of the shell.
Basically the further you get from the exact center of the sphere, you’re going to have more gravity from the closer edge pulling you, and less gravity from the further edge offsetting that. So there’ll be a gradient of increasing gravity as you get further away from the center
Not quite. If by “edge” you mean the surface of the earth, then the force of gravity from the closer edge will always exactly offset the gravity from the farther edge. So if the earth were hollow, then you would experience zero gravity at any point in the hollow portion.
Of course, the earth is not hollow. And any mass under you (i.e. closer to the center than you are) will not be offset, and all of it will pull you towards the center. As you move further away from the center, more of the earth’s mass will be closer to the center than you and therefore the force of gravity will increase.
Excellent point, well made!
So if the original poster wanted to have a 10km sphere in the center of the earth of zero gravity (earth gravity at least), then all they have to do is hollow out 10km and they are good to go?
Nah, there’s only one exact center of gravity for a given body of mass. You can’t just make a vacuum and have gravity equalizer throughout it.
Yes you can, as long as you are inside a perfectly spherical shell.
The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells. This proves that the force from any spherically symmetric mass distribution on a mass inside its radius is zero.
From: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html#wtls
Thank you for replying.
This feels very close to answering the question in a way my brain can interpret it. So, going outward makes complete sense to me but the area at the center, the way I under your answer is, yes, the area or zone will increase proportional to its mass?
This may be asking too much, but, have any idea the size of that low gravity zone of earth bs our Sun?
I can’t answer that question for you. Because you’re using a relative term. Only the exact center will have no gravity. Anything outside the exact center will have some gravity. So you have to define what negligible it means.
So once you define low gravity. You can do the math to figure out the size of that zone of low gravity.
I highly recommend doing the math anyway. Follow along with a YouTube example or a written example on gravitational attraction of a sphere. It’s really good calculus. Then you can you know put it into octave and get the exact answer for yourself. Just plugging in numbers for the relative density of the Earth and the mass and the sizes. These will be approximate of course. Because nothing is perfectly uniformly dense so it’s just a rule of thumb anyway
I guess I don’t know enough about the equations necessary to solve for a gradient of area at the exact center, equal to it’s surrounding mass.
All the same, thank you for replying. Seems like the area might not be as large as I had supposed.
Depends what you’re trying to do. If you want to balance something so it never moves you can only use the exact center. If you want something to stay relatively in the center for a period of hours then you’re going to have a much larger area. If you’re okay with minutes it’s going to be much much larger area. If it needs to be stable for years in the area is smaller. Gravity is going to apply a force of acceleration and on an object, and if there’s nothing resisting that acceleration things will just fall off the center. You know imagine trying to balance something on top of a cone.
My original question stemmed from thinking about the possible different area sizes of low gravity within different size stars - and if that area was gradient.
deleted by creator
deleted by creator
Inside a sphere of constant density, gravity is linearly related to distance from the center.
So for example the Earth has a radius of ~4000 miles. Assuming it has constant density, a 200 pound man would be weightless at its center, weigh 0.2 pounds at 4 miles from the center, weigh 2 pounds at 40 miles from the center, weigh 100 pounds at halfway to the surface, and so on.
So for the Sun, taking its density/pressure into account, will the same gravity gradient exist but on a much larger scale?
Thank you
A linear relationship would exist if the sun were uniform in density, but it isn’t.
Though there is still a nonlinear change in gravity as you approach the center of the sun.
So the larger the star, given that most (or all) aren’t uniform, there will come a gradient of gravity at its center that one can’t even call it low gravity - it’s heavy material is simply churning too much for their to be a stable center of gravity?
I think the best way to visualize it is that when you are inside a star, you are effectively “standing” on a smaller star. Everything behind you can theoretically be ignored. When you are very close to the center, you are standing on a very tiny star.
So instead of the hole density from one side to the other, I only have the density from the center to its surface, am I understanding that correctly?
I’m not sure what you mean by “surface”.
Imagine you are standing on the surface of Earth, and you weighed 200 pounds.
Now imagine Earth were magically transported to the center of the sun, completely replacing an equal volume of solar core. Inside the very middle of the sun, standing on planet earth, you would still weigh 200 pounds. The gravity of all the solar mass surrounding the Earth would cancel out.
If you traveled upwards, to the surface of the sun, your weight would increase. At the sun’s surface, you would weigh 5400 pounds.
If you had a planet that was hollow in the center*, the entire hollow region would have zero gravity. You could have a thin-skinned planet with the entire interior an empty weightless void. I doubt any planets like this actually exist.
* Assuming radial symmetry. If you can represent the planet as concentric spherical shells then you’re good.
If you had a planet that was hollow in the center*, the entire hollow region would have zero gravity. You could have a thin-skinned planet with the entire interior an empty weightless void. I doubt any planets like this actually exist.
- Assuming radial symmetry. If you can represent the planet as concentric spherical shells then you’re good.
I thought this was wrong, but it is true:
- https://physics.stackexchange.com/questions/150238/gravitational-field-intensity-inside-a-hollow-sphere
- https://en.wikipedia.org/wiki/Shell_theorem
Brb fixing my other comments.
Yeah it’s a pretty counter intuitive result. I’d expect a greater pull of gravity towards the nearer side, but it turns out to be exactly cancelled out by the greater mass on the further side.
E: oops, looking at your edited comment, I should stress this is only for hollow bodies. Your comment pre-edit was correct for non-hollow bodies. If you’re part way to the middle of a planet, you can think of the planet as two sections, a small sphere for the part that’s below you, and a larger hollow shell for the rest. You experience no gravity from the outer shell, so only feel gravity of the smaller mass below. 10m from the earth’s center, you feel equivalent gravity to if you were on a 10m radius iron sphere.
A center of gravity is a single point and couldn’t be expanded to fill a planet’s interior, even if the space was only 1m^3
You’re right but that was not the point. The comment just explained that at any point inside a hollow sphere gravity forces cancel out so that effectively there is no gravity.
