gravitic force in planetoids

[Chaigidel]

does the gravitic force increase or decrease as you approach the center of planetoids?

remember not pressure im talking about the influence of gravitic forces.

I cant seem to find anything on it, just wanted to put it out there

[cankles]

increases. If the moon were any closer it would be pulled into us. At least I think so.

[Gaidin]

increases. If the moon were any closer it would be pulled into us. At least I think so.

The moon is currently being pulled into us. That's the cause of the orbit.

blah it has to increase relating to density

Volume has nothing to do with it. Only mass.

f = Gm1m2/r*r

Where r is the distance between the bodies.

[Chaigidel]

blah it has to increase relating to density-- I just got confused because of rotating space stations-- thinking that all our force came from the spin, but its totally attached to that mass bro

[Redem]

The assumption in that formula is that we are dealing with point masses. If we're talking about approaching the centre of a body of mass, then the mass above you will be pulling in a different direction from the mass below you, until it exactly cancels out when you are in the centre of mass. i.e. if you dug a tunnel through a solid planetoid, the force of gravity would lessen as you neared the centre.

[Chaigidel]

so it would actually lessen in the center? interesting

[Simetrical]

As you approach a sphere of uniform density from the outside, gravitational force will increase. It will peak at the surface and decrease as you dig below the surface, as follows: at any given distance below the surface, the gravity you experience is equal to that exerted by anything that's further below the surface. So if you're 1000 miles below the Earth's surface, the top 1000 miles of the Earth will exert no net force on you. And if you're at the center of the Earth, there's nothing further below the surface than you are, so the gravity is zero.

One way to look at it uses this fact: a hollow spherical shell of uniform density (and any thickness) exerts no gravitational force on its interior. So if the Earth was hollow, there would be gravity inside it. This occurs because the different parts of the shell exactly cancel one another out. Using this, suppose I've dug 1000 miles beneath the Earth's surface. Now each part of the Earth exerts some gravity on me. Consider it in two parts: the parts more than 1000 miles below the surface (the center), and the parts above. Notice that the parts above me form a spherical shell, and I'm in the interior of it, right at the edge. Therefore, that exerts no force on me. Only the center, below me, exerts force.

This is probably not so easy to follow without diagrams, I'm afraid.

(Technical note: the above examples using the Earth might be taken to imply that it's of uniform density, which it's probably not to any acceptable approximation. But its density can be taken as spherically symmetric about its center, which is good enough for this result.)

[chriscase]

As you approach a sphere of uniform density from the outside, gravitational force will increase. It will peak at the surface and decrease as you dig below the surface, as follows: at any given distance below the surface, the gravity you experience is equal to that exerted by anything that's further below the surface. So if you're 1000 miles below the Earth's surface, the top 1000 miles of the Earth will exert no net force on you. And if you're at the center of the Earth, there's nothing further below the surface than you are, so the gravity is zero.

One way to look at it uses this fact: a hollow spherical shell of uniform density (and any thickness) exerts no gravitational force on its interior. So if the Earth was hollow, there would be gravity inside it. This occurs because the different parts of the shell exactly cancel one another out. Using this, suppose I've dug 1000 miles beneath the Earth's surface. Now each part of the Earth exerts some gravity on me. Consider it in two parts: the parts more than 1000 miles below the surface (the center), and the parts above. Notice that the parts above me form a spherical shell, and I'm in the interior of it, right at the edge. Therefore, that exerts no force on me. Only the center, below me, exerts force.

This is probably not so easy to follow without diagrams, I'm afraid.

(Technical note: the above examples using the Earth might be taken to imply that it's of uniform density, which it's probably not to any acceptable approximation. But its density can be taken as spherically symmetric about its center, which is good enough for this result.)

IIRC there is a Gaussian surface integral for this too.

[ajm317]

IIRC there is a Gaussian surface integral for this too.

Yes.

[Simetrical]

Good enough for which result? If the density of the "outer layers" is very low, then the force of gravity may continue to increase as you penetrate deeper. For example if you take the atmosphere to be part of the Earth (a reasonable thing to do) then the force of gravity is higher at sea level than it is on top of Mt. Everest. For Earth this seems like a trivial point, maximum gravity is almost certainly on the planets surface, but if one considers planets like Jupiter things become less clear.

The result that the total gravitational force is precisely due to that part of the sphere closer to the center than you applies to any spherically-symmetric density distribution, as I realized. However, I assumed without thinking about it that this implied decreasing gravitational pull as one moved toward the center. As you point out, this is of course wrong. I wasn't accounting for the decreasing distance to the center of what remains.

IIRC there is a Gaussian surface integral for this too.

Well, of course you can always integrate it, for arbitrary mass distributions.

[Sphere]

Someone actually did some research on this. One interesting conclusion was that if you were to fall through a tunnel between any two places on the Earths surface, it would take 43 min (IIRC) to reach the other side, no matter the distance between the two points. Google it if you want more details.

[Ruin]

As you move towards a planet the attraction to you increases exponentially by a factor of two. The formula is attraction=GM/r^2 to find the attraction at any point.

(G=Universal Constant of Gravitation (6.67x10^-11), M=mass of planet, r=radius. The mass of the Earth is 5.98x10^24 kg)

This dictates that as you move towards it, every time you halve this distance you quadruple the force of attraction you are experiencing towards it, until you reach the Earth's surface.

When you begin to burrow into the surface you experience less attraction. This is because only the matter below you will attract you in that direction, all area to your sides and above you will pull in their respective directions. This effect increases until you are at the very center, and are experiencing equal forces in all directions, as there is an equal volume of matter in each direction to attract you.

Sorry if I repeated any or all of sim's points.

[ajm317]

As you move towards a planet the attraction to you increases exponentially by a factor of two.

When someone says something "increases exponentially" they mean it goes like e^r. Nobody says "exponentially by a factor of two." More appropriate language here would be something like "the attraction goes as the inverse square of the distance."

Technical note: the above examples using the Earth might be taken to imply that it's of uniform density, which it's probably not to any acceptable approximation. But its density can be taken as spherically symmetric about its center, which is good enough for this result.

Good enough for which result? If the density of the "outer layers" is very low, then the force of gravity may continue to increase as you penetrate deeper. For example if you take the atmosphere to be part of the Earth (a reasonable thing to do) then the force of gravity is higher at sea level than it is on top of Mt. Everest. For Earth this seems like a trivial point, maximum gravity is almost certainly on the planets surface, but if one considers planets like Jupiter things become less clear.

If we're talking about the result that the force is 0 at the center, then yeah.

[Ruin]

When someone says something "increases exponentially" they mean it goes like e^r. Nobody says "exponentially by a factor of two." More appropriate language here would be something like "the attraction goes as the inverse square of the distance."

Sorry, it should have been increasing by a factor of two, that was my mistake. I'm used to seeing graphs of that shape and judging them for exponentiallity.

[Ruin]

I'm pretty certain if you were to fall through a tunnel from one end of the Earth to the other then you would eventually end up right in the middle, completely unmoving. As you fall towards the center you will reach maximum velocity, as you pass it the gravitational pull reverses, and you will slow, then fall in the other direction, in smaller and smaller distances.

[Chaigidel]

save for pressures caused by the matter around the core-- you probably wouldnt float but be crushed into something the size of a peanut.

but in place of gravity alone very interestingly yes--- what is that theoretical sphere around a black hole? -- I mean because a singularity is a gravitic phenomena, but if your in the center of a black hole would you also be under the influencce of 0 g?

[Ruin]

Technically no, you would be under the influence of huge amounts of g in all directions, it just so happens they would cancel one another out at the center of your mass. You would be torn to shreds, but you'd stay in the same place!

[Simetrical]

Someone actually did some research on this. One interesting conclusion was that if you were to fall through a tunnel between any two places on the Earths surface, it would take 43 min (IIRC) to reach the other side, no matter the distance between the two points. Google it if you want more details.

Obviously not correct. If I fall through a tunnel that leads from my current location to, say, one inch away, it's not going to take 43 minutes to get there. It may be that it would take that long to get to the other side of the Earth.

I'm pretty certain if you were to fall through a tunnel from one end of the Earth to the other then you would eventually end up right in the middle, completely unmoving. As you fall towards the center you will reach maximum velocity, as you pass it the gravitational pull reverses, and you will slow, then fall in the other direction, in smaller and smaller distances.

Yes, it would just be harmonic oscillation, like with a spring. With no damping you'd oscillate forever; with damping you'd end up in the middle, in equilibrium. If there's too much friction you'd never get close enough to the other side to be able to jump out, even on the first pass.

but in place of gravity alone very interestingly yes--- what is that theoretical sphere around a black hole? -- I mean because a singularity is a gravitic phenomena, but if your in the center of a black hole would you also be under the influencce of 0 g?

Whatever point on your body was exactly in the center would be under the influence of exactly 0 g. Parts of you that were, say, one millimeter to the side would not necessarily be under that influence. If the black hole were precisely spherically symmetric and you hollowed out a precisely spherical cavity in the middle, you would experience 0 g as long as you remained in that cavity, yes. But if it were non-symmetric by even, say, 0.0001%, you would almost certainly experience enough gravity to instantly kill you, although it would be almost nothing compared to the full strength of the black hole's pull.

Of course, as far as I'm aware it's considered possible that the black hole really has a singularity at the middle, a single point where all its mass is concentrated (or practically all). If this is the case, you wouldn't be able to go inside it at all, so this discussion doesn't apply.

Technically no, you would be under the influence of huge amounts of g in all directions, it just so happens they would cancel one another out at the center of your mass. You would be torn to shreds, but you'd stay in the same place!

Well, it depends on the exact configuration of the mass.

[chriscase]

I think the point of the spherical symmetry is that all the mass outside the sphere on which the point in question sits is effectively cancelled out.

EDIT: Sorry if I'm reiterating but I'm not sure everyone got that. The field calculation happens at individual points. A macro object under the influence of a field with high force gradients across the body would of course be subject to structural stress, but that's a different question than the field calculation.

[Sphere]

Google strikes again. Using only gravity it would indeed take you 43 minutes to "fall" 2 inches.

http://www.npl.washington.edu/AV/altvw32.html

If a subway car is put into the Manhattan-to-Sydney subway tunnel, the car behaves like the bob of a pendulum. It bobs back and fourth between the two extreme points of its travel at a rate which is independent of the mass of the subway car and depends only on the mass and diameter of the Earth. Assuming that the Earth has a uniform mass density, that there is no friction or air resistance in the tunnel, and that the subway car is moving solely under the influence of gravity with no other forces, the time required for a one-way Manhattan-to-Sydney trip in the subway car is an amazing 43 minutes. (Flying non-stop from JFK Airport to Sydney by 747 at 500 mph would require about 25 hours.)

Sydney and New York City lie almost along a diameter through the center of the Earth. What about other locations, say Seattle to Moscow, Chicago to Houston, or Tokyo to London. It turns out that under the above assumptions the result is always the same. The transit time from anywhere to anywhere is 43 minutes. Such a subway system would, with no net expenditure of fuel or energy, would provide 43 minute transport between any two point on the Earth's surface.