# Planet internal gravity

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inside  planets................not  @ the  center  but lets say like here

Shown  in the picture

Is  the gravity less than  on the  surface as  there  is some mass  above and  also  less mass below your feet

Or  is  the  gravity more  because your are  closer the the center of mass ?

Thanx

Edited by Saber
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Depends on from which side you take the measurement.

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If the planet is spherically symmetric then only the planet mass between the body and the planet center attracts the body. Outer part of the planet cancels.

So, the short answer, less gravity.

* "between the body and the planet center" means a sphere centered in the planet center

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Sorry, I need to clarify. This answer assumes that the density is the same throughout the planet depth. In reality it is not so, of course. The answer depends on how the density changes with depth and might be different at different depths.

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3 hours ago, Saber said:

inside  planets................not  @ the  center  but lets say like here

Shown  in the picture

Is  the gravity less than  on the  surface as  there  is some mass  above and  also  less mass below your feet

Or  is  the  gravity more  because your are  closer the the center of mass ?

Thanx

Look up Newton’s shell theorem. That explains why the mass at a radius greater than the location of the red dot has no gravitational effect on it.

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Sorry, I need to clarify. This answer assumes that the density is the same throughout the planet depth.

No, you had it right the first time. The derivation of the shell theorem that exchemist just mentioned applies to each (spherically symmetric) shell individually, regardless of its density.

For constant density, the gravitational field is linear in the radius:

$\displaystyle{a(r) = \frac{G m(r)}{r^2}}$

$\displaystyle{a(r) = \frac{G \rho (\frac{4}{3} \pi r^3)}{r^2}}$

$a(r) = G \rho (\frac{4}{3} \pi r)$

$a(r) = G \rho (\frac{4}{3} \pi R) \displaystyle{ \left(\frac{r}{R}\right)}$

$\displaystyle{a(r) = \frac{GM}{R^2}\left(\frac{r}{R}\right)}$

$\displaystyle{a(r) = g\left(\frac{r}{R}\right)}$

Edited by Lorentz Jr
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3 hours ago, Lorentz Jr said:

The derivation of the shell theorem that exchemist just mentioned applies to each (spherically symmetric) shell individually, regardless of its density.

This is right. So, why I thought the clarification is needed? Just consider the extreme case where the whole mass is near the planet center and the rest of the planet is massless. As the body goes from the surface toward the center it just gets closer to the same mass. The gravitational force increases.

In the other extreme case where the whole mass is in the surface shell and the planet is empty inside, the body will become weightless as soon as it crosses the surface.

Ergo, the answer depends on the density distribution with the depth.

Also, take a case where the entire mass is in the shell halfway down. As the body gets closer to that depth the gravity increases, and when it crosses that shell the gravity drops to 0.

Ergo, the answer depends on the depth.

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Genady is correct the density and distribution of the mass makes a difference.

There is nothing wrong with your maths, but note you initial conditions.

4 hours ago, Lorentz Jr said:

each (spherically symmetric) shell individually, regardless of its density.

For constant density, the gravitational field is linear in the radius:

However there is more to this than meets the eye since gravity, whether considered as force or as an acceleration is a vector, not a scalar.

The distribition and density of the mass affects the direction of gravity.

So the geometric centre and the centr of gravity no longer conincide.

It is interesting to note the the discovery of the massiveness of the Himalaya was discovered by Everest from deductions on observations variations of alignment of plumb bobs on the plains beneath.

Yet another effect has not been mentioned and is oft forgot.

The reason why gravity is greater at the poles than the equator , due to centrigugal effects of our spinning planet.

Once again this also has an effect on direction.

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the answer depends on the depth.

You said "the short answer, less gravity" and "This answer assumes that the density is the same throughout the planet depth."

So it looked like you were saying "less gravity assumes that the density is the same throughout the planet depth."

The boundary between increasing gravity and decreasing gravity (i.e acceleration as a function of r) is constant gravity:

$\displaystyle{a(r) = \frac{Gm(r)}{r^2} = g}$

$Gm = g r^2$

$G dm = 2 g r dr$

$\displaystyle{\rho(r) = \frac{dm}{dV} = \frac{dm}{ 4 \pi r^2 dr}}$

$\displaystyle{\rho(r) = \frac{dm/dr }{ 4\pi r^2 } = \frac{2gr}{4\pi r^2 G}}$

So $\displaystyle{\rho(r) = \frac{k}{r}}$, where $\displaystyle{k = \frac{g }{ 2 \pi G}}$.

In other words, the acceleration will be g all the way down as long as the density is inversely proportional to the distance from the center. It will decrease with r (ie. increase with depth) if the density depends on r more sensitively than 1/r, and will increase with r (decrease with depth) if it depends less.

Edited by Lorentz Jr
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28 minutes ago, Lorentz Jr said:

it looked like you were saying...

Fair enough. I should've said instead something like, "In that answer I've assumed..."

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Okay, that makes sense.

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"The gravity anomaly at a location on the Earth's surface is the difference between the observed value of gravity and the value predicted by a theoretical model. If the Earth were an ideal oblate spheroid of uniform density, then the gravity measured at every point on its surface would be given precisely by a simple algebraic expression. However, the Earth has a rugged surface and non-uniform composition, which distorts its gravitational field."

ps. Gravity anomaly is widely used in e.g. mining industry.

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4 hours ago, studiot said:

Yet another effect has not been mentioned and is oft forgot.

The reason why gravity is greater at the poles than the equator , due to centrigugal effects of our spinning planet.

This has an effect on the normal force, which is what we perceive, but not on gravity.

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According  to  the  shell theorem  inside  shells   there is no  gravity .....right    or better say  gravitational  forces   end up  cancelling  each  other...........

In a  planet  thats  not a  shell...................but  @  the  very very  center...............can  we  assume  its a  thick  thick   shell  ( thickness of the  shell = R  of the planet )

therefore @ the  center of planets  there is no  gravity ........ ?

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4 hours ago, Saber said:

In a  planet  thats  not a  shell...................but  @  the  very very  center...............can  we  assume  its a  thick  thick   shell  ( thickness of the  shell = R  of the planet )

therefore @ the  center of planets  there is no  gravity ........ ?

Yes, that's reasonable. I believe it would be an extension of Newton's original formulation for thin shells, but it's simple enough. You can think of any spherically symmetric planet as an onion, with lots of concentric shells. 🙂

Edited by Lorentz Jr
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On 1/14/2023 at 4:16 PM, swansont said:

This has an effect on the normal force, which is what we perceive, but not on gravity.

Yes indeed, strictly so. Also since the centrifugal force is imaginary, we really should instigate a more complicated analysis.

However the effect still affects the direction of gravity since the rotation causing this apparent reduction in the magnitude of gravity is not in the plane of the gravitational attraction, so the effect depends upon latitude and the forces necessary to calculate the normal force must be added vectorially.

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