Fractional Reduction of Earth's Gravity at Halfway to Center

[Kaxa2000]

You dig a hole half way to the center of the earth. You lower an object to the bottom
of the hole. By what fraction has the force of the Earth’s gravity on the object been reduced
relative to that at the surface? Assume that the Earth is a sphere and has uniform density

 

[Feldoh]

Assuming the mass of the hole is very, very small compared to the mass of the Earth http://en.wikipedia.org/wiki/Shell_theorem might help.

 

[Kaxa2000]

Would the force of Earth's gravity be reduced by 1/2?

 

[denverdoc]

Would the force of Earth's gravity be reduced by 1/2?

No. Did you look at the reference provided? Or looked at it and get totally intimidated--which is OK. Just say so. The answer is far less than what you suggest. Go back to the reference cited.

 

[Kaxa2000]

I used the "Inside a solid sphere of constant density the gravitational force varies linearly with distance from the center, becoming zero at the center of mass."

from the reference to make my previous suggestion and yes I did get intimidated when I scrolled down from there. Can someone direct me to a section in the reference which I should look at?

 

[ehild]

Would the force of Earth's gravity be reduced by 1/2?

Yes, correct.

ehild

 

[Kaxa2000]

Is there an equation that shows this?

 

[Kaxa2000]

What about ignoring all the layers of the Earth above the hole and just using the mass left to calculate the gravitational pull of the earth?

 

[ehild]

What about ignoring all the layers of the Earth above the hole and just using the mass left to calculate the gravitational pull of the earth?

Yes, this is the method.

ehild

 

[Borek]

Would the force of Earth's gravity be reduced by 1/2?

Yes, correct.

What about ignoring all the layers of the Earth above the hole and just using the mass left to calculate the gravitational pull of the earth?

Yes, this is the method.

This is inconsistent. If you ignore everything above, you are left with 1/8 volume, that means 1/8 mass, that means 1/8 force, not 1/2. What am I missing?

 

[Stonebridge]

What about ignoring all the layers of the Earth above the hole and just using the mass left to calculate the gravitational pull of the earth?

This wouldn't work because you would then be on the surface of a planet with a radius half that of earth. The resultant acceleration there would depend as usual on r squared.
Inside the solid sphere, you would be subject to a gravitational force in all directions, including back towards the surface. This is why, when you do the maths, the force inside is less (depends on r not r squared) than if you were on the surface.

 

[Stonebridge]

This is inconsistent. If you ignore everything above, you are left with 1/8 volume, that means 1/8 mass, that means 1/8 force, not 1/2. What am I missing?

Nothing. The poster was incorrect.

 

[ehild]

Assuming homogeneous density for the Earth, the mass M_r inside a sphere of radius r<R:

M_r=\frac{r^3}{R^3}M.

The gravitational force on a point mass m at distance r is the same as if all the mass of this sphere was concentrated in the centre:

G=\gamma\frac{mM_r}{r^2}=\gamma\frac{m\frac{r^3}{R^3}M}{r^2}=\gamma\frac{mM}{R^2}\frac{r}{R}

Which is r/R times the gravitational force at the surface. So Kaxa was right.

ehild

 

[Borek]

You can't at the same time assume homogeneous density and mass concetrated in the centre.

 

[ehild]

You can't at the same time assume homogeneous density and mass concetrated in the centre.

Yes, we can, as far as gravitational or electric field is the question. It is taught so even in high-school Physics.

http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation#Bodies_with_spatial_extent

ehild

 

[Borek]

OK, I have misread your post, I have missed that you are now referring only to the part of the mass that is inside r. No problem with that part being in the center, somehow I thought you were referring to the whole mass.

And I know what I have missed earlier - mass is 1/8, but square of the distance change is 1/4, this gives 1/2 as the answer to the original question. All fits now.

 

[D H]

You can't at the same time assume homogeneous density and mass concetrated in the centre.

You can show, however, that for any point outside a spherically symmetric mass distribution^1[/color], the gravitational acceleration of an object at that point toward the mass is exactly the same as the gravitational acceleration toward an equivalent point mass^2[/color]. How to show this? Newton did it with his shell theorem. An even easier way is to compute the gravitational potential and take the gradient. Even easier, use Gauss' law for gravity. Another consequence of Newton's shell theorem is that the gravitational force toward a hollow shell of mass with a spherically symmetric mass distribution is identically zero throughout the interior of the shell.

Back to the problem at hand, (artificially) split the Earth into a solid sphere and a spherical shell, with boundary between the two being the spherical surface that intersects the object lowered down the hole. The outer shell does nothing, and the inner sphere acts just like a point mass.

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^1[/color] A spherically symmetric mass distribution is one in which density is purely a function of distance from the center of the distribution. A uniform density is a special case of a spherically symmetric mass distribution.

^2[/color] "Equivalent point mass" means a point mass with the same mass as the mass distribution in question and located at the mass distribution's center of mass.

 

[Kaxa2000]

Can someone explain the gamma symbol in the gravitation equation ehild posted ?

 

[D H]

That would be the universal gravitational constant, which is more often represented as big G (the alternative is to use gamma). He used big G for gravitational force.