Yet another physics problem...best answer possibility!?

Question

At what distance from the center of The Earthwould the acceleration due to gravity be equal to half of its value at the surface of the Earth, g=9.81 m/s^2? Please show how you get your answer/your thought process or whatever. Thank you in advance!

Answer 1

well, we know that F = ma, and with gravity, F = mg

What causes this force? gravitational attraction, which is defined by:

F = Gm1m2/r^2

So, we can say that mg=Gm1m2/r^2, where r is the distance between 2 objects (1 and 2).

now, lets solve it for g

g=Gm/r^2

this shows us that g varies inversely with the SQUARE of the distance. So the distance between the earth and an object would need to be r*(sqrt2)

Answer 2

You can simplify this a lot by using the shell theorem, see reference. If you are inside a uniform spherical shell of matter, the net gravitational force on you due to that shell is zero.

If you are outside a uniform sphere of matter, its gravitational attraction is identical to that of a point source at its center with the same total mass.

So figure out how deep you need to go before half of the mass of earth is in the sphere beneath you. Volume is proportional to the cube of the radius, so if you were at 57.7% of the radius of the earth from the center (3682 km), half the mass of theearth would be beneath you, and the acceleration of gravity would be half of what it is on the surface.

For a more accurate answer, account for the fact that the density of the earth is not uniform.

Answer 3

The force of grvity doesn't behave like 1/r^2 inside the Earth. It would have to be infinite at the centre...
it grows like r. I will try to show that.
Let r be your distance from the centre of the Earth, R - Earth's radius.
First of all, once you are inside the Earth, , all the layers above youhave no contribution to the force. The ones on your side of the centre cancel out with the ones on the far side. It is not obvious, but true.
Let us assume the Earth has uniform density, rho. Let V(r) be the volume of a sphere with radius r.
V(r)=(4*PI*r^3)/3
the force the sphere with radius r (Earth without outer layers) exerts on you isF(r)= G*M(r)*m/r^2 where M(r) is the mass of the sphere.
M(r)=V(r)*rho= (4*rho*PI*r^3)/3 ,so
F(r)=(4*G*rho*PI*m*r^3)/3/r^2 ,which we can simplify.
F(r)=(4*G*rho*PI*m*r)/3=
=((4*G*rho*PI*m)/3)*r=
=constant*r
so if at ground level the force exerted is F, F/2 is exerted halfway down the Earth
Hope it helps, contact me if you want an explanation as to why the outer layers don't count

Answer 4

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