Gravitation vertical mine shaft question

[homomorphism]

Homework Statement

Show that, at the bottom of a vertical mine shaft dug to depth D, the measured value of $$g$$ will be

$$g = g_{s}\left(1-\frac{D}{R}\right)$$$$g_{s}$$ being the surface value. Assume that the Earth is a uniform sphere of radius $$R$$.

Homework Equations

$$F = \frac{GMm}{r^{2}}$$

$$V_{S} = \frac{4}{3}\pi r^{3}$$

The Attempt at a Solution

I thought you could just plug in $$(R-D)$$ in the force equation but when I looked at the solution they did something with a ratio of masses that looked like this:

$$\frac{M(r)}{\frac{4}{3}\pi r^{3}} = \frac{M}{\frac{4}{3}\pi R^{3}}$$

where $$M$$ is the total mass.

Then the solution went on to this:

$$F = \frac{GM_{E}m}{r^{2}}\left(\frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi R^{3}}\right)$$

I don't really understand these last two steps. Can someone please explain what is happening here, and why are they doing a ratio of masses (related by density) and then multiplying by this ratio?

 

[cryptoguy]

Right, they are figuring out what part of the Earth contributes to the gravity because as you can imagine, if you are a distance D in a mine shaft, the mass of the ground above you is not going to attract you towards the center of the Earth while the ground below you will. So they used a volume ratio to represent the mass of the Earth still "below" you aka between you and the center

 

[dynamicsolo]

The Attempt at a Solution

I thought you could just plug in $$(R-D)$$ in the force equation

Unfortunately, it's not quite that simple, since that equation expresses the force outside a sphere of mass M and radius R.

What the solution is doing, which crytoguy is saying in another way, is to compare the gravitational acceleration, g_s, at the Earth's surface, to a faked-up planet which has the same average density of Earth, but is smaller in radius by an amount D.

The value for Earth's surface gravity is $$g_s = \frac{GM}{R^2}$$; you can use a similar equation for the alternative planet's surface gravity, $$g = \frac{GM(r)}{(R-D)^2}$$, with r = R - D (I am referring to the solution's manual's notation in part of this).

Putting these pieces together will lead to the desired equation.