(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

[tex]g = g_{s}\left(1-\frac{D}{R}\right)[/tex]

[tex]g_{s}[/tex] being the surface value. Assume that the Earth is a uniform sphere of radius [tex]R[/tex].

2. Relevant equations

[tex]F = \frac{GMm}{r^{2}}[/tex]

[tex]V_{S} = \frac{4}{3}\pi r^{3}[/tex]

3. The attempt at a solution

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

[tex]\frac{M(r)}{\frac{4}{3}\pi r^{3}} = \frac{M}{\frac{4}{3}\pi R^{3}}[/tex]

where [tex]M[/tex] is the total mass.

Then the solution went on to this:

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

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?

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# Homework Help: Gravitation vertical mine shaft question

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