Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of $R/2$ where $R$ is the radius of the Earth. If $F(r)$ is the gravitational force of the Earth on a point mass at a distance $r$, what is: \[\frac{F(R)}{F(2R)}?\]
- 32
- 8
- 4
- 2
- 1
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Using the same logic, (1/2)^2 is 1/4. Use basic math, cross out the k's and R's, and tada: Just 1/4.
Yup that's correct.
Careful, the gravitational force is not inverse square inside the earth, if it were you would have a nearly infinite force at the center of the earth and the planet would implode.
Ooops. I guess I wasn't thinking very straight. The correct answer is 2 isn't it?
By Gauss's Law for Gravity,
\[g\cdot A_{enclosed} = 4\pi G\rho V_{enclosed}.\] (Some will note that I have reduced it to a form that is useful for this question.) At a radius of $r$, the (surface) area enclosed is: \[A_{enclosed} = 4\pi r^2.\] Likewise, the volume enclosed is: \[V_{enclosed} = \frac{4}{3}\pi r^3.\] Finally, the density of the Earth, $\rho$, is: \[\rho = \frac{M}{\frac{4}{3}\pi R^3}\] where $M$ and $R$ are the radius and mass of the Earth respectively.
Putting it all in, we have: \[g\cdot (4\pi r^2) = 4\pi G\frac{M}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3.\] After some simplification, we have \[g = \frac{GM}{2R^2}\] at $r = \frac{R}{2}$. Recall that at $r=R$, \[g=\frac{GM}{R^2}.\] Dividing the two expressions for $g$ above gives 2 as the proper answer.
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