- 32
- 8
- 4
- 2
- 1

Solution :

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: \[\frac{F(R)}{F(2R)}?\]

**Choice 3 is the answer.**

A very simple question today.

Recall that the gravitational force between two objects is given by \[F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}\] where $k$ is a constant.

Hence we have: \[\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.\]

On the same exam, they also asked for \[\frac{F(R)}{F\left(\frac{R}{2}\right)}.\] Try it if you want some extra practice.

- 32
- 8
- 4
- 2
- 1

Solution :

A very simple question today.

Recall that the gravitational force between two objects is given by \[F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}\] where $k$ is a constant.

Hence we have: \[\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.\]

On the same exam, they also asked for \[\frac{F(R)}{F\left(\frac{R}{2}\right)}.\] Try it if you want some extra practice.

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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

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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