# Question #56ea6

Mar 24, 2016

The acceleration at a distance of $2 R$ will be one quarter of that observed at a distance of $R$.

#### Explanation:

The question in your photo is asking about gravity a short distance above the surface of the earth. 60 miles sounds like a lot, but the earth has a radius of 4000 miles.

The gravitational force is calculated using this equation:
$F = \frac{G M m}{R} ^ 2$
Where $M$ is the mass of the earth and $m$ is the mass of some smaller object. And $G$ is the universal gravitational constant.

Comparing $F$ at the surface where $R = 4000$ to $F$ just 60 miles above the surface makes $R = 4060$. The question in your image is trying to get you to see that this is a small change.

In the Mars example we must change R to twice the value of R. So we are comparing these two equations:
${F}_{s u r f a c e} = \frac{G M m}{R} ^ 2$
${F}_{2 R} = \frac{G M m}{2 R} ^ 2$
A little algebra turns that second equation into:
${F}_{2 R} = \frac{G M m}{4 {R}^{2}}$
Or, writing that a little more clearly:
${F}_{2 R} = \frac{1}{4} \frac{G M m}{R} ^ 2$
And it should be clear that the gravitational force at this distance is one quarter what it is at the surface.

To be completely clear, this $F$ is a force between two specific masses. If $m$ is small compared to $M$ (and most things are small compared to planets), we usually just think of this the force per unit mass. The acceleration, usually denoted as little $g$ is related to the force in the usual way:
$F = m g$
So the acceleration will be:
$g = \frac{F}{m}$
But it should be easy to show that this is in the same proportions at the differences in F.