Question

Question #56ea6

Answer 1

The acceleration at a distance of `2R` 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 = (GMm)/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_(surface) = (GMm)/R^2`
`F_(2R) = (GMm)/(2R)^2`
A little algebra turns that second equation into:
`F_(2R) = (GMm)/(4R^2)`
Or, writing that a little more clearly:
`F_(2R) =1/4 (GMm)/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 = mg`
So the acceleration will be:
`g = F/m`
But it should be easy to show that this is in the same proportions at the differences in F.