Question #56ea6

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.