Gravity on Mars

## Question:

Hi,
I had a question in one of my assignments that I found
interesting:

find acceleration of a falling object on mars, given that the
radius of mars is 1/2 of Earth and the mass of mars is 1/8 of
Earth.

I tried to use G=(f
d^{2})/(m1*m2) for each of them and equate
G's. I couldn't get to the answer.

please help me!

## Answer:

I would think of the problem in this way. If Mars were the same
mass as the Earth the acceleration at the surface would be 4
times as great as Earth's because the radius is 1/2 of
Earth's and the acceleration is inversely proportional to the
square of the radius. The acceleration is directly proportional
to the mass of the planet so on the surface of Mars instead of
being 4 times as great it is 4/8 times as great or 1/2.
There is nothing wrong with your approach. In the expression
G=(f d^{2})/(m1*m2) let's
replace f/m2 with A, using Newton's third law just to clean
up the notation a bit then on Earth
G=d^{2}*Ae/Me and on Mars
G=(1/2*d)^{2}*Am/(1/8*Me). Setting G=G we get
d^{2}*Ae/Me=(1/2)^{2}*d^{2}*Am/(1/8*Me).
Dividing both sides of this equation by d^{2} and
multiplying both sides by Me we get
Ae=8/4*Am or Am=1/2*Ae.

I hope this helps.

J. D. Jones