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 d²)/(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 d²)/(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=d²*Ae/Me and on Mars G=(1/2*d)²*Am/(1/8*Me). Setting G=G we get d²*Ae/Me=(1/2)²*d²*Am/(1/8*Me). Dividing both sides of this equation by d² and multiplying both sides by Me we get Ae=8/4*Am or Am=1/2*Ae.

I hope this helps.

J. D. Jones