How do you use circular motion equations to solve this problem?

Mars has a mass of one tenth that of Earth and a diameter half that of Earth. The acceleration of a falling body near the surface of Mars is...?



Please don't give me an answer, give me a process. Thank you~How do you use circular motion equations to solve this problem?Go to the equation for the gravitational force:



F= m(Mars)*m(object)*G / r(Mars)^2



Assuming that its the only force acting on the object, and knowing that the sum of the forces on an object equals the mass of the object times the acceleration of the object, we can set these equal so that we have:



m(Mars)*m(object)*G / r(Mars)^2 = m(object)*a(object)



Now, notice that the mass of the object is found on both sides of the equation - so we can factor it out to get:



m(Mars)*G / r(Mars)^2 = a(object)



We know what G is - its just a constant. The mass of Mars is 1/10 = 0.1 times the mass of Earth, so let's say:



0.1m(Earth)*G / r(Mars)^2 = a(object)



Now we use the fact that the diameter of Mars is half that of Earth. This means that the radius of Mars is half that of Earth. r(Mars) = 0.5*r(Earth) But! It's squared. We have to square the 0.5 as well so we get r(Mars)^2 = 0.25*r(Earth)^2. Now we have:



0.1m(Earth)*G / (0.25*r(Earth)^2) = a(object)



Look at this carefully. We know that the acceleration of an object close to the Earth's surface is 9.8 m/s^2. That is:



m(Earth)*G / r(Earth)^2 = 9.8 m/s^2



In our equation above, we have an extra 0.1/0.25 = 0.4 out front. This is fraction should be multiplied by 9.8 m/s^2 to get the answer!



0.4*9.8 m/s^2 = 3.92 m/s^2



Hope that made sense!How do you use circular motion equations to solve this problem?Newton's law of gravitation will give you a force, Newton's second law of motion will allow you to use this force to find an acceleration.



To adjust the mass of Mars and the radius of Mars for the equation... Mars is (Earth's Mass/ 10) and the radius is (Earth's radius/ 2).



Assume a 1 kg mass for the second mass in the equation and the force is equal to the acceleration. You must divide by one to make sure you cancel out the proper units to end up with an answer in m/s^2How do you use circular motion equations to solve this problem?Let's keep it simple. The gravitation equation is:

g = GM/r^2

We are given:

g(earth) = G*M(earth)/r(earth)^2 = 9.8 m/s^2

M(mars)/M(earth) = 0.1

r(mars)/r(earth) = 0.5

And we can deduce:

g(mars) = g(earth) * M(mars)/M(earth) / (r(mars)/r(earth))^2

Is that sufficient?