- #1
Balsam
- 226
- 8
Homework Statement
Imagine another planet with an acceleration of 10m/s^2 at its equator when ignoring the rotation of the planet. The radius of the planet is 6.2 x 10^6m. An object dropped at the equator yields an acceleration of 9.70m/s^2. Determine the length of one day on this planet.
r=6.2x10^6m
g=9.70m/s^2
a=10m/s^2--> I don't know why we were given the first acceleration, I don't think it's equal to ac.
Homework Equations
I used Fc=4pi^2mr/T^2 and isolated for T.
The Attempt at a Solution
I used the above equation, plugging in mg for Fc because I think the force of gravity is the centripetal force. The 'm' variables canceled out since I just divided them out. Then, I plugged all of my other given values in and solved for T. Since T is the time it takes in seconds to complete one revolution, I divided my result by 3600 to get the time it takes in hours to complete one revolution. My final answer was about 1.3hours. However, the book's answer is 7.9hours. I don't know where I went wrong