After a spacewalk, a 7-kg tool is left 54 m from the center of gravity of a 21-metric ton space station, orbiting along with it. How much closer to the space station will the tool drift in two hours due to the gravitational attraction of the space station?
Gravitational Constant : 6.674 X 10^-11 -->G
The Attempt at a Solution
1 - Finding gravitational force =G x 7 x 21,000 / 54^2 = Mt ( tool mass ) a
2 - Finding acceleration = (G x 7 x 21 , 000 / 54 ^2) x Mt = a
3 - Finding distance = Vf = Vi(0) + at ----> Vf = at
Final answer----> Vf^2 = Vi(0) + 2a ( d) ----> Vf^2 / 2a = d
This is just the procedure to solving the answer. I'll appreciate it, If someone can kindly point out my mistakes by showing work in response.