The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:

where

is the orbital radius of the first satellite

is the orbital radius of the second satellite

is the orbital period of the first satellite

is the orbital period of the second satellite
If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
One way to do it is she could right down the data that she got
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
Therefore, clearing the normal force we have:
The friction is 565N.
(565 / 0.8) = 706.25N. weight.
Answer: a) work done = 3946429.5 J
b) work done = 943.22 nutritional calories
Explanation: