An artificial satellite orbits the Earth in a week. At what height above the Earth’s surface this satellite rotates.

Since it is not known at what speed the satellite is moving, we will use the known astronomical values and compare the speed of the satellite with the movement of the moon. We know the following data about the Moon:
R1 = 384000 km
T1 = 27 days
About the satellite
T2 = 7 days
R2 we are looking for
Using Kepler’s Law
(R2) ^ 3 / (R1) ^ 3 = (T2) ^ 2 / (T1) ^ 2
(R2) ^ 3 = (R1) ^ 3 * (T2) ^ 2 / (T1) ^ 2
R2 = R1 * cubic root (T2) ^ 2 / (T1) ^ 2
R2 = 384000 * cubic root (7) ^ 2 / (27) ^ 2 = 99454 km
Satellite altitude minimum thickness of the Earth’s surface = 99454-6370 = 93,086 km
Answer 93086km