Question

Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the Earth’s surface.

Answer 1

`T_2 = 1.40 hours`

Step-by-step explanation:

As per kepler's law of time period we know that

square of time period is proportional to the cube of the distance

so here we can say

`(T_1^2)/(T_2^2) = (r_1^3)/(r_2^3)`

so we know that

for moon

`T_1 = 27.4 days`

`r_1 = 384,400 km`

`r_2 = 6370 km`

now from above formula we have

`(27.4^2)/(T_2^2) = (384,400^3)/(6370^3)`

`T_2 = 0.058 days`

`T_2 = 1.40 hours`