91Ó°ÊÓ

Satellite in a circular orbit about the Earth with a radius is equal to one-half the radius of the Moon’s orbit.

Using Kepler’s third law, find theperiod of revolution of satellite in lunar months from the given radius and the ratio of Earth’s radius to Moon’s radius. According to Kepler’s third law, the squares of the orbital periods of the planets are directly proportional to the cubes of the semi major axes of their orbits.

Formula is as follow:

${\left(\frac{T_s}{T_M}\right)}^2={\left(\frac{R_s}{R_M}\right)}^3$

where, T is corresponding period and R is corresponding radius.

Now,

${\left(\frac{T_s}{T_M}\right)}^2={\left(\frac{R_s}{R_M}\right)}^3$

As,

$\frac{R_s}{R_M}=\left(\frac{1}{2}\right)$

Putting given values in formula,

${\left(\frac{T_s}{1lunarmonth}\right)}^2={\left(\frac{1}{2}\right)}^3$

$T_s=0.35lunarmonth$

Hence, the period of revolution of satellite in lunar is $0.35lunarmonths$

Therefore,using Kepler’s third law of gravitation or law of period, the time period of the satellite can be found.