91Ó°ÊÓ

The period of an artificial satellite can be determined by the using Kepler’s third law.

The mathematical form of the third law is given as,

\({T^2} \propto {R^3}\)

Here, \(T\) is the time period and \(R\) is the radius.

The time period of the Moon is \(\left( {{T_m}} \right) = 27.4{\rm{ days }}\).

Standard value:

The orbital radius of the Moon is \({R_m} = 3.84 \times {10^8}\;{\rm{m}}\).

The radius of the Earth \({R_e} = 6.4 \times {10^6}\;{\rm{m}}\).

The expression for the Kepler’s third law is given as,

\({\left( {\frac{{{T_m}}}{{{T_s}}}} \right)^2} = {\left( {\frac{{{R_e}}}{{{R_m}}}} \right)^3}\)

Substitute the values in the above equation,

\(\begin{aligned}{\left( {\frac{{{T_m}}}{{{T_s}}}} \right)^2} &= {\left( {\frac{{{R_M}}}{{{R_e}}}} \right)^3}\\{T_s} &= \left( {\frac{{{{\left( {6.4 \times {{10}^6}\;{\rm{m}}} \right)}^3} \times {{\left( {27.4\;{\rm{days}} \times \frac{{{\rm{24}}\;{\rm{hours}}}}{{1\;{\rm{day}}}}} \right)}^2}}}{{{{\left( {3.84 \times {{10}^8}\;{\rm{m}}} \right)}^3}}}} \right)\\{T_s} &= 2\;{\rm{hours}}\end{aligned}\)

Thus, the time period of an artificial satellite is \(2\;{\rm{hours}}\).