91Ó°ÊÓ

The satellite undergoes centripetal as well as gravitational force when it revolves around the Earth.

By using Conservation of energy, the centripetal force becomes equal to the gravitational force,

\({F_c} = {F_g}\)

The time period of the Moon is,

\(\begin{aligned}T &= 27.4\;{\rm{days}}\\ &= {\rm{27}}{\rm{.4}}\;{\rm{days}} \times 24\;{\rm{hours}} \times 60\;\sec \times 60\;\sec \\ &= 2367360\;{\rm{s}}\end{aligned}\)

Thus, the time period of the Moon is \(2367360\;{\rm{s}}\).

By using Conservation of energy,

\(\begin{aligned}{F_c} &= {F_g}\\\frac{{m{v^2}}}{r} &= \frac{{GMm}}{{{r^2}}}\\\frac{{m{{\left( {\frac{{2\pi r}}{T}} \right)}^2}}}{r} &= \frac{{GMm}}{{{r^2}}}\\M &= \frac{{4{\pi ^2}{r^3}}}{{GT}}\end{aligned}\) ... (i)

Here, \(G\) is the gravitational constant, \(M\) is the mass of the planet and \(m\) is the mass of the satellite.

Substitute the value in equation 1,

\(\begin{aligned}M &= \frac{{4{\pi ^2}{{\left( {3.84 \times {{10}^8}\;{\rm{m}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}} \times {{\left( {27.4\;{\rm{days}} \times {\rm{24 hr}} \times {\rm{3600 s}}} \right)}^2}}}\\ &= 5.98 \times {10^{24}}\;{\rm{kg}}\end{aligned}\)

Thus, the mass of the Earth is \(5.98 \times {10^{24}}\;{\rm{kg}}\).