Question

Determine the mass of the Earth from the known period and distance of the Moon.

Short answer

The time period of the Moon is \(2367360\;{\rm{s}}\) and the mass of the Earth is \(5.98 \times {10^{24}}\;{\rm{kg}}\).

Step-by-step solution

Concept

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,

\({F_c} = {F_g}\)

Finding time period of the Moon

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 \(2367360\;{\rm{s}}\).

Calculation of mass of the Earth

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 \(5.98 \times {10^{24}}\;{\rm{kg}}\).