Question

Taking the moon's period of revolution about the earth as 30 days. Calculate its distance from the earth. ($G = 6.7 \times 10\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^2}$ and mass of the earth=$6 \times {10^{24}}\,{\text{kg}}$)

Answer 1

Use the expression for Kepler’s third law of planetary motion. This expression gives the relation between the time period of the planet or any astronomical object around the sun or other planet, radius of orbit of motion, universal gravitational constant and mass of the sun or planet which are being orbited.

Formula used:
The expression for Kepler’s third law of planetary motion is
${T^2} = \dfrac{{4{\pi ^2}{R^3}}}{{GM}}$ …… (1)
Here, $T$ is the time period of a planet around the sun or other planet, $R$ is the radius of orbit of the planet, $G$ is the universal gravitational constant and $M$ is the mass of the sun or planet around which the other planet is moving.

Complete step by step solution:
We have given that the time period of the moon around the earth is 30 days.
$T = 30\,{\text{days}}$

The mass of the earth is $6 \times {10^{24}}\,{\text{kg}}$ and the universal gravitational constant is $6.7 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^2}$.
$M = 6 \times {10^{24}}\,{\text{kg}}$
$G = 6.7 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^2}$

Convert the unit of time period of the moon around the sun in the SI system of units.
$T = \left( {30\,{\text{days}}} \right)\left( {\dfrac{{24\,{\text{hr}}}}{{1\,{\text{day}}}}} \right)\left( {\dfrac{{60\,{\text{min}}}}{{1\,{\text{hr}}}}} \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)$
$\Rightarrow T = 2.592 \times {10^6}\,{\text{s}}$

Hence, the time period of the moon around the sun is $2.592 \times {10^6}\,{\text{s}}$.

We can determine the radius of the orbit or distance of the moon from the earth using equation (1).

Rearrange equation (1) for the distance of the moon from the earth.
$R = {\left( {\dfrac{{GM{T^2}}}{{4{\pi ^2}}}} \right)^{\dfrac{1}{3}}}$

Substitute $6.7 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^2}$ for $G$, $6 \times {10^{24}}\,{\text{kg}}$ for $M$, $2.592 \times {10^6}\,{\text{s}}$ for $T$ and $3.14$ for $\pi$ in the above equation.
$R = {\left[ {\dfrac{{\left( {6.7 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^2}} \right)\left( {6 \times {{10}^{24}}\,{\text{kg}}} \right){{\left( {2.592 \times {{10}^6}\,{\text{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}} \right]^{\dfrac{1}{3}}}$
$\Rightarrow R = 4.09 \times {10^8}\,{\text{m}}$
$\Rightarrow R = \left( {4.09 \times {{10}^8}\,{\text{m}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{km}}}}{{1\,{\text{m}}}}} \right)$
$\Rightarrow R = 4.09 \times {10^5}\,{\text{km}}$

Hence, the distance between the moon and the earth is $4.09 \times {10^5}\,{\text{km}}$.

Note:
The students may think that Kepler's third law is for the planet revolving around the sun, then how it can be used for the moon revolving around the earth. But the Kepler’s third law of time period is applicable for any astronomical object revolving around any other astronomical object. So, it can be used for the moon revolving around the earth.