Question

Why does the moon revolve around earth not sun?

Answer 1

Hint: We know that the gravitational force of attraction keeps the moon revolving around the earth, and the earth revolving around the Sun. However, recall that for the moon to escape the bounds of the earth’s gravitational force and fall under the influence of the Sun’s gravity, it must possess a certain velocity. This is what is known as the escape velocity.

Complete step by step answer:
Determine a relation for the escape velocity by equating the centripetal force of attraction between the moon and the earth and the gravitational force acting between them and arrive at a numerical value for the escape velocity. Keeping in mind that the revolutionary speed of the moon is $1.02km{s}^{-1}$, compare this with the calculated escape velocity to arrive at a reasonable explanation.
Complete step by step answer: The revolution of the moon around the earth is due to gravitational force. The sun also exerts gravitational force on the moon, which is almost twice as large as the gravitational force of earth. But even after this the moon revolves around the earth. To understand this, we need to know the concept of escape velocity.
Escape velocity of an object can be defined as the velocity required for an object to overcome the gravitational attraction of the planet or any star.
The mathematical expression for escape velocity can be given as,
$V_{escape}=\sqrt{{\displaystyle \frac{2GM}{R}}}$
Where G is the gravitational constant, M is the mass of earth and R is the distance of the object from the centre of earth.
Now, $\begin{array}{rl}& G=6.67\times {10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}\\ & M=5.97\times {10}^{24}kg\end{array}$
Distance of moon from earth,
$R=3.84\times {10}^{5}km$
So, escape velocity of moon from earth’s surface is,
$\begin{array}{rl}& V_{\text{escape}}=\sqrt{{\displaystyle \frac{2GM}{R}}}\\ & \Rightarrow V_{\text{escape}}=\sqrt{{\displaystyle \frac{2\times 6.67\times {10}^{-11}\times 5.97\times {10}^{24}}{3.84\times {10}^8}}}\\ & \Rightarrow V_{\text{escape}}=\sqrt{{\displaystyle \frac{79.63\times {10}^{13}}{3.84\times {10}^8}}}\\ & \Rightarrow V_{\text{escape}}=\sqrt{{\displaystyle \frac{79.63\times {10}^{13}}{3.84\times {10}^8}}}\\ & \Rightarrow V_{\text{escape}}=\sqrt{20.73\times {10}^5}\\ & \Rightarrow V_{\text{escape}}=1.44\times {10}^{3}m{s}^{-1}\\ & \Rightarrow V_{\text{escape}}=1.44km{s}^{-1}\end{array}$
So, the velocity of moon to escape earth and revolve around the sun is approximately, $V_{escape}=1.44km{s}^{-1}$
But the velocity of moon around earth is $V=1.02km{s}^{-1}$
Which is lower than the escape velocity needed for the moon to go out of earth’s gravitational pull.
Because of this the moon revolves around earth not sun.

Note:
One may think that the moon revolves around the earth because the distance of the moon from the earth is very less than the distance from the sun. but this is not the case. Even with this distance the gravitational pull of the sun is more than the pull of earth. Only because of this escape velocity case the moon is revolving around earth.