Question

If the mass of earth were 4 times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be approximately.
A. \[56{\rm{ days}}\]
B. \[28{\rm{ days}}\]
C. \[14{\rm{ days}}\]
D. \[7{\rm{ days}}\]

Answer 1

Hint: From the concept of third law of Kepler, we can say that the relationship between cubic value of semi-major axis of elliptical path of the sun is directly proportional to the quadratic value of time period of oscillation. We will use the equation of this law to find out the new time period of oscillation of the moon.

Complete step by step answer:
It is given that the mass of the earth is increased by four times of the present mass, the mass of the moon is reduced to its half value. Keeping the orbital radius of the moon the same as present, we have to calculate the new time period of revolution of the moon.
Let us consider the present mass of earth is M and the orbital radius of the moon is R.
From the concept of third law of Kepler, we can write the time period of revolution of the mass at
present.
\[T = \dfrac{{4{\pi ^2}{R^3}}}{{GM}}\]
Here G is the Newtonian constant.
Let \[R'\] is the new orbital radius of the moon and \[M'\] is the new mass of the moon.
Write the expression of the new time period of the revolution of the moon.
\[T' = \dfrac{{4{\pi ^2}{R^3}}}{{GM'}}\]
Substitute \[4M\] for \[M'\]in the above expression.
 \[\begin{array}{c}
T' = \dfrac{{4{\pi ^2}{R^3}}}{{G\left( {4M} \right)}}\\
 = \dfrac{1}{4}\left( {\dfrac{{4{\pi ^2}{R^3}}}{{GM}}} \right)
\end{array}\]
Substitute T for \[\left( {\dfrac{{4{\pi ^2}{R^3}}}{{GM}}} \right)\] in the above expression to find out the new time period of revolution in terms of its present value.
\[T' = \dfrac{1}{4}T\]
We know that the present time period of oscillation of the moon in normal condition is equal to \[28{\rm{ days}}\]. Therefore substitute \[28{\rm{ days}}\] for T in the above expression.
\[\begin{array}{c}
T' = \dfrac{1}{4}\left( {28{\rm{ days}}} \right)\\
 = 7{\rm{ }}{\rm{days}}
\end{array}\]
Therefore, based on above expression we can say that the new time period of oscillation of moon is equal to \[7{\rm{ days}}\] when the mass of earth is increased by four times and mass of moon is reduced to its half value keeping orbital radius constant.

So, the correct answer is “Option D”.

Note:
Remember that the particle speed is related to distance and particle velocity is related to the displacement that’s why their same value gives the same amount of distance and displacement. For achieving this condition, the particles move in a straight line.