Question

A stationary object is released from a point P a distance 3R from the center of the moon which has radius R and mass M. Which one of the following expressions gives the speed of the object on hitting the moon?
A. ${{\left( \dfrac{2GM}{3R} \right)}^{\dfrac{1}{2}}}$
B. ${{\left( \dfrac{4GM}{3R} \right)}^{\dfrac{1}{2}}}$
C. ${{\left( \dfrac{2GM}{R} \right)}^{\dfrac{1}{2}}}$
D. ${{\left( \dfrac{GM}{R} \right)}^{\dfrac{1}{2}}}$

Answer 1

Hint: Use the conservation of energy theorem. The total energy of the system at point P and at the surface will be the same. First, calculate the total energy at point P and then at the surface. Both should be the same. Total energy is the summation of kinetic energy and potential energy.

Formula Used:
Potential Energy of an object in a gravitational system is given by,
$U=-\dfrac{GMm}{r}$

Where,
$G$ is the Gravitational constant
$M$ is the mass of the first object
$m$ is the mass of the second object
$r$ is the distance between the two objects.

The kinetic energy of an object is given by,
$KE=\dfrac{1}{2}m{{v}^{2}}$

Where,
$m$ is the mass of the object,
$v$ is the velocity of the object.

Complete step by step solution:
Let’s assume that the mass of the object is m.

It is given that the mass of the moon is M, and the radius of the moon is R.

Above the surface of the moon, we can consider it to be a point mass at the center (Using Gauss’s Law).

So, we can assume that the distance between the two objects is,
3R.

Hence, the potential energy at point P,
${{U}_{P}}=-\dfrac{GMm}{3R}$

The kinetic energy at point P is zero, as the object is stationary.

Hence, the total energy at point P is,
${{E}_{P}}=-\dfrac{GMm}{3R}$

Now, let us consider the total energy at the surface.

The potential energy of the object at the surface is given by,
${{U}_{surface}}=-\dfrac{GMm}{R}$

Let’s assume the required velocity = v.
So, the kinetic energy at the surface is given by,
$K{{E}_{surface}}=\dfrac{1}{2}m{{v}^{2}}$

Hence, the total energy of the object on the surface is given by,
${{E}_{surface}}=-\dfrac{GMm}{R}+\dfrac{1}{2}m{{v}^{2}}$

The law of conservation of energy states that the total energy must remain the same in the lossless system.

Hence, we can say that:

Total energy at point P = Total energy at the surface
$\Rightarrow {{E}_{P}}={{E}_{surface}}$
$\Rightarrow -\dfrac{GMm}{3R}=-\dfrac{GMm}{R}+\dfrac{1}{2}m{{v}^{2}}$
$\Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{2}{3}\dfrac{GMm}{R}$
$\Rightarrow {{v}^{2}}=\dfrac{4}{3}\dfrac{GMm}{R}$
$\Rightarrow v=\sqrt{\dfrac{4GMm}{3R}}$

Hence, the velocity of the object before hitting the surface will be,
$v=\sqrt{\dfrac{4GMm}{3R}}$
So, the correct choice is (B).

Note: In this case, we have assumed there are no external effects on the two objects. If there were any other forces in the system, we had to take the effect of those gravitational fields as well.