Question

Moon is revolving in a circular orbit of radius $60R$ ($R=$ radius of earth). Assume that the radius of the moon is $\dfrac{R}{4}$. If the moon is stopped for an instant and then released, it will fall towards the earth. Ignoring atmospheric friction, the velocity of the moon just before it strikes the earth is (Take $g=$ acceleration due to gravity at the surface of earth)

• A. $\sqrt{\dfrac{59}{30}\,gR}$
• B. $\sqrt{\dfrac{118}{61}\,gR}$
• C. $\sqrt{\dfrac{7gR}{10}}$
• D. $\sqrt{\dfrac{47}{30}\,gR}$

Hint:

When gravity varies with distance, use conservation of energy with $GM=gR^2$ instead of constant-$g$ equations.

Answer 1

The Correct Option is D

Step 1: Express the gravitational constant $GM$ in terms of $g$ and $R$. \[ g=\frac{GM}{R^2}\ \Rightarrow\ GM=gR^2 \]
Step 2: Identify the initial and final distances of the moon’s centre from the centre of the earth. \[ r_i = 60R \] Since the moon has radius $\dfrac{R}{4}$, it strikes the earth when the distance between centres is: \[ r_f = R+\frac{R}{4}=\frac{5R}{4} \]
Step 3: Use conservation of mechanical energy. Initial velocity is zero, so: \[ \frac{1}{2}mv^2 = GMm\left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]
Step 4: Substitute the values: \[ \frac{1}{2}v^2 = gR^2\left(\frac{1}{\frac{5R}{4}}-\frac{1}{60R}\right) \]
Step 5: Simplify: \[ \frac{1}{2}v^2 = gR\left(\frac{4}{5}-\frac{1}{60}\right) = gR\left(\frac{48-1}{60}\right) = gR\cdot\frac{47}{60} \]
Step 6: Hence, \[ v^2=\frac{47}{30}gR \quad \Rightarrow \quad v=\sqrt{\frac{47}{30}\,gR} \]