Question

Two planets have masses $M$ and $16M$and their radii are $a$ and $2a$, respectively. The separation between the centres of the planets is $10a$. A body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet, the minimum firing speed needed is:

• A. $2\sqrt{\frac{GM}{a}}$
• B. $\sqrt{\frac{G{M}^2}{ma}}$
• C. $\frac{3}{2}\sqrt{\frac{5GM}{a}}$
• D. $4\sqrt{\frac{GM}{a}}$

Answer 1

The correct option is C

$\frac{3}{2}\sqrt{\frac{5GM}{a}}$

Step 1: Given

1. Mass of the two planets are $M$ and $16M$and their radii are $a$ and $2a$ respectively.
2. The distance between the centers of the planets is $10a$.
3. Mass of the fired particle $m$.
4. The net force is zero at a given point $A$.

Step 2: To find

The minimum firing speed of the body so that it can reach the surface of the smaller planet.

Step 3: Solution

Let the length of the distance between the point $A$ and $B$.

The net force of the gravitation is zero at $A$.

$F_{net}=0$

$$\begin{gathered} \Rightarrow \frac{[G(16M)(m\left)\right]}{x^2}-\frac{[G(M)(m)]}{{(10a–x)}^2}=0 \\ Here, \\ G=Gravitationalcons\tan t \\ x=Dis\tan cebetweenAandB \\ \Rightarrow \frac{[G(16M)(m\left)\right]}{x^2}=\frac{[G(M)(m)]}{{(10a–x)}^2} \\ \Rightarrow \frac{{(10a–x)}^2}{x^2}=\frac{1}{16} \\ \Rightarrow \frac{100a^2+x^2-20ax}{x^2}=\frac{1}{16} \\ \Rightarrow x=8a \end{gathered}$$

Now by conserving energy between $A$and $B$.

$$\begin{gathered} \frac{1}{2}m{v}^2-\frac{G\left(16M\right)m}{2a}-\frac{GMm}{8a}=-\frac{G16Mm}{8a}-\frac{GMm}{2a}\left[herevistherequiredvelocity\right] \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{6GMm}{a}-\frac{3GMm}{8a} \\ \Rightarrow v=\frac{3}{2}\sqrt{\frac{5GM}{a}} \end{gathered}$$

Hence, the correct option is $C$.