Question

Planets $A$ has a mass $M$and radius $R$. Planet $B$ has half the mass and half the radius of planet $A$. If the escape velocities from the planets $A$ and $B$ are $V_A$ and $V_B$ respectively, then surfaces is $\frac{V_A}{V_B}=\frac{n}{4}$, the value of $n$ is :

• A. $3$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is C

$4$

Step 1: Given Data:

Mass of planet A $=M$

The radius of planet A $=R$

Mass of planet B $=\frac{M}{2}$

The radius of planet B $=\frac{R}{2}$

Escape velocities of planet A and planet B is $V_A$ and $V_B$ respectively

It is given that,

$\frac{V_A}{V_B}=\frac{n}{4}$

Step 2: Formula Used:

The escape velocity is defined as-

Escape velocity$\left(V_e\right)=\sqrt{\frac{2GM}{R}}$

Where is the gravitational constant

$MandR$ are mass of the planet and radius of the planet respectively

Step 3: Calculation of velocity:

According to the formula the escape velocity of planet A is expressed as-

$V_A=\sqrt{\frac{2GM}{R}}$…….(1)

The escape velocity of planet B is expressed as-

$$\begin{gathered} V_B=\sqrt{\frac{2G{\displaystyle \frac{M}{2}}}{{\displaystyle \frac{R}{2}}}} \\ {V}_B=\sqrt{\frac{2GM}{R}}..........\left(2\right) \end{gathered}$$

On dividing equation (1) by equation (2) we get-

$\frac{V_A}{V_B}=1$

It is given that,

$\frac{V_A}{V_B}=\frac{n}{4}$

So,

$$\begin{gathered} \frac{n}{4}=1 \\ n=4 \end{gathered}$$

Thus, the value of $n$ is $4$.

Hence option C is the correct answer.