JEE Physics 2025 Question with Solution

Given: Earth has mass $8$ times and radius $2$ times that of the planet, and escape velocity from Earth is $11.2 \, \text{km/s}$.

Find: Escape velocity from the planet.

Escape velocity is given by

$$v_{escape} = \sqrt{\frac{2GM}{R}}$$

From the given comparison,

$$\frac{M_{planet}}{M_{earth}} = \frac{1}{8}, \quad \frac{R_{planet}}{R_{earth}} = \frac{1}{2}$$

Now compare the escape velocities:

$$\frac{v_{escape,\,planet}}{v_{escape,\,earth}} = \sqrt{\frac{M_{planet}R_{earth}}{M_{earth}R_{planet}}}$$

Substituting the ratios,

$$\frac{v_{escape,\,planet}}{v_{escape,\,earth}} = \sqrt{\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{2}\right)}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$

Therefore,

$$v_{escape,\,planet} = \frac{1}{2} \times 11.2 = 5.6 \, \text{km/s}$$

Therefore, the escape velocity from the planet is $5.6 \, \text{km/s}$. The correct option is A.

Given: $v_{escape} \propto \sqrt{\frac{M}{R}}$, with $M_{earth} = 8M_{planet}$ and $R_{earth} = 2R_{planet}$.

Find: $v_{escape,\,planet}$.

Using proportionality,

$$\frac{v_{escape,\,earth}}{v_{escape,\,planet}} = \sqrt{\frac{M_{earth}/R_{earth}}{M_{planet}/R_{planet}}}$$

Substitute the given relations:

$$\frac{v_{escape,\,earth}}{v_{escape,\,planet}} = \sqrt{\frac{8/2}{1}} = \sqrt{4} = 2$$

So,

$$v_{escape,\,planet} = \frac{v_{escape,\,earth}}{2} = \frac{11.2}{2} = 5.6 \, \text{km/s}$$

Hence, the required escape velocity is $5.6 \, \text{km/s}$.