JEE Physics 2025 Question with Solution

Given: $m = 1000 \, \text{kg}$, $M = 6 \times 10^{24} \, \text{kg}$, $R = 6.4 \times 10^6 \, \text{m}$, $h = 270 \times 10^3 \, \text{m}$, and $G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$.

Find: The value of kinetic energy in the form _____ x $10^{10}$ J.

For a satellite in circular orbit, gravitational force provides the centripetal force.

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

So,

$$v^2 = \frac{GM}{r}$$

where $r = R + h$.

The kinetic energy is

$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r}$$

Now compute the orbital radius:

$$r = R + h = 6.4 \times 10^6 + 0.27 \times 10^6 = 6.67 \times 10^6 \, \text{m}$$

Substitute the values:

$$K = \frac{1}{2} \times 1000 \times \frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.67 \times 10^6}$$

Simplifying,

$$\frac{GM}{r} = \frac{6.67 \times 6}{6.67} \times 10^{-11+24-6} = 6 \times 10^7$$

Hence,

$$K = \frac{1}{2} \times 1000 \times 6 \times 10^7 = 3 \times 10^{10} \, \text{J}$$

Therefore, the required value is $3$.

Given: $m = 1000 \, \text{kg}$, $M = 6 \times 10^{24} \, \text{kg}$, $R = 6.4 \times 10^6 \, \text{m}$, $h = 270 \times 10^3 \, \text{m}$.

Find: Kinetic energy of the satellite.

Use the direct formula for a satellite in circular orbit:

$$KE = \frac{GMm}{2r} = \frac{GMm}{2(R+h)}$$

Now,

$$KE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4 \times 10^6 + 2.7 \times 10^5)}$$

With $R+h = 6.67 \times 10^6 \, \text{m}$,

$$KE = 3 \times 10^{10} \, \text{J}$$

Therefore, the required value is $3$.