JEE Physics 2026 Question with Solution

Given: Satellite mass is $100 \, \text{kg}$. Initial orbit radius is $r_1 = 1.5R_E$ and final orbit radius is $r_2 = 3R_E$. Also, $R_E = 6 \times 10^6 \, \text{m}$ and $g = 10 \, \text{m/s}^2$.

Find: The value of $a$ in the energy expression $a \times 10^6 \, \text{J}$.

For a satellite in a circular orbit, the total mechanical energy is

$$E = -\frac{GMm}{2r}$$

Using $g = \frac{GM}{R_E^2}$, we get

$$GM = gR_E^2$$

So the energy can be written as

$$E = -\frac{gR_E^2 m}{2r}$$

Initial energy for $r_1 = 1.5R_E$ is

$$E_1 = -\frac{gR_E^2 m}{2(1.5R_E)} = -\frac{gR_E m}{3}$$

Final energy for $r_2 = 3R_E$ is

$$E_2 = -\frac{gR_E^2 m}{2(3R_E)} = -\frac{gR_E m}{6}$$

Energy supplied is the increase in total energy:

$$\Delta E = E_2 - E_1$$ $$\Delta E = \left(-\frac{gR_E m}{6}\right) - \left(-\frac{gR_E m}{3}\right)$$ $$\Delta E = gR_E m\left(-\frac{1}{6} + \frac{1}{3}\right) = \frac{gR_E m}{6}$$

Substituting the values:

$$\Delta E = \frac{(10)(6 \times 10^6)(100)}{6}$$ $$\Delta E = 1000 \times 10^6 \, \text{J}$$

Comparing with $a \times 10^6 \, \text{J}$, we get $a = 1000$.

Therefore, the correct option is A.

Given: The satellite is moved from a lower circular orbit to a higher circular orbit.

Find: Whether energy must be supplied and how much.

The total energy of a satellite in orbit is negative. When the orbital radius increases, the total energy becomes less negative. That means the satellite gains energy, so external energy must be supplied.

Using

$$E = -\frac{GMm}{2r}$$

a larger value of $r$ gives a numerically smaller negative value. Hence $E_2 > E_1$, and the required supplied energy is positive:

$$\Delta E = E_2 - E_1 > 0$$

Evaluating this gives

$$\Delta E = \frac{gR_E m}{6} = 1000 \times 10^6 \, \text{J}$$

So the required value is $a = 1000$.