JEE Physics 2023 Question with Solution

Given: A body of mass $m$ is moved from the Earth's surface to height $h = 2R_e$.

Find: Increase in gravitational potential energy.

The gravitational potential energy at distance $R$ from the center of the Earth is

$$U = -\frac{GM_e m}{R}$$

At the surface,

$$U_i = -\frac{GM_e m}{R_e}$$

At height $2R_e$ above the surface, the distance from the center is $R_e + 2R_e = 3R_e$. Therefore,

$$U_f = -\frac{GM_e m}{3R_e}$$

Now,

$$\Delta U = U_f - U_i$$ $$\Delta U = -\frac{GM_e m}{3R_e} - \left(-\frac{GM_e m}{R_e}\right)$$ $$\Delta U = \frac{2GM_e m}{3R_e}$$

Using

$$g = \frac{GM_e}{R_e^2}$$

we get

$$\Delta U = \frac{2}{3} mgR_e$$

Therefore, the increase in potential energy is $\frac{2}{3} mgR_e$. The correct option is C.

Given: Initial position at Earth's surface and final height $2R_e$ above the surface.

Find: The increase in potential energy in terms of $m$, $g$, and $R_e$.

First write the initial and final distances from Earth's center:

• Initial distance = $R_e$
• Final distance = $R_e + 2R_e = 3R_e$

Potential energy formula:

$$U = -\frac{GM_e m}{R}$$

So,

$$U_i = -\frac{GM_e m}{R_e}, \qquad U_f = -\frac{GM_e m}{3R_e}$$

Hence,

$$\Delta U = U_f - U_i = -\frac{GM_e m}{3R_e} + \frac{GM_e m}{R_e}$$ $$\Delta U = \frac{2GM_e m}{3R_e}$$

Now substitute

$$GM_e = gR_e^2$$

Then,

$$\Delta U = \frac{2}{3} \cdot \frac{gR_e^2 m}{R_e}$$ $$\Delta U = \frac{2}{3} mgR_e$$

Therefore, the required increase in potential energy is $\frac{2}{3} mgR_e$.