JEE Physics 2023 Question with Solution

Given: A body is released from rest from a height equal to the earth's radius, so its initial distance from the centre of the earth is $2R$ and its final distance is $R$.

Find: The velocity with which it strikes the earth's surface.

Use conservation of mechanical energy.

Gravitational potential energy at a distance $r$ from the centre is

$$U=-\frac{GMm}{r}$$

So,

$$U_i=-\frac{GMm}{2R}, \qquad U_f=-\frac{GMm}{R}$$

The body starts from rest, so initial kinetic energy is zero. At the surface, if the speed is $v$, then

$$-\frac{GMm}{2R}=-\frac{GMm}{R}+\frac{1}{2}mv^2$$

Therefore,

$$\frac{1}{2}mv^2=\frac{GMm}{R}-\frac{GMm}{2R}=\frac{GMm}{2R}$$

Hence,

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

Using

$$g=\frac{GM}{R^2}$$

we get

$$GM=gR^2$$

Substituting,

$$v^2=\frac{gR^2}{R}=gR$$

So,

$$v=\sqrt{gR}$$

Therefore, the correct option is D.

Given: The body falls from $r=2R$ to $r=R$.

Find: The impact speed.

For large heights, gravitational potential energy must be used in the form $-\frac{GMm}{r}$.

The loss in potential energy is

$$\Delta U=\left(-\frac{GMm}{R}\right)-\left(-\frac{GMm}{2R}\right)=-\frac{GMm}{2R}$$

So the gain in kinetic energy is

$$\frac{GMm}{2R}$$

Thus,

$$\frac{1}{2}mv^2=\frac{GMm}{2R}$$

which gives

$$v^2=\frac{GM}{R}=gR$$

Hence,

$$v=\sqrt{gR}$$

Therefore, the correct option is D.

For problems involving heights comparable to the earth's radius, constant-$g$ formulas are not valid.