Gravitation Previous Year Questions (JEE Main 2026)

For a planet in a circular orbit, the gravitational force provides the centripetal force:

$$\frac{GMm}{r^2} = \frac{mr\,\omega^2}{1} = mr\left(\frac{2\pi}{T}\right)^2$$

Solving for the time period:

$$T^2 = \frac{4\pi^2 r^3}{GM} \quad\Rightarrow\quad T \propto \sqrt{\frac{r^3}{M}}$$

Taking the ratio of $P_2$ to $P_1$:

$$\frac{T_2}{T_1} = \sqrt{\frac{r_2^{\,3}/M_2}{r_1^{\,3}/M_1}} = \sqrt{\frac{(2R)^3/(4M)}{R^3/(2M)}}$$$$\frac{T_2}{T_1} = \sqrt{\frac{8R^3/(4M)}{R^3/(2M)}} = \sqrt{\frac{2R^3/M}{0.5\,R^3/M}} = \sqrt{4} = 2$$

Answer: B ($2$)

The acceleration due to gravity at height $h$ above the surface is:

$$g_h = \frac{g}{\left(1 + \dfrac{h}{R}\right)^2}$$

Setting $g_h = \dfrac{g}{9}$:

$$\frac{g}{\left(1 + \dfrac{h}{R}\right)^2} = \frac{g}{9} \quad\Rightarrow\quad \left(1 + \frac{h}{R}\right)^2 = 9$$

Taking the positive root:

$$1 + \frac{h}{R} = 3 \quad\Rightarrow\quad \frac{h}{R} = 2 \quad\Rightarrow\quad h = 2R$$

Answer: C ($2R$)

A body falling from infinity reaches the surface with the escape speed (energy conservation gives the same magnitude):

$$v = \sqrt{2gR}$$

Substituting $g = 9.8\,\text{m/s}^2$ and $R = 6400\,\text{km} = 6.4 \times 10^6\,\text{m}$:

$$v = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.2544 \times 10^8}$$$$v = 1.12 \times 10^4\,\text{m/s} = 11.2\,\text{km/s}$$

The kinetic energy gained by the $m = 1\,\text{kg}$ body:

$$k = \frac{1}{2}mv^2 = \frac{1}{2}(1)(1.2544 \times 10^8) = 6.27 \times 10^7\,\text{J}$$

Answer: A ($11.2\,\text{km/s};\ 6.27 \times 10^7\,\text{J}$)

At a depth $d$ below the surface:

$$g_d = g\left(1 - \frac{d}{R}\right)$$

At a height $h$ above the surface (small $h$, first-order approximation):

$$g_h \approx g\left(1 - \frac{2h}{R}\right)$$

The value of $g$ is larger at depth than at height, so the change on moving from the deep point up to the high point is:

$$\Delta g = g_d - g_h = g\left[\left(1 - \frac{d}{R}\right) - \left(1 - \frac{2h}{R}\right)\right] = g\left(\frac{2h - d}{R}\right)$$

With $d = h = 16\,\text{km}$ and $R = 6400\,\text{km}$:

$$\Delta g = g\left(\frac{2(16) - 16}{6400}\right) = g\left(\frac{16}{6400}\right)$$

Expressing as a percentage of $g$:

$$\alpha = \frac{16}{6400} \times 100 = 0.25\%$$

Answer: B ($0.25$)

The gravitational potential energy at a distance $r$ from the earth’s centre is:

$$U = -\frac{GM m}{r}$$

At the surface ($r = R_e$) and at height $h = 2R_e$ (so $r = R_e + 2R_e = 3R_e$):

$$\Delta U = U_{\text{final}} - U_{\text{initial}} = -\frac{GMm}{3R_e} - \left(-\frac{GMm}{R_e}\right)$$$$\Delta U = \frac{GMm}{R_e}\left(1 - \frac{1}{3}\right) = \frac{2}{3}\cdot\frac{GMm}{R_e}$$

Using $g = \dfrac{GM}{R_e^2}$, i.e. $\dfrac{GM}{R_e} = gR_e$:

$$\Delta U = \frac{2}{3}\,m g R_e$$

Answer: D ($\dfrac{2}{3}mgR_e$)