Gravitation Previous Year Questions — JEE Main Physics
22 past-year Gravitation questions from JEE Main (Physics) — page 4.
Q46 (2024)
The mass of the moon is $1/144$ times the mass of a planet and its diameter $1/16$ times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:
- $\frac{v}{3}$
- $\frac{v}{4}$
- $\frac{v}{12}$
- $\frac{v}{6}$
### Related Formula
$$v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$$
### Core Logic
For the planet: $v = \sqrt{\frac{2GM_p}{R_p}}$
For the moon: $M_m = \frac{M_p}{144}$ and $R_m = \frac{R_p}{16}$.
### Step 1: Setup the Ratio
$$v_m = \sqrt{\frac{2G M_m}{R_m}}$$
$$v_m = \sqrt{\frac{2G \left(\frac{M_p}{144}\right)}{\left(\frac{R_p}{16}\right)}}$$
$$v_m = \sqrt{\frac{2G M_p}{R_p} \times \frac{16}{144}}$$
### Step 2: Simplification
$$v_m = \sqrt{\frac{2G M_p}{R_p}} \times \sqrt{\frac{1}{9}}$$
$$v_m = v \times \frac{1}{3} = \frac{v}{3}$$
### Pattern Recognition
Escape velocity scales as $\sqrt{M/R}$. If $M$ scales by $x$ and $R$ scales by $y$, velocity scales by $\sqrt{x/y}$. Here, $\sqrt{(1/144)/(1/16)} = \sqrt{16/144} = \sqrt{1/9} = 1/3$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q1987 (2024)
Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\frac{2\sqrt{2} + 1}{32}\right)\frac{\mathrm{Gm}^2}{\mathrm{L}^2}$, the length of the sides of the square is
- $\frac{\mathrm{L}}{2}$
- $4 L$
- $3L$
- $2 L$
### Related Formula
$$F = \frac{G m_1 m_2}{r^2}$$
### Core Logic
{{SOL_IMG1}}
Let the side length of the square be $a$. Considering one corner mass, it experiences forces from the adjacent two masses (distance $a$) and the diagonally opposite mass (distance $\sqrt{2}a$).
The forces from the two adjacent masses are at $90^\circ$ to each other:
$$F = \frac{Gm^2}{a^2}$$
The resultant of these two is $\sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$, directed along the diagonal.
### Step 2: Total Force Equation
The force from the diagonal mass is:
$$F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$$
Total resultant force $F_{\text{net}} = \sqrt{2}F + F'$:
$$F_{\text{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$$
$$F_{\text{net}} = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$$
Equating this to the given force value:
$$\left(\frac{2\sqrt{2} + 1}{32}\right)\frac{Gm^2}{L^2} = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$$
$$\frac{1}{32 L^2} = \frac{1}{2 a^2}$$
$a^2 = 16 L^2$
$a = 4L$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q44 (2026)
Initially a satellite of 100 kg is in a circular orbit of radius $1.5R_{E}$. This satellite can be moved to a circular orbit of radius $3R_{E}$ by supplying $\alpha \times 10^{6}J$ of energy. The value of $\alpha$ is ____.
(Take Radius of Earth $R_{E} = 6 \times 10^{6}\text{ m}$ and $g = 10\text{ m/s}^{2}$)
- 150
- 500
- 100
- 1000
### Related Formula
$$E = \frac{-GM_E m}{2r}$$
$$\Delta E = E_f - E_i$$
$$g = \frac{GM_E}{R_E^2} \implies GM_E = g R_E^2$$
### Core Logic
Energy of a satellite in a circular orbit is given as $E = \frac{-GM_E m}{2r}$ where r is the radius of the circular orbit.
Required energy to be supplied $\Delta E = E_f - E_i$:
$$\Delta E = \left( \frac{-GM_E m}{2(3R_E)} \right) - \left( \frac{-GM_E m}{2(1.5R_E)} \right)$$
$$\Delta E = \frac{-GM_E m}{6R_E} + \frac{GM_E m}{3R_E} = \frac{GM_E m}{6R_E}$$
### Step 1: Evaluate Delta E
Substitute $GM_E = g R_E^2$ into the expression:
$$\Delta E = \frac{(g R_E^2) m}{6R_E} = \frac{1}{6} m g R_E$$
$$= \frac{1}{6} \times 100 \times 10 \times (6 \times 10^6)$$
$$= 1000 \times 10^6\text{ J}$$
Comparing with $\alpha \times 10^6\text{ J}$, we get $\alpha = 1000$.
### Pattern Recognition
Orbital transition energy is always $\Delta E = \frac{GMm}{2} (\frac{1}{r_i} - \frac{1}{r_f})$. Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q30 (2026)
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is \_\_\_\_ m/s.
- 1000
- $200\sqrt{5}$
- $100\sqrt{10}$
- $1000\sqrt{2}$
### Related Formula
$$V_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \rho \frac{4\pi R^3}{3}}{R}} \implies V_e \propto \sqrt{\rho} \times R$$
### Core Logic
Given density and radius of B are 10% ($0.1$) of planet A:
$$\frac{(V_e)_B}{(V_e)_A} = \sqrt{\frac{\rho_B}{\rho_A}} \times \frac{R_B}{R_A} = \sqrt{0.1} \times 0.1 = \frac{1}{10\sqrt{10}}$$
$$(V_e)_B = \frac{10 \times 1000}{10\sqrt{10}} = 100\sqrt{10} \text{ m/sec}$$
### Pattern Recognition
Sees: Escape velocity scaling with density and radius.
Shortcut: Express escape velocity in terms of density $\rho$ and radius $R$, then take ratios.
Check: Matches option (3). ✓
### Chapter Mix
Class 11 Physics: Gravitation
Q35 (2026)
Net gravitational force at the centre of a square is found to be $F_{1}$ when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of 3M and 4M are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is \_\_\_\_.
{{IMG1}}
- 2
- 3
- 1
- $2\sqrt{5}$
### Related Formula
$$F = \frac{G m_1 m_2}{r^2}$$
### Core Logic
{{SOL_IMG1}}
Initial configuration:
$$F_1 = 2\sqrt{2} \frac{Gmm_0}{r^2}$|
New configuration (after interchanging 3M and 4M):
$$F_2 = \sqrt{10} \frac{Gmm_0}{r^2}$Ratio:$\frac{F_1}{F_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}} \implies \alpha = 2$$
### Pattern Recognition
Sees: Gravitational force vector sum at center of square.
Shortcut: Compute net vector resultants for both mass configurations and take ratio.
Check: Matches option (1). ✓
### Chapter Mix
Class 11 Physics: Gravitation
