Gravitation Previous Year Questions — JEE Main Physics
22 past-year Gravitation questions from JEE Main (Physics).
Q25 (2025)
A satellite of mass $1000\mathrm{kg}$ is launched to revolve around the earth in an orbit at a height of $270\mathrm{km}$ from the earth's surface. Kinetic energy of the satellite in this orbit is \_ \times 10^{10}\mathrm{J}$ .
(Mass of earth $= 6\times 10^{24}\mathrm{kg},$
Radius of earth $= 6.4 \times 10^{6} \mathrm{~m}$ ,
Gravitational constant $= 6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{kg}^{-2}$)
### Related Formula
1. Orbital speed ($v_0$) of a satellite at distance $r$ from earth's center:
$$v_0 = \sqrt{\frac{G M_e}{r}}$$
2. Orbital Radius:
$r = R_e + h$
3. Kinetic Energy of the orbiting satellite:
$$\mathrm{KE} = \frac{1}{2} m v_0^2 = \frac{G M_e m}{2(R_e + h)}$$
### Core Logic
Given parameters:
- Mass of satellite $m = 1000 \ \mathrm{kg} = 10^3 \ \mathrm{kg}$
- Orbit altitude $h = 270 \ \mathrm{km} = 0.27 \times 10^6 \ \mathrm{m}$
- Earth Radius $R_e = 6.4 \times 10^6 \ \mathrm{m}$
- Earth Mass $M_e = 6 \times 10^{24} \ \mathrm{kg}$
- Gravitational constant $G = 6.67 \times 10^{-11} \ \mathrm{N \cdot m^2 / kg^2}$
### Step 1: Calculate kinetic energy
First, compute the orbital radius $r$:
$$r = R_e + h = 6.4 \times 10^6 \ \mathrm{m} + 0.27 \times 10^6 \ \mathrm{m} = 6.67 \times 10^6 \ \mathrm{m}$$
Substitute $r = 6.67 \times 10^6 \ \mathrm{m}$ into the kinetic energy equation:
$$\mathrm{KE} = \frac{G M_e m}{2 r}$$
$$\mathrm{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^3}{2 \times 6.67 \times 10^6}$$
Notice that the value $6.67$ cancels out directly:
$$\mathrm{KE} = \frac{6 \times 10^{16}}{2 \times 10^6} = 3 \times 10^{10} \ \mathrm{J}$$
Thus, the kinetic energy coefficient is $3$.
### Pattern Recognition
Sees: Kinetic energy of a satellite orbiting at an altitude above earth's surface.
Trap: Doing long division calculation for $6.67/2$. Check for clean cancellations in formulas first!
Shortcut: Notice that $R_e + h = 6.4 \times 10^6 + 0.27 \times 10^6 = 6.67 \times 10^6$, which matches the value of the Gravitational constant $G = 6.67 \times 10^{-11}$ perfectly. This clean cancellation leaves behind simple integer math to give $3 \times 10^{10} \mathrm{~J}$ immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q24 (2025)
Two planets, A and B are orbiting a common star in circular orbits of radii $R_{\mathrm{A}}$ and $R_{\mathrm{B}}$ , respectively, with $R_{\mathrm{B}} = 2R_{\mathrm{A}}$ . The planet B is $4\sqrt{2}$ times more massive than planet A. The ratio $\left(\frac{L_{\mathrm{B}}}{L_{\mathrm{A}}\mathrm{}}\right)$ of angular momentum $(L_{\mathrm{B}})$ of planet B to that of planet $\mathrm{A}(L_{\mathrm{A}})$ is closest to integer ______.
### Related Formula
$$v_0 = \sqrt{\frac{GM_{\text{star}}}{R}}$$
$$L = m v_0 R = m \sqrt{G M_{\text{star}} R}$$
### Core Logic
The orbital angular momentum scales as $L \propto m \sqrt{R}$, where $m$ is the mass of the orbiting planet and $R$ is its orbital radius.
Setting up the ratio for planet B to planet A:
$$\frac{L_B}{L_A} = \left(\frac{m_B}{m_A}\right) \cdot \sqrt{\frac{R_B}{R_A}}$$
Substitute the relative constraints provided by the text:
- $m_B = 4\sqrt{2} m_A$
- $R_B = 2 R_A$
$$\frac{L_B}{L_A} = (4\sqrt{2}) \times \sqrt{2} = 4 \times 2 = 8$$
### Pattern Recognition
Orbital velocity goes down as $1/\sqrt{R}$, but angular momentum features an explicit distance product multiplier ($m v R$), shifting the baseline radius factor to a clean numerator scaling profile: $\sqrt{R}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q21 (2025)
Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time $T=4$ seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ________ seconds.
### Related Formula
By Dimensional Analysis or Scaling of Kepler's Third Law:
$$T \propto m^x G^y a^z$$
### Core Logic
Let's perform a dimensional matching to express the collision time $T$ in terms of physical scaling variables $m$, $G$, and $a$:
$$[T] = [M]^x [M^{-1}L^3T^{-2}]^y [L]^z$$
Equating dimensions on both sides:
- Mass ($M$): $x - y = 0 \implies x = y$
- Length ($L$): $3y + z = 0 \implies z = -3y$
- Time ($T$): $-2y = 1 \implies y = -1/2$
Solving these equations:
$$x = -1/2, \quad y = -1/2, \quad z = 3/2$$
### Step 1: Scaling Formula of Time
Therefore, the scaling relationship for time $T$ is:
$$T \propto m^{-1/2} G^{-1/2} a^{3/2} \implies T \propto \sqrt{\frac{a^3}{m}}$$
Let's write the ratio for two cases:
$$\frac{T_2}{T_1} = \sqrt{\left(\frac{a_2}{a_1}\right)^3 \cdot \left(\frac{m_1}{m_2}\right)}$$
### Step 2: Calculating Final Time
Given values:
- $a_1 = a$, $a_2 = 2a$
- $m_1 = m$, $m_2 = 2m$
- $T_1 = 4\mathrm{~s}$
$$\frac{T_2}{4} = \sqrt{\left(\frac{2a}{a}\right)^3 \cdot \left(\frac{m}{2m}\right)} = \sqrt{2^3 \cdot \frac{1}{2}} = \sqrt{4} = 2$$
$$T_2 = 4 \times 2 = 8\mathrm{~seconds}$$
### Pattern Recognition
Kepler's Third Law / free-fall collapse scaling: whenever a orbit or a direct gravitational collapse scale is involved, the time scales as $T \propto \sqrt{\frac{R^3}{GM}}$. Thus doubling $R$ multiplies time by $\sqrt{8}$ and doubling $M$ divides time by $\sqrt{2}$. The combination results in a clean doubling of time: $\sqrt{8}/\sqrt{2} = 2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Class 11 Physics: Units and Measurements: Dimensional Analysis
Q9 (2025)
An object is kept at rest at a distance of 3R above the earth's surface where R is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume $M=$ mass of earth, $G=$ Universal gravitational constant)
- $\sqrt{\frac{GM}{2R}}$
- $\sqrt{\frac{GM}{R}}$
- $\sqrt{\frac{3GM}{R}}$
- $\sqrt{\frac{2GM}{R}}$
### Related Formula
Conservation of Total Mechanical Energy:
$E_i = E_f$
$$U_i + K_i = U_f + K_f$$
### Core Logic
The initial distance from the center of the earth is $r = R + 3R = 4R$.
Initial mechanical energy:
$$E_i = -\frac{GMm}{4R} + \frac{1}{2}mv^2$$
To just escape to infinity, the final mechanical energy at infinity must be at least zero:
$E_f = 0$
### Step 1: Apply Energy Conservation
Setting initial energy equal to zero:
$$-\frac{GMm}{4R} + \frac{1}{2}mv^2 = 0$$
$$\frac{1}{2}v^2 = \frac{GM}{4R} \implies v = \sqrt{\frac{GM}{2R}}$$ {{SOL_IMG1}}
### Pattern Recognition
Be extremely careful with the phrase 'above the earth's surface'. Distance from center $r = R + h$. Escape condition always sets net mechanical energy $\ge 0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q914 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The kinetic energy needed to project a body of mass m from earth surface to infinity is $\frac{1}{2}mgR$, where R is the radius of earth.
Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
In the light of the above statements, choose the correct answer from the option given below
- A is False but R is true
- Both A and R are true and R is the correct explanation of A
- A is true but R is false
- Both A and R are true but R is NOT the correct explanation of A
### Related Formula
Escape kinetic energy requirement formulation:
$$\text{KE}_{\text{escape}} = \frac{GMm}{R} = mgR$$
Gravitational potential energy field definition:
$$U = -\frac{GMm}{r}$$
At $r \to \infty$, $U_{\text{max}} = 0$.
### Core Logic
* **Assertion Check:** The minimum work required to project an object from Earth's surface ($r = R$) to infinity ($r \to \infty$) equals the change in gravitational potential energy:
$$\Delta U = U(\infty) - U(R) = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} = mgR$$
Therefore, the required kinetic energy is $mgR$. The statement asserts it is $\frac{1}{2}mgR$ (which corresponds to orbital kinetic energy near Earth's surface). Hence, **Assertion A is false**.
* **Reason Check:** Since the gravitational force is attractive, potential energy is negative everywhere in the field and reaches its maximum value asymptotically at infinity ($U_{\text{max}} = 0$). Hence, **Reason R is true**.
### Pattern Recognition
Escape energy from the surface is $mgR$, while circular orbital kinetic energy near the surface is $\frac{1}{2}mgR$. Because gravitational potential energy is defined with zero at infinity, every bound state has $U < 0$, making zero the absolute maximum potential energy.
### Evaluation Rubric / Model Answer
Option A: A is false but R is true
### Chapter Mix
Class 11 Physics: Gravitation
