Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released…

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More Gravitation Previous-Year Questions — Page 2

Q25 jee_main_2025_02_april_evening Satellite Motion and Orbital Energy

A satellite of mass

1000kg

is launched to revolve around the earth in an orbit at a height of

270km

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}$)

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Orbital speed ( $v₀$ $v_0$) of a satellite at distance $r$ $r$ from earth's center:

v₀ = \sqrt((G Mₑ)/(r))

Orbital Radius:

$r = Rₑ + h$ $r = R_e + h$

Kinetic Energy of the orbiting satellite:

KE = (1)/(2) m v₀² = (G Mₑ m)/(2(Rₑ + h))

Core Logic

Given parameters:

Mass of satellite $m = 1000 kg = 10³ kg$
Orbit altitude $h = 270 km = 0.27 \times 10⁶ m$
Earth Radius $Rₑ = 6.4 \times 10⁶ m$ $R_e = 6.4 \times 10^6 \ \mathrm{m}$
Earth Mass $Mₑ = 6 \times 10²⁴ kg$ $M_e = 6 \times 10^{24} \ \mathrm{kg}$
Gravitational constant $G = 6.67 \times 10⁻¹¹ N \cdot m² / kg²$

Step 1: Calculate kinetic energy

First, compute the orbital radius $r$ $r$:

r = Rₑ + h = 6.4 \times 10⁶ m + 0.27 \times 10⁶ m = 6.67 \times 10⁶ m

Substitute $r = 6.67 \times 10⁶ m$ $r = 6.67 \times 10^6 \ \mathrm{m}$ into the kinetic energy equation:

KE = (G Mₑ m)/(2 r)

KE = 6.67 \times 10⁻¹¹ \times 6 \times 10²⁴ \times 10³2 \times 6.67 \times 10⁶

Notice that the value $6.67$ $6.67$ cancels out directly:

KE = 6 \times 10¹⁶2 \times 10⁶ = 3 \times 10¹⁰ J

Thus, the kinetic energy coefficient is $3$ $3$.

Pattern Recognition

Sees: Kinetic energy of a satellite orbiting at an altitude above earth's surface. Trap: Doing long division calculation for $6.67/2$ $6.67/2$. Check for clean cancellations in formulas first! Shortcut: Notice that $Rₑ + h = 6.4 \times 10⁶ + 0.27 \times 10⁶ = 6.67 \times 10⁶$ , which matches the value of the Gravitational constant $G = 6.67 \times 10⁻¹¹$ $G = 6.67 \times 10^{-11}$ perfectly. This clean cancellation leaves behind simple integer math to give $3 \times 10¹⁰ ~J$ $3 \times 10^{10} \mathrm{~J}$ immediately.

Chapter Mix

Class 11 Physics: Gravitation

Q24 jee_main_2025_29_jan_evening Kepler's Laws and Planetary Motion

Two planets, A and B are orbiting a common star in circular orbits of radii

RA

and

RB

, respectively, with

RB = 2RA

. The planet B is

4\sqrt(2)

times more massive than planet A. The ratio

( LBLA)

of angular momentum

(LB)

of planet B to that of planet

A(LA)

is closest to integer ______.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

v₀ = GMₛₜₐᵣR

L = m v₀ R = m G Mₛₜₐᵣ R

Core Logic

The orbital angular momentum scales as $L \propto m \sqrt(R)$ $L \propto m \sqrt{R}$, where $m$ $m$ is the mass of the orbiting planet and $R$ $R$ is its orbital radius.

Setting up the ratio for planet B to planet A:

(LB)/(LA) = ((mB)/(mA)) \cdot \sqrt((RB)/(RA))

Substitute the relative constraints provided by the text:

$mB = 4\sqrt(2) mA$ $m_B = 4\sqrt{2} m_A$
$RB = 2 RA$ $R_B = 2 R_A$

(LB)/(LA) = (4\sqrt(2)) \times \sqrt(2) = 4 \times 2 = 8

Pattern Recognition

Orbital velocity goes down as $1/\sqrt(R)$ $1/\sqrt{R}$, but angular momentum features an explicit distance product multiplier ( $m v R$ $m v R$), shifting the baseline radius factor to a clean numerator scaling profile: $\sqrt(R)$ $\sqrt{R}$.

Chapter Mix

Class 11 Physics: Gravitation

Q9 jee_main_2025_04_april_evening Escape Velocity

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)

A. $\sqrt((GM)/(2R))$ $\sqrt{\frac{GM}{2R}}$
B. $\sqrt((GM)/(R))$ $\sqrt{\frac{GM}{R}}$
C. $\sqrt((3GM)/(R))$ $\sqrt{\frac{3GM}{R}}$
D. $\sqrt((2GM)/(R))$ $\sqrt{\frac{2GM}{R}}$

Solution

Related Formula

Conservation of Total Mechanical Energy: $Eᵢ = Ef$ $E_i = E_f$

Uᵢ + Kᵢ = Uf + Kf

Core Logic

The initial distance from the center of the earth is $r = R + 3R = 4R$ $r = R + 3R = 4R$. Initial mechanical energy:

Eᵢ = -(GMm)/(4R) + (1)/(2)mv²

To just escape to infinity, the final mechanical energy at infinity must be at least zero: $Ef = 0$ $E_f = 0$

Step 1: Apply Energy Conservation

Setting initial energy equal to zero:

-(GMm)/(4R) + (1)/(2)mv² = 0

$(1)/(2)v² = (GM)/(4R) v = \sqrt((GM)/(2R))$

Escape projection trajectory from height 3R

Pattern Recognition

Be extremely careful with the phrase 'above the earth's surface'. Distance from center $r = R + h$ $r = R + h$. Escape condition always sets net mechanical energy $\geq 0$ $\ge 0$.

Chapter Mix

Class 11 Physics: Gravitation

Q jee_main_2025_04_april_morning Escape Velocity and Potential Energy

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

(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. A is False but R is true
B. Both A and R are true and R is the correct explanation of A
C. A is true but R is false
D. Both A and R are true but R is NOT the correct explanation of A

Solution

Related Formula

Escape kinetic energy requirement formulation:

KEescape = (GMm)/(R) = mgR

Gravitational potential energy field definition:

U = -(GMm)/(r)

At $r \to \infty$ $r \to \infty$, $Umax = 0$ $U_{\text{max}} = 0$.

Core Logic

Assertion Check: The minimum work required to project an object from Earth's surface ( $r = R$ $r = R$) to infinity ( $r \to \infty$ $r \to \infty$) equals the change in gravitational potential energy:

Δ U = U(\infty) - U(R) = 0 - (-(GMm)/(R)) = (GMm)/(R) = mgR

Therefore, the required kinetic energy is $mgR$ $mgR$. The statement asserts it is $(1)/(2)mgR$ $\frac{1}{2}mgR$ (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 ( $Umax = 0$ $U_{\text{max}} = 0$). Hence, Reason R is true.

Pattern Recognition

Escape energy from the surface is $mgR$ $mgR$, while circular orbital kinetic energy near the surface is $(1)/(2)mgR$ $\frac{1}{2}mgR$. Because gravitational potential energy is defined with zero at infinity, every bound state has $U < 0$ $U < 0$, 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

Solution & Explanation

Related Formula

Core Logic

Step 1: Scaling Formula of Time

Step 2: Calculating Final Time

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate kinetic energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Energy Conservation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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