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…

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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

Q21 jee_main_2025_03_april_morning Gravitational Interaction and Scaling

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.

Numerical Answer. Answer: 8 to 8

Solution

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 $T$ $T$ in terms of physical scaling variables $m$ $m$, $G$ $G$, and $a$ $a$:

[T] = [M]^x [M⁻¹L³T⁻²]^y [L]^z

Equating dimensions on both sides:

Mass ( $M$ $M$): $x - y = 0 x = y$ $x - y = 0 \implies x = y$
Length ( $L$ $L$): $3y + z = 0 z = -3y$ $3y + z = 0 \implies z = -3y$
Time ( $T$ $T$): $-2y = 1 y = -1/2$ $-2y = 1 \implies y = -1/2$

Solving these equations:

x = -1/2, y = -1/2, z = 3/2

Step 1: Scaling Formula of Time

Therefore, the scaling relationship for time $T$ $T$ is:

T \propto m-1/2 G-1/2 a3/2 T \propto \sqrt((a³)/(m))

Let's write the ratio for two cases:

(T₂)/(T₁) = \sqrt(((a₂)/(a₁))³ \cdot ((m₁)/(m₂)))

Step 2: Calculating Final Time

Given values:

$a₁ = a$ $a_1 = a$, $a₂ = 2a$ $a_2 = 2a$
$m₁ = m$ $m_1 = m$, $m₂ = 2m$ $m_2 = 2m$
$T₁ = 4~s$ $T_1 = 4\mathrm{~s}$

(T₂)/(4) = \sqrt(((2a)/(a))³ \cdot ((m)/(2m))) = \sqrt(2³ \cdot (1)/(2)) = \sqrt(4) = 2

T₂ = 4 \times 2 = 8~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 $T \propto \sqrt((R³)/(GM))$ $T \propto \sqrt{\frac{R^3}{GM}}$. Thus doubling $R$ $R$ multiplies time by $\sqrt(8)$ $\sqrt{8}$ and doubling $M$ $M$ divides time by $\sqrt(2)$ $\sqrt{2}$. The combination results in a clean doubling of time: $\sqrt(8)/\sqrt(2) = 2$ $\sqrt{8}/\sqrt{2} = 2$.

Chapter Mix

Class 11 Physics: Gravitation Class 11 Physics: Units and Measurements: Dimensional Analysis

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

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=$M=$ mass of earth, G=$G=$ Universal gravitational constant)

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Energy Conservation

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: Scaling Formula of Time

Step 2: Calculating Final Time

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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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 $M=$ $M=$ mass of earth, $G=$ $G=$ Universal gravitational constant)