A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-

Related Formula

(dA)/(dt) = (L)/(2m)$$\frac{dA}{dt} = \frac{L}{2m}$$ [cite: 684]

Core Logic

Kepler's Second Law state that the areal velocity (dA)/(dt)$\frac{dA}{dt}$ is directly proportional to the angular momentum L$L$ of the planet[cite: 55, 684]. Because the gravitational force between the Sun and the planet acts strictly along the line joining their centers (a central force field), its torque τ = r × F = 0$\vec{\tau} = \vec{r} \times \vec{F} = 0$[cite: 56, 686]. Since torque is zero, the angular momentum L$L$ remains completely constant over time[cite: 686]. As a consequence, (dA)/(dt) = constant$\frac{dA}{dt} = \text{constant}$[cite: 55, 684]. Both statements are true and (R) is the correct explanation of (A)[cite: 63].

Pattern Recognition

Areal velocity constancy is a direct geometric manifestation of the conservation of angular momentum under any central force field[cite: 55, 56, 686].

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Surface gravitational acceleration:

Core Logic

Since diameter drops to 1/3$1/3$, the radius R'$R'$ also scales down to 1/3$1/3$ its original value (R' = R/3$R' = R/3$), while mass M$M$ remains constant.

New acceleration value g'$g'$ calculation:

The scaling factor is 9.

Pattern Recognition

Gravity follows an inverse-square law with respect to radius. Shrinking the radius by a factor of n$n$ increases surface gravity by n²$n^2$ if the mass is unchanged.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

The expression for escape velocity from a spherical planetary body is given by:

Core Logic

Let the planet's mass be MP$M_P$ and its radius be RP$R_P$. According to the problem statement :

• Earth's mass, ME = 8 MP (MP)/(ME) = (1)/(8)$M_E = 8 M_P \implies \frac{M_P}{M_E} = \frac{1}{8}$
• Earth's radius, RE = 2 RP (RE)/(RP) = 2$R_E = 2 R_P \implies \frac{R_E}{R_P} = 2$

Taking the ratio of escape velocities :

Given that vE = 11.2 km/s$v_E = 11.2 \text{ km/s}$:

Pattern Recognition

Setting up quick ratios prevents substitution mistakes. For any planetary variant, notice how scaling properties scale inside the root operator directly.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Variation of g$g$ with height:

Time period of a simple pendulum:

Core Logic

Given height h = 2R$h = 2R$, the effective acceleration due to gravity becomes:

For a second's pendulum, the time period is defined exactly as T = 2~s$T = 2\mathrm{~s}$.

Step 1: Calculate Length

Substitute T = 2~s$T = 2\mathrm{~s}$ and g' = (g)/(9)$g' = \frac{g}{9}$ into the time period formula:

Squaring both sides:

Since g = π² ~m/s²$g = \pi^2 \mathrm{~m/s^2}$:

Pattern Recognition

Second's pendulum always has T = 2~s$T = 2\mathrm{~s}$. Height 2R$2R$ from the surface means a total distance of 3R$3R$ from the center, which yields a (1)/(9)$\frac{1}{9}$ drop in gravity.

Chapter Mix

Class 11 Physics: Gravitation Class 11 Physics: Oscillations