Gravitation Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

At what depth below Earth’s surface is \( g \) reduced to \( 8.33 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(d) = g_0 (1 - d/R_E) \).

\( 8.33 = 9.8 (1 - d/R_E) \).

\( 1 - d/R_E = 0.85 \).

\( d/R_E = 0.15 \).

\( d = 0.15 \times 6.4 \times 10^6 = 9.6 \times 10^5 \, \text{m} \).

9.4 × 10⁵ m
9.6 × 10⁵ m
9.8 × 10⁵ m
1.0 × 10⁶ m
2

What is the escape speed from a planet with mass \( 3 \times 10^{24} \, \text{kg} \) and radius \( 4 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( v_e = \sqrt{\frac{2 G M}{R}} \).

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 3 \times 10^{24}}{4 \times 10^6}} \).

\( v_e = \sqrt{\frac{4.002 \times 10^{14}}{4 \times 10^6}} = \sqrt{1.001 \times 10^8} \).

\( v_e \approx 1.0 \times 10^4 \, \text{m/s} = 10 \, \text{km/s} \).

8 km/s
10 km/s
12 km/s
14 km/s
2

Which of the following statements is incorrect about Kepler’s second law?

Kepler’s second law (equal areas in equal times) implies varying speed (faster at perihelion, slower at aphelion) due to angular momentum conservation. Option 2 is incorrect as speed is not constant.

It implies angular momentum conservation
Planets move at constant speed
Area swept is constant over time
It applies to elliptical orbits
2

A projectile is launched at \( 7 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

\( \frac{1}{2} v_i^2 - \frac{v_e^2}{2} = -\frac{v_e^2}{2} \frac{R_E}{r} \).

\( 24.5 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{38.22} \approx 1.64 \).

\( r = 1.64 \times 6.4 \times 10^6 \approx 1.05 \times 10^7 \, \text{m} \).

1.0 × 10⁷ m
1.1 × 10⁷ m
1.2 × 10⁷ m
1.3 × 10⁷ m
2

Why does Newton’s law of gravitation apply universally to all objects?

Newton’s law (\( F = G \frac{m_1 m_2}{r^2} \)) is universal because it depends only on mass and distance, properties common to all objects, and \( G \) is a universal constant, applicable everywhere.

It varies with location
It depends on universal properties
It is specific to Earth
It ignores small masses
2

What is the gravitational force on a \( 8 \, \text{kg} \) mass \( 5 \, \text{m} \) from the center of a spherical shell of mass \( 400 \, \text{kg} \) and radius \( 3 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Outside shell: \( F = \frac{G M m}{r^2} \).

\( F = \frac{6.67 \times 10^{-11} \times 400 \times 8}{5^2} \).

\( F = \frac{2.135 \times 10^{-8}}{25} \approx 8.54 \times 10^{-10} \, \text{N} \).

8.3 × 10⁻¹⁰ N
8.5 × 10⁻¹⁰ N
8.7 × 10⁻¹⁰ N
8.9 × 10⁻¹⁰ N
2

A \( 9 \, \text{kg} \) mass is moved from \( 9 R_E \) to \( 18 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( \Delta V = -G M_E m \left(\frac{1}{r_2} - \frac{1}{r_1}\right) \).

\( r_1 = 5.76 \times 10^7 \, \text{m} \), \( r_2 = 1.152 \times 10^8 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 9 \left(\frac{1}{1.152 \times 10^8} - \frac{1}{5.76 \times 10^7}\right) \).

\( \Delta V = -3.602 \times 10^{15} (-8.681 \times 10^{-9}) \approx 3.13 \times 10^7 \, \text{J} \).

3.0 × 10⁷ J
3.1 × 10⁷ J
3.2 × 10⁷ J
3.3 × 10⁷ J
2

A projectile is launched at \( 2 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

\( \frac{1}{2} v_i^2 - \frac{v_e^2}{2} = -\frac{v_e^2}{2} \frac{R_E}{r} \).

\( 2 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{60.72} \approx 1.033 \).

\( r = 1.033 \times 6.4 \times 10^6 \approx 6.61 \times 10^6 \, \text{m} \).

6.5 × 10⁶ m
6.6 × 10⁶ m
6.7 × 10⁶ m
6.8 × 10⁶ m
2

Why does the gravitational force on a body inside Earth decrease linearly with depth?

Inside a uniform Earth, only the mass within radius \( r < R_E \) contributes to gravity, and this mass decreases with \( r^3 \), while the force depends on \( 1/r^2 \). The net effect is a linear decrease: \( g(d) = g (1 - d/R_E) \).

Due to Earth’s rotation
Due to decreasing contributing mass
Due to atmospheric pressure
Due to the shell theorem
2

A satellite orbits a planet at \( 6 \times 10^7 \, \text{m} \) from its center with a period of 15 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( M = \frac{4\pi^2 r^3}{G T^2} \).

\( T = 15 \times 3600 = 54000 \, \text{s} \), \( T^2 = 2.916 \times 10^9 \, \text{s}^2 \).

\( r^3 = (6 \times 10^7)^3 = 2.16 \times 10^{23} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 2.16 \times 10^{23}}{6.67 \times 10^{-11} \times 2.916 \times 10^9} \).

\( M = \frac{8.51 \times 10^{24}}{1.948 \times 10^{-1}} \approx 4.37 \times 10^{25} \, \text{kg} \).

4.2 × 10²⁵ kg
4.3 × 10²⁵ kg
4.4 × 10²⁵ kg
4.5 × 10²⁵ kg
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0