Gravitation Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What does the zero gravitational potential at infinity imply about gravitational systems?

\( V = 0 \) at \( r \to \infty \) as a convention, implying that all bound systems (e.g., orbits) have negative potential energy, requiring external work to unbind them to infinity.

All systems have positive energy
Bound systems have negative energy
Energy is constant everywhere
Force is zero at infinity
2
F

A planet orbits the Sun with a period of 9 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

Kepler’s third law: \( \frac{T_p^2}{T_E^2} = \frac{a_p^3}{a_E^3} \).

\( T_E = 1 \, \text{year} \), \( T_p = 9 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{9^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 81 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 81 \times 3.375 \times 10^{33} = 2.73375 \times 10^{35} \).

\( a_p = (2.73375 \times 10^{35})^{1/3} \approx 6.49 \times 10^{11} \, \text{m} \).

6.3 × 10¹¹ m
6.4 × 10¹¹ m
6.5 × 10¹¹ m
6.6 × 10¹¹ m
3

A planet’s orbital period around the Sun is 8 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

Kepler’s third law: \( \frac{T_p^2}{T_E^2} = \frac{a_p^3}{a_E^3} \).

\( T_E = 1 \, \text{year} \), \( T_p = 8 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{8^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 64 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 64 \times 3.375 \times 10^{33} = 2.16 \times 10^{35} \).

\( a_p = (2.16 \times 10^{35})^{1/3} \approx 6.0 \times 10^{11} \, \text{m} \).

5.8 × 10¹¹ m
6.0 × 10¹¹ m
6.2 × 10¹¹ m
6.4 × 10¹¹ m
2

A satellite’s orbit changes from \( 2 R_E \) to \( 4 R_E \) from Earth’s center. What is the change in its potential energy? (\( m = 200 \, \text{kg}, M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 2 R_E = 1.28 \times 10^7 \, \text{m} \), \( r_2 = 4 R_E = 2.56 \times 10^7 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200 \left(\frac{1}{2.56 \times 10^7} - \frac{1}{1.28 \times 10^7}\right) \).

\( \Delta V = -8.004 \times 10^{16} (-3.906 \times 10^{-8}) \approx 3.13 \times 10^9 \, \text{J} \).

2.8 × 10⁹ J
3.0 × 10⁹ J
3.1 × 10⁹ J
3.5 × 10⁹ J
3

A body weighs \( 98 \, \text{N} \) on Earth’s surface. What is its weight at a depth of \( R_E/2 \)? (\( R_E = 6.4 \times 10^6 \, \text{m}, g = 9.8 \, \text{m/s}^2 \))

\( g(d) = g \left(1 - \frac{d}{R_E}\right) \).

At \( d = R_E/2 \), \( g(d) = 9.8 \left(1 - \frac{R_E/2}{R_E}\right) = 9.8 \times 0.5 = 4.9 \, \text{m/s}^2 \).

Mass: \( m = \frac{W}{g} = \frac{98}{9.8} = 10 \, \text{kg} \).

Weight at depth: \( W = m g(d) = 10 \times 4.9 = 49 \, \text{N} \).

45 N
49 N
55 N
60 N
2

A body weighs \( 196 \, \text{N} \) on Earth’s surface. What is its weight at a depth \( d = R_E/4 \)? (\( g = 9.8 \, \text{m/s}^2 \))

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

\( d = R_E/4 \), \( g(d) = 9.8 (1 - 1/4) = 9.8 \times 3/4 = 7.35 \, \text{m/s}^2 \).

Mass: \( m = 196/9.8 = 20 \, \text{kg} \).

Weight: \( W = 20 \times 7.35 = 147 \, \text{N} \).

145 N
147 N
149 N
151 N
2

What is the escape speed from a planet with mass \( 1.2 \times 10^{24} \, \text{kg} \) and radius \( 3 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 1.2 \times 10^{24}}{3 \times 10^6}} \).

\( v_e = \sqrt{5.336 \times 10^7} \approx 7.3 \times 10^3 \, \text{m/s} = 7.3 \, \text{km/s} \).

7.1 km/s
7.2 km/s
7.3 km/s
7.4 km/s
3

What is the gravitational force on a \( 10 \, \text{kg} \) mass outside a spherical shell of mass \( 500 \, \text{kg} \) and radius \( 2 \, \text{m} \), at \( 3 \, \text{m} \) from the center? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 500 \times 10}{3^2} \).

\( F = \frac{3.335 \times 10^{-7}}{9} \approx 3.71 \times 10^{-8} \, \text{N} \).

3.5 × 10⁻⁸ N
3.7 × 10⁻⁸ N
3.9 × 10⁻⁸ N
4.1 × 10⁻⁸ N
2

What is the minimum speed to escape from \( 4 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{4 R_E}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} \).

\( v_e = \sqrt{3.136 \times 10^7} \approx 5.6 \times 10^3 \, \text{m/s} = 5.6 \, \text{km/s} \).

5.4 km/s
5.6 km/s
5.8 km/s
6.0 km/s
2

A satellite near Earth has a period of 81 minutes. What is its period at \( h = 10 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

\( T^2 \propto (R_E + h)^3 \).

\( T_0^2 = k R_E^3 \), \( h = 10 R_E \), \( r = 11 R_E \).

\( T^2 = k (11 R_E)^3 = 1331 k R_E^3 \).

\( T = T_0 \sqrt{1331} = 81 \times 36.48 \approx 2955 \, \text{min} \).

2940 min
2950 min
2960 min
2970 min
3

What is the minimum speed to escape from \( 9 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{9 R_E}} = \sqrt{\frac{2 \times 9.8 \times 6.4 \times 10^6}{9}} \).

\( v_e = \sqrt{1.395 \times 10^7} \approx 3.73 \times 10^3 \, \text{m/s} \approx 3.7 \, \text{km/s} \).

3.6 km/s
3.7 km/s
3.8 km/s
3.9 km/s
2

Which of the following statements is correct about Kepler’s first law?

Kepler’s first law (law of orbits) describes planetary orbits as ellipses with the Sun at one focus, not circles, contradicting option 4.

Planets move in straight lines
The Sun is at the center of circular orbits
Orbits are elliptical with varying speeds
Orbits are circular with the Sun at the center
3

What is the gravitational force on a \( 11 \, \text{kg} \) mass \( 9 \, \text{m} \) from the center of a spherical shell of mass \( 500 \, \text{kg} \) and radius \( 7 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 500 \times 11}{9^2} \).

\( F = \frac{3.6685 \times 10^{-8}}{81} \approx 4.53 \times 10^{-10} \, \text{N} \).

4.4 × 10⁻¹⁰ N
4.5 × 10⁻¹⁰ N
4.6 × 10⁻¹⁰ N
4.7 × 10⁻¹⁰ N
2

Why is the gravitational potential energy of a system of multiple masses the sum of pairwise interactions?

Gravitational potential energy follows the superposition principle, where the total energy is the sum of the potential energies of all possible pairs (\( V = -\frac{G m_1 m_2}{r} \)) due to the linear nature of gravitational force.

Due to conservation of energy
Due to the superposition principle
Due to Kepler’s laws
Due to Earth’s gravitational field
2

What is the kinetic energy of a \( 400 \, \text{kg} \) satellite at \( 6 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( K = \frac{G M_E m}{2 r} \).

\( r = 6 R_E = 3.84 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 400}{2 \times 3.84 \times 10^7} \).

\( K = \frac{1.601 \times 10^{17}}{7.68 \times 10^7} \approx 2.08 \times 10^9 \, \text{J} \).

2.0 × 10⁹ J
2.1 × 10⁹ J
2.2 × 10⁹ J
2.3 × 10⁹ J
2

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