Gravitation Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How much energy is required to move a \( 600 \, \text{kg} \) satellite from \( 10 R_E \) to \( 20 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 6.4 \times 10^7 \, \text{m} \), \( r_2 = 1.28 \times 10^8 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 600 \left(\frac{1}{1.28 \times 10^8} - \frac{1}{6.4 \times 10^7}\right) \).

\( \Delta E = -2.401 \times 10^{17} (-7.8125 \times 10^{-9}) \approx 1.88 \times 10^9 \, \text{J} \).

1.8 × 10⁹ J
1.9 × 10⁹ J
2.0 × 10⁹ J
2.1 × 10⁹ J
2

What is the escape speed from a planet with \( g = 4.9 \, \text{m/s}^2 \) and radius \( 3.2 \times 10^6 \, \text{m} \)?

\( v_e = \sqrt{2 g R} \).

\( v_e = \sqrt{2 \times 4.9 \times 3.2 \times 10^6} = \sqrt{3.136 \times 10^7} \).

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

5.0 km/s
5.6 km/s
6.0 km/s
6.5 km/s
2

A body is projected from Earth with \( 13 \, \text{km/s} \). What is its speed far away? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (13)^2 - (11.2)^2 = 169 - 125.44 = 43.56 \).

\( v_f = \sqrt{43.56} \approx 6.6 \, \text{km/s} \).

6.0 km/s
6.6 km/s
7.0 km/s
7.5 km/s
2

Why is the gravitational potential energy negative in a bound system?

Gravitational potential energy as \( V = -\frac{G M m}{r} \), with zero at infinity. For a bound system (e.g., orbit), the negative value indicates that work must be done to separate the objects to infinity, reflecting gravitational attraction.

The force is repulsive
Work is required to reach infinity
Energy is lost to friction
The system is unbound
2

Why does a satellite in a circular orbit have negative total energy?

The total energy \( E = -\frac{G M m}{2 r} \) is negative because the potential energy (\( -\frac{G M m}{r} \)) is negative and twice the magnitude of the positive kinetic energy (\( \frac{G M m}{2 r} \)). This negative value indicates a bound system.

Kinetic energy is zero
Potential energy exceeds kinetic energy
It is moving away from Earth
Gravitational force is repulsive
2

What is the gravitational potential due to Earth at \( 3.84 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( U = -\frac{G M_E}{r} \).

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

\( U = -1.0417 \times 10^7 \, \text{J/kg} \).

-1.0 × 10⁷ J/kg
-1.1 × 10⁷ J/kg
-1.2 × 10⁷ J/kg
-1.3 × 10⁷ J/kg
1

Why does Kepler’s first law indicate that planetary orbits are not perfectly circular?

Kepler’s first law (law of orbits) describes planetary orbits as ellipses with the Sun at one focus, not circles (a special case of an ellipse). This deviation from circular motion reflects the real dynamics of gravitational interaction.

The Sun is not at the center
Orbits are elliptical with the Sun at one focus
Planets move at constant speed
Gravitational force varies linearly
2

What ensures that a satellite remains in a stable circular orbit?

A stable circular orbit requires the gravitational force (\( \frac{G M_E m}{r^2} \)) to equal the centripetal force (\( \frac{m v^2}{r} \)), balancing the forces to maintain constant radius and speed.

Equal gravitational and frictional forces
Balance of gravitational and centripetal forces
Constant kinetic energy
Zero potential energy
2

Three \( 2 \, \text{kg} \) masses form an equilateral triangle of side \( 1 \, \text{m} \). What is the potential energy? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

3 pairs: \( V = -3 \frac{G m^2}{r} \).

\( V = -3 \frac{6.67 \times 10^{-11} \times 2 \times 2}{1} \).

\( V = -3 \times 2.668 \times 10^{-10} = -8.004 \times 10^{-10} \, \text{J} \).

-7.8 × 10⁻¹⁰ J
-8.0 × 10⁻¹⁰ J
-8.2 × 10⁻¹⁰ J
-8.4 × 10⁻¹⁰ J
2

A moon orbits a planet with a period of 13 days and radius \( 1.0 \times 10^9 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

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

\( T = 13 \times 86400 = 1.1232 \times 10^6 \, \text{s} \).

\( T^2 = 1.262 \times 10^{12} \, \text{s}^2 \).

\( r^3 = (1.0 \times 10^9)^3 = 1.0 \times 10^{27} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 10^{27}}{6.67 \times 10^{-11} \times 1.262 \times 10^{12}} \).

\( M = \frac{3.947 \times 10^{27}}{8.418 \times 10^1} \approx 4.69 \times 10^{25} \, \text{kg} \).

4.6 × 10²⁵ kg
4.7 × 10²⁵ kg
4.8 × 10²⁵ kg
4.9 × 10²⁵ kg
2

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