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  4. Gravitation Chapter-Wise Test 3

Gravitation Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the primary source of centripetal force for an Earth satellite?

For a satellite in circular orbit, the centripetal force (\( F = \frac{m v^2}{r} \)) is provided by Earth’s gravitational force (\( F = \frac{G M_E m}{r^2} \)), which keeps it in orbit.

Earth’s magnetic field
Gravitational force
Frictional force
Centrifugal force
2

Three masses of \( 4 \, \text{kg} \) each are at the vertices of an equilateral triangle with side \( 3 \, \text{m} \). What is the net force on one mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Force between two masses: \( F = G \frac{m_1 m_2}{r^2} = 6.67 \times 10^{-11} \frac{4 \times 4}{3^2} = 1.185 \times 10^{-10} \, \text{N} \).

Two forces act at 60°: \( F_R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^\circ} \).

\( F_R = \sqrt{(1.185 \times 10^{-10})^2 (1 + 1 + 1)} = 1.185 \times 10^{-10} \sqrt{3} \).

\( F_R \approx 2.05 \times 10^{-10} \, \text{N} \).

1.8 × 10⁻¹⁰ N
2.0 × 10⁻¹⁰ N
2.2 × 10⁻¹⁰ N
2.4 × 10⁻¹⁰ N
3

What is the kinetic energy of a \( 500 \, \text{kg} \) satellite at \( 7 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 7 R_E = 4.48 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 500}{2 \times 4.48 \times 10^7} \).

\( K = \frac{2.001 \times 10^{17}}{8.96 \times 10^7} \approx 2.23 \times 10^9 \, \text{J} \).

2.1 × 10⁹ J
2.2 × 10⁹ J
2.3 × 10⁹ J
2.4 × 10⁹ J
3

What is the kinetic energy of a \( 150 \, \text{kg} \) satellite at \( 3 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 3 R_E = 1.92 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 150}{2 \times 1.92 \times 10^7} \).

\( K = \frac{6.003 \times 10^{16}}{3.84 \times 10^7} \approx 1.56 \times 10^9 \, \text{J} \).

1.4 × 10⁹ J
1.5 × 10⁹ J
1.6 × 10⁹ J
1.7 × 10⁹ J
3

What does the negative total energy of a bound system indicate?

Negative total energy (\( E = KE + PE < 0 \)) means the system is bound, requiring external energy to separate the objects to infinity, where \( E = 0 \), typical for orbits.

It can escape
It is bound
It has no kinetic energy
It is at infinity
2

What condition must an object’s energy satisfy to remain bound to Earth?

For an object to remain bound, its total mechanical energy (\( E = KE + PE \)) must be negative, indicating it lacks sufficient energy to reach infinity where \( E = 0 \).

Total energy must be positive
Total energy must be negative
Kinetic energy must be zero
Potential energy must be positive
2

A satellite near Earth has a period of 86 minutes. What is its period at \( h = 5 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( h = 5 R_E \), \( r = 6 R_E \).

\( T^2 = k (6 R_E)^3 = 216 k R_E^3 \).

\( T = T_0 \sqrt{216} = 86 \times 14.7 \approx 1264 \, \text{min} \).

1240 min
1250 min
1260 min
1270 min
4

What is the escape speed from a planet with mass \( 1.8 \times 10^{24} \, \text{kg} \) and radius \( 3.5 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 1.8 \times 10^{24}}{3.5 \times 10^6}} \).

\( v_e = \sqrt{6.861 \times 10^7} \approx 8.28 \times 10^3 \, \text{m/s} \approx 8.3 \, \text{km/s} \).

8.1 km/s
8.2 km/s
8.3 km/s
8.4 km/s
3

A planet moves in an elliptical orbit around the Sun with a semi-major axis of \( 2.25 \times 10^{11} \, \text{m} \). If its orbital period is 2 years, what is the mass of the Sun? (Take \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( 1 \, \text{year} = 3.156 \times 10^7 \, \text{s} \))

Using Kepler’s third law: \( T^2 = \frac{4\pi^2}{G M_s} a^3 \).

Rearrange for \( M_s \): \( M_s = \frac{4\pi^2 a^3}{G T^2} \).

\( T = 2 \times 3.156 \times 10^7 = 6.312 \times 10^7 \, \text{s} \).

\( a = 2.25 \times 10^{11} \, \text{m} \).

\( T^2 = (6.312 \times 10^7)^2 = 3.984 \times 10^{15} \, \text{s}^2 \).

\( a^3 = (2.25 \times 10^{11})^3 = 1.139 \times 10^{33} \, \text{m}^3 \).

\( M_s = \frac{4 \times (3.14)^2 \times 1.139 \times 10^{33}}{6.67 \times 10^{-11} \times 3.984 \times 10^{15}} \).

\( M_s = \frac{4.49 \times 10^{34}}{2.657 \times 10^5} \approx 1.69 \times 10^{30} \, \text{kg} \).

1.5 × 10³⁰ kg
1.7 × 10³⁰ kg
2.0 × 10³⁰ kg
2.5 × 10³⁰ kg
2

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

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

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

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

\( 100 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 100 \times 3.375 \times 10^{33} = 3.375 \times 10^{35} \).

\( a_p = (3.375 \times 10^{35})^{1/3} \approx 6.96 \times 10^{11} \, \text{m} \).

6.8 × 10¹¹ m
6.9 × 10¹¹ m
7.0 × 10¹¹ m
7.1 × 10¹¹ m
3

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