Gravitation Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Three masses of \( 8 \, \text{kg} \) each form an equilateral triangle with side \( 6 \, \text{m} \). 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: \( F = G \frac{m^2}{r^2} = 6.67 \times 10^{-11} \frac{8 \times 8}{6^2} = 1.185 \times 10^{-10} \, \text{N} \).

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

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

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

1.9 × 10⁻¹⁰ N
2.0 × 10⁻¹⁰ N
2.1 × 10⁻¹⁰ N
2.2 × 10⁻¹⁰ N
3

What is the gravitational force on a \( 7 \, \text{kg} \) mass \( 6 \, \text{m} \) from the center of a spherical shell of mass \( 300 \, \text{kg} \) and radius \( 4 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 300 \times 7}{6^2} \).

\( F = \frac{1.401 \times 10^{-8}}{36} \approx 3.89 \times 10^{-10} \, \text{N} \).

3.7 × 10⁻¹⁰ N
3.8 × 10⁻¹⁰ N
3.9 × 10⁻¹⁰ N
4.0 × 10⁻¹⁰ N
3

A \( 600 \, \text{kg} \) satellite orbits Earth at \( 8 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 8 R_E = 5.12 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 600}{2 \times 5.12 \times 10^7} \).

\( E = -\frac{2.401 \times 10^{17}}{1.024 \times 10^8} \approx -2.34 \times 10^9 \, \text{J} \).

-2.2 × 10⁹ J
-2.3 × 10⁹ J
-2.4 × 10⁹ J
-2.5 × 10⁹ J
3

A satellite orbits at \( 3 R_E \) from Earth’s center. What is its speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v = \sqrt{\frac{g R_E^2}{r}} \).

\( r = 3 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{3}} \).

\( v = \sqrt{2.09 \times 10^7} \approx 4.57 \times 10^3 \, \text{m/s} \).

4.2 × 10³ m/s
4.5 × 10³ m/s
4.6 × 10³ m/s
4.8 × 10³ m/s
3

A \( 8 \, \text{kg} \) mass is moved from \( 8 R_E \) to \( 16 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 5.12 \times 10^7 \, \text{m} \), \( r_2 = 1.024 \times 10^8 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 8 \left(\frac{1}{1.024 \times 10^8} - \frac{1}{5.12 \times 10^7}\right) \).

\( \Delta V = -3.202 \times 10^{15} (-9.766 \times 10^{-9}) \approx 3.13 \times 10^7 \, \text{J} \).

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

At what height above Earth’s surface is \( g \) reduced to \( 6.12 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(h) = \frac{g_0}{(1 + h/R_E)^2} \).

\( 6.12 = \frac{9.8}{(1 + h/R_E)^2} \).

\( (1 + h/R_E)^2 = 1.601 \).

\( 1 + h/R_E = \sqrt{1.601} \approx 1.265 \).

\( h/R_E = 0.265 \).

\( h = 0.265 \times 6.4 \times 10^6 \approx 1.7 \times 10^6 \, \text{m} \).

1.6 × 10⁶ m
1.7 × 10⁶ m
1.8 × 10⁶ m
1.9 × 10⁶ m
2

A \( 5 \, \text{kg} \) mass is moved from \( R_E \) to \( 2 R_E \) from Earth’s center. What is the work done? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 6.4 \times 10^6 \, \text{m} \), \( r_2 = 1.28 \times 10^7 \, \text{m} \).

\( W = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 5 \left(\frac{1}{1.28 \times 10^7} - \frac{1}{6.4 \times 10^6}\right) \).

\( W = -2.001 \times 10^{15} (-7.8125 \times 10^{-8}) \approx 1.56 \times 10^8 \, \text{J} \).

1.4 × 10⁸ J
1.5 × 10⁸ J
1.6 × 10⁸ J
1.7 × 10⁸ J
3

A body is launched from Earth at \( 12.5 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

\( v_f^2 = v_i^2 - v_e^2 \).

\( v_f^2 = (12.5)^2 - (11.2)^2 = 156.25 - 125.44 = 30.81 \).

\( v_f = \sqrt{30.81} \approx 5.55 \, \text{km/s} \).

5.4 km/s
5.5 km/s
5.6 km/s
5.7 km/s
2

A \( 700 \, \text{kg} \) satellite orbits Earth at \( 10 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 10 R_E = 6.4 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 700}{2 \times 6.4 \times 10^7} \).

\( E = -\frac{2.801 \times 10^{17}}{1.28 \times 10^8} \approx -2.19 \times 10^9 \, \text{J} \).

-2.1 × 10⁹ J
-2.2 × 10⁹ J
-2.3 × 10⁹ J
-2.4 × 10⁹ J
2

A body is launched from Earth at \( 11.5 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

\( v_f^2 = v_i^2 - v_e^2 \).

\( v_f^2 = (11.5)^2 - (11.2)^2 = 132.25 - 125.44 = 6.81 \).

\( v_f = \sqrt{6.81} \approx 2.61 \, \text{km/s} \).

2.5 km/s
2.6 km/s
2.7 km/s
2.8 km/s
2

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