Gravitation Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the kinetic energy of a \( 300 \, \text{kg} \) satellite at \( 5 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 5 R_E = 3.2 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 300}{2 \times 3.2 \times 10^7} \).

\( K = \frac{1.201 \times 10^{17}}{6.4 \times 10^7} \approx 1.88 \times 10^9 \, \text{J} \).

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

A satellite near Earth has a period of 82 minutes. What is its period at \( h = 9 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( h = 9 R_E \), \( r = 10 R_E \).

\( T^2 = k (10 R_E)^3 = 1000 k R_E^3 \).

\( T = T_0 \sqrt{1000} = 82 \times 31.62 \approx 2593 \, \text{min} \).

2580 min
2590 min
2600 min
2610 min
2

Two masses \( 3 \, \text{kg} \) and \( 6 \, \text{kg} \) are \( 6 \, \text{m} \) apart. What is the gravitational potential at a point \( 2 \, \text{m} \) from the \( 3 \, \text{kg} \) mass on the line joining them? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to \( 6 \, \text{kg} \): \( 6 - 2 = 4 \, \text{m} \).

\( U = -\frac{G m_1}{r_1} - \frac{G m_2}{r_2} \).

\( U = -6.67 \times 10^{-11} \left(\frac{3}{2} + \frac{6}{4}\right) \).

\( U = -6.67 \times 10^{-11} (1.5 + 1.5) = -2.001 \times 10^{-10} \, \text{J/kg} \).

-1.8 × 10⁻¹⁰ J/kg
-2.0 × 10⁻¹⁰ J/kg
-2.2 × 10⁻¹⁰ J/kg
-2.4 × 10⁻¹⁰ J/kg
2

What is the minimum speed to escape from \( 8 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{8 R_E}} = \sqrt{\frac{2 \times 9.8 \times 6.4 \times 10^6}{8}} \).

\( v_e = \sqrt{1.568 \times 10^7} \approx 3.96 \times 10^3 \, \text{m/s} \approx 4.0 \, \text{km/s} \).

3.9 km/s
4.0 km/s
4.1 km/s
4.2 km/s
2

Why is the value of \( g \) maximum at Earth’s surface?

At the surface. Above, \( g \) decreases with \( 1/(R_E + h)^2 \); below, only inner mass contributes, reducing \( g \) linearly as \( g(d) = g (1 - d/R_E) \). Thus, \( g \) peaks at \( r = R_E \).

All mass contributes at the surface
Earth’s core increases it
Atmosphere enhances it
It is zero at the surface
1

What is the escape speed from a planet with mass \( 4.5 \times 10^{24} \, \text{kg} \) and radius \( 6 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.5 \times 10^{24}}{6 \times 10^6}} \).

\( v_e = \sqrt{1.0005 \times 10^8} \approx 1.00 \times 10^4 \, \text{m/s} = 10.0 \, \text{km/s} \).

9.8 km/s
9.9 km/s
10.0 km/s
10.1 km/s
3

Four \( 8 \, \text{kg} \) masses form a square of side \( 8 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 8 \, \text{m} \), 2 diagonals: \( r = 8\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{8} - 2 \frac{G m^2}{8\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 64 \left(\frac{4}{8} + \frac{2}{8\sqrt{2}}\right) \).

\( V = -4.269 \times 10^{-9} (0.5 + 0.177) \approx -2.89 \times 10^{-9} \, \text{J} \).

-2.8 × 10⁻⁹ J
-2.9 × 10⁻⁹ J
-3.0 × 10⁻⁹ J
-3.1 × 10⁻⁹ J
2

What is the gravitational potential energy of a \( 2 \, \text{kg} \) mass at a height of \( 6.4 \times 10^6 \, \text{m} \) above Earth’s surface? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance from center: \( r = R_E + h = 6.4 \times 10^6 + 6.4 \times 10^6 = 1.28 \times 10^7 \, \text{m} \).

\( V = -\frac{G M_E m}{r} = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2}{1.28 \times 10^7} \).

\( V = -\frac{8.004 \times 10^{14}}{1.28 \times 10^7} = -6.25 \times 10^7 \, \text{J} \).

-6.0 × 10⁷ J
-6.25 × 10⁷ J
-6.5 × 10⁷ J
-7.0 × 10⁷ J
2

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

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

\( r_1 = 3.2 \times 10^7 \, \text{m} \), \( r_2 = 6.4 \times 10^7 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 5 \left(\frac{1}{6.4 \times 10^7} - \frac{1}{3.2 \times 10^7}\right) \).

\( \Delta V = -2.001 \times 10^{15} (-1.5625 \times 10^{-8}) \approx 3.13 \times 10^7 \, \text{J} \).

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

What is the minimum speed to escape from \( 7 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{7 R_E}} = \sqrt{\frac{2 \times 9.8 \times 6.4 \times 10^6}{7}} \).

\( v_e = \sqrt{1.79 \times 10^7} \approx 4.23 \times 10^3 \, \text{m/s} \approx 4.2 \, \text{km/s} \).

4.1 km/s
4.2 km/s
4.3 km/s
4.4 km/s
2

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