Gravitation Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A satellite’s period is 100 minutes at \( h = 0 \). What is its period at \( h = 2 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( T^2 = k (3 R_E)^3 = 27 k R_E^3 \).

\( T = T_0 \sqrt{27} = 100 \times 5.196 \approx 520 \, \text{min} \).

500 min
510 min
520 min
530 min
3

Two masses \( 9 \, \text{kg} \) and \( 18 \, \text{kg} \) are \( 18 \, \text{m} \) apart. What is the gravitational potential at a point \( 8 \, \text{m} \) from the \( 9 \, \text{kg} \) mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to \( 18 \, \text{kg} \): \( 18 - 8 = 10 \, \text{m} \).

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

\( U = -6.67 \times 10^{-11} \left(\frac{9}{8} + \frac{18}{10}\right) \).

\( U = -6.67 \times 10^{-11} (1.125 + 1.8) \approx -1.95 \times 10^{-10} \, \text{J/kg} \).

-1.9 × 10⁻¹⁰ J/kg
-2.0 × 10⁻¹⁰ J/kg
-2.1 × 10⁻¹⁰ J/kg
-2.2 × 10⁻¹⁰ J/kg
1

What is the gravitational force on a \( 9 \, \text{kg} \) mass \( 7 \, \text{m} \) from the center of a spherical shell of mass \( 350 \, \text{kg} \) and radius \( 5 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 350 \times 9}{7^2} \).

\( F = \frac{2.101 \times 10^{-8}}{49} \approx 4.29 \times 10^{-10} \, \text{N} \).

4.1 × 10⁻¹⁰ N
4.2 × 10⁻¹⁰ N
4.3 × 10⁻¹⁰ N
4.4 × 10⁻¹⁰ N
3

What is the kinetic energy of a \( 600 \, \text{kg} \) satellite at \( 8 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

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

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

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

At what depth below Earth’s surface is \( g \) reduced to \( 7.35 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(d) = g_0 (1 - d/R_E) \).

\( 7.35 = 9.8 (1 - d/R_E) \).

\( 1 - d/R_E = 0.75 \).

\( d/R_E = 0.25 \).

\( d = 0.25 \times 6.4 \times 10^6 = 1.6 \times 10^6 \, \text{m} \).

1.5 × 10⁶ m
1.6 × 10⁶ m
1.7 × 10⁶ m
1.8 × 10⁶ m
2

A satellite orbits Earth at \( 9 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 9 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{9}} \).

\( v = \sqrt{6.964 \times 10^6} \approx 2.64 \times 10^3 \, \text{m/s} \).

2.5 × 10³ m/s
2.6 × 10³ m/s
2.7 × 10³ m/s
2.8 × 10³ m/s
2

Two masses \( 11 \, \text{kg} \) and \( 22 \, \text{kg} \) are \( 22 \, \text{m} \) apart. What is the gravitational potential at a point \( 10 \, \text{m} \) from the \( 11 \, \text{kg} \) mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to \( 22 \, \text{kg} \): \( 22 - 10 = 12 \, \text{m} \).

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

\( U = -6.67 \times 10^{-11} \left(\frac{11}{10} + \frac{22}{12}\right) \).

\( U = -6.67 \times 10^{-11} (1.1 + 1.833) \approx -1.96 \times 10^{-10} \, \text{J/kg} \).

-1.9 × 10⁻¹⁰ J/kg
-2.0 × 10⁻¹⁰ J/kg
-2.1 × 10⁻¹⁰ J/kg
-2.2 × 10⁻¹⁰ J/kg
2

A satellite near Earth has a period of 87 minutes. What is its period at \( h = 6 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( h = 6 R_E \), \( r = 7 R_E \).

\( T^2 = k (7 R_E)^3 = 343 k R_E^3 \).

\( T = T_0 \sqrt{343} = 87 \times 18.52 \approx 1611 \, \text{min} \).

1590 min
1600 min
1610 min
1620 min
4

What is the gravitational potential due to Earth at \( 3.2 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( U = -1.25 \times 10^7 \, \text{J/kg} \).

-1.2 × 10⁷ J/kg
-1.3 × 10⁷ J/kg
-1.4 × 10⁷ J/kg
-1.5 × 10⁷ J/kg
1

Why does the gravitational force act as if all mass is at the center for a uniform sphere?

For a uniform sphere, symmetry ensures that the net gravitational force on an external point is equivalent to that of a point mass at the center, as forces from all parts combine accordingly.

Mass is concentrated at the center
Symmetry simplifies the force
Density varies radially
The sphere is hollow
2

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