Gravitation Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Energy required to move a \( 300 \, \text{kg} \) satellite from \( 3 R_E \) to \( 6 R_E \) is? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 3 R_E = 1.92 \times 10^7 \, \text{m} \), \( r_2 = 6 R_E = 3.84 \times 10^7 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 300 \left(\frac{1}{3.84 \times 10^7} - \frac{1}{1.92 \times 10^7}\right) \).

\( \Delta E = -1.201 \times 10^{17} (-2.604 \times 10^{-8}) \approx 3.13 \times 10^9 \, \text{J} \).

3.0 × 10⁹ J
3.1 × 10⁹ J
3.2 × 10⁹ J
3.3 × 10⁹ J
2

A \( 900 \, \text{kg} \) satellite orbits Earth at \( 14 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 14 R_E = 8.96 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 900}{2 \times 8.96 \times 10^7} \).

\( E = -\frac{3.602 \times 10^{17}}{1.792 \times 10^8} \approx -2.01 \times 10^9 \, \text{J} \).

-1.9 × 10⁹ J
-2.0 × 10⁹ J
-2.1 × 10⁹ J
-2.2 × 10⁹ J
2

A \( 7 \, \text{kg} \) mass is moved from \( 7 R_E \) to \( 14 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 4.48 \times 10^7 \, \text{m} \), \( r_2 = 8.96 \times 10^7 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 7 \left(\frac{1}{8.96 \times 10^7} - \frac{1}{4.48 \times 10^7}\right) \).

\( \Delta V = -2.801 \times 10^{15} (-1.116 \times 10^{-8}) \approx 3.13 \times 10^7 \, \text{J} \).

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

The time period of a satellite near Earth’s surface is 84 minutes. What is its time period at a height \( h = R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

\( T^2 = k (R_E + h)^3 \), where \( k = \frac{4\pi^2}{G M_E} \).

For \( h = 0 \), \( T_0 = 84 \, \text{min} \), \( T_0^2 = k R_E^3 \).

For \( h = R_E \), \( r = 2 R_E \), \( T^2 = k (2 R_E)^3 = 8 k R_E^3 \).

\( T = \sqrt{8} T_0 = 2.828 \times 84 \approx 237 \, \text{min} \).

200 min
237 min
250 min
270 min
2

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

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

\( r = 2 R_E = 2 \times 6.4 \times 10^6 = 1.28 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 500}{2 \times 1.28 \times 10^7} \).

\( K = \frac{2.001 \times 10^{17}}{2.56 \times 10^7} \approx 7.82 \times 10^9 \, \text{J} \).

7.0 × 10⁹ J
7.5 × 10⁹ J
7.8 × 10⁹ J
8.0 × 10⁹ J
3

Three masses of \( 12 \, \text{kg} \) each form an equilateral triangle with side \( 9 \, \text{m} \). 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: \( F = G \frac{m^2}{r^2} = 6.67 \times 10^{-11} \frac{12 \times 12}{9^2} = 1.185 \times 10^{-10} \, \text{N} \).

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

How much energy is required to move a \( 100 \, \text{kg} \) satellite from \( 4 R_E \) to \( 8 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 2.56 \times 10^7 \, \text{m} \), \( r_2 = 5.12 \times 10^7 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 100 \left(\frac{1}{5.12 \times 10^7} - \frac{1}{2.56 \times 10^7}\right) \).

\( \Delta E = -4.002 \times 10^{16} (-1.953 \times 10^{-8}) \approx 7.82 \times 10^8 \, \text{J} \).

7.6 × 10⁸ J
7.8 × 10⁸ J
8.0 × 10⁸ J
8.2 × 10⁸ J
2

What is the escape speed from a planet with mass \( 3.6 \times 10^{24} \, \text{kg} \) and radius \( 5 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 3.6 \times 10^{24}}{5 \times 10^6}} \).

\( v_e = \sqrt{9.607 \times 10^7} \approx 9.80 \times 10^3 \, \text{m/s} \approx 9.8 \, \text{km/s} \).

9.6 km/s
9.7 km/s
9.8 km/s
9.9 km/s
3

Why is the gravitational force on a point mass outside a hollow spherical shell equivalent to that of a point mass at its center?

For a uniform spherical shell, the gravitational forces from all parts of the shell combine such that the net force on an external point acts as if the shell’s entire mass is concentrated at its center, due to symmetry.

The shell has no thickness
Symmetry causes forces to cancel radially
Mass is distributed evenly at the center
The force is zero outside
2

Why does the gravitational potential energy approach zero as the distance between two masses increases?

\( V = -\frac{G M m}{r} \), with the convention that \( V = 0 \) at \( r \to \infty \). As \( r \) increases, \( \frac{1}{r} \) decreases, making \( V \) less negative and approaching zero.

The force becomes repulsive
The force decreases with distance
It is defined as zero at infinity
Kinetic energy increases
3

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