Gravitation Chapter-Wise Test 22

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why is the escape speed independent of the mass of the object being launched?

Escape speed as \( v_e = \sqrt{\frac{2 G M}{R}} \). The mass \( m \) of the object cancels out in the energy conservation equation (\( \frac{1}{2} m v_e^2 = \frac{G M m}{R} \)), making \( v_e \) dependent only on the planet’s mass \( M \) and radius \( R \).

Mass cancels out in the energy equation
Gravity affects all masses equally
It depends on the launch angle
It is due to Earth’s rotation
1

How much energy is required to move a \( 700 \, \text{kg} \) satellite from \( 11 R_E \) to \( 22 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 7.04 \times 10^7 \, \text{m} \), \( r_2 = 1.408 \times 10^8 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 700 \left(\frac{1}{1.408 \times 10^8} - \frac{1}{7.04 \times 10^7}\right) \).

\( \Delta E = -2.801 \times 10^{17} (-7.102 \times 10^{-9}) \approx 1.99 \times 10^9 \, \text{J} \).

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

A moon orbits a planet with a period of 7 days and radius \( 5 \times 10^8 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

\( M = \frac{4\pi^2 r^3}{G T^2} \).

\( T = 7 \times 86400 = 6.048 \times 10^5 \, \text{s} \).

\( T^2 = 3.658 \times 10^{11} \, \text{s}^2 \).

\( r^3 = (5 \times 10^8)^3 = 1.25 \times 10^{25} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 1.25 \times 10^{25}}{6.67 \times 10^{-11} \times 3.658 \times 10^{11}} \).

\( M = \frac{4.93 \times 10^{25}}{2.44 \times 10^1} \approx 2.02 \times 10^{24} \, \text{kg} \).

1.9 × 10²⁴ kg
2.0 × 10²⁴ kg
2.1 × 10²⁴ kg
2.2 × 10²⁴ kg
2

What is the minimum speed to escape from \( 3 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{3 R_E}} = \sqrt{\frac{2 \times 9.8 \times 6.4 \times 10^6}{3}} \).

\( v_e = \sqrt{4.181 \times 10^7} \approx 6.47 \times 10^3 \, \text{m/s} \approx 6.5 \, \text{km/s} \).

6.2 km/s
6.4 km/s
6.5 km/s
6.7 km/s
3

Which of the following statements is incorrect about \( g \) variation?

\( g \) decreases with altitude (option 1 correct), depth (option 2 correct), and is zero at the center (option 4 correct). Option 3 is incorrect as \( g \) decreases, not increases, below the surface.

It decreases with altitude
It decreases with depth
It increases below the surface
It is zero at Earth’s center
3

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

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

\( v_e = \sqrt{1.254 \times 10^7} \approx 3.54 \times 10^3 \, \text{m/s} \approx 3.5 \, \text{km/s} \).

3.4 km/s
3.5 km/s
3.6 km/s
3.7 km/s
2

A planet has a moon with an orbital radius of \( 5 \times 10^8 \, \text{m} \) and a period of 10 days. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

\( T^2 = \frac{4\pi^2}{G M} R^3 \).

\( M = \frac{4\pi^2 R^3}{G T^2} \).

\( T = 10 \times 86400 = 8.64 \times 10^5 \, \text{s} \).

\( T^2 = (8.64 \times 10^5)^2 = 7.465 \times 10^{11} \, \text{s}^2 \).

\( R^3 = (5 \times 10^8)^3 = 1.25 \times 10^{25} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 1.25 \times 10^{25}}{6.67 \times 10^{-11} \times 7.465 \times 10^{11}} \).

\( M = \frac{4.93 \times 10^{26}}{4.979 \times 10^1} \approx 9.9 \times 10^{24} \, \text{kg} \).

8.5 × 10²⁴ kg
9.0 × 10²⁴ kg
9.9 × 10²⁴ kg
10.5 × 10²⁴ kg
3

Why did Newton conclude that the gravitational force decreases with the square of the distance?

Newton observed that the centripetal acceleration of the Moon (\( a_m = \frac{V^2}{R_m} \)) was much smaller than the acceleration due to gravity on Earth’s surface (\( g \)), and their ratio matched the inverse square of their distances from Earth’s center (\( \frac{g}{a_m} \approx \frac{R_m^2}{R_E^2} \)).

Due to Kepler’s third law
Due to the Moon’s orbit compared to Earth’s gravity
Due to the Sun’s influence
Due to experimental data from Cavendish
2

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

\( r = R_E + \frac{R_E}{2} = 1.5 R_E \).

\( v = \sqrt{\frac{G M_E}{r}} = \sqrt{\frac{g R_E^2}{1.5 R_E}} = \sqrt{\frac{g R_E}{1.5}} \).

\( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{1.5}} = \sqrt{4.181 \times 10^7} \).

\( v \approx 6.47 \times 10^3 \, \text{m/s} \).

6.0 × 10³ m/s
6.5 × 10³ m/s
7.0 × 10³ m/s
7.5 × 10³ m/s
2

Why does an object require more energy to escape from a higher altitude than to reach that altitude?

Escape energy from altitude \( h \) is \( \frac{G M m}{R + h} \), while energy to reach \( h \) from the surface is \( G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) \), which is less. Escape requires overcoming all potential energy to infinity.

Air resistance increases
Escape needs energy to infinity
Gravity increases with height
Mass changes with altitude
2

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