Gravitation Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 3 \, \text{kg} \) mass is moved from \( 4 R_E \) to \( 8 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 3 \left(\frac{1}{5.12 \times 10^7} - \frac{1}{2.56 \times 10^7}\right) \).

\( \Delta V = -1.201 \times 10^{15} (-1.953 \times 10^{-8}) \approx 2.34 \times 10^7 \, \text{J} \).

2.2 × 10⁷ J
2.3 × 10⁷ J
2.4 × 10⁷ J
2.5 × 10⁷ J
3

A satellite orbits a planet at \( 2 \times 10^6 \, \text{m} \) from its center with a period of 2 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( T = 2 \times 3600 = 7200 \, \text{s} \), \( T^2 = 5.184 \times 10^7 \, \text{s}^2 \).

\( r^3 = (2 \times 10^6)^3 = 8 \times 10^{18} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 8 \times 10^{18}}{6.67 \times 10^{-11} \times 5.184 \times 10^7} \).

\( M = \frac{3.155 \times 10^{20}}{3.458 \times 10^{-3}} \approx 9.12 \times 10^{22} \, \text{kg} \).

8.5 × 10²² kg
9.0 × 10²² kg
9.1 × 10²² kg
9.5 × 10²² kg
3

What ensures that the area swept by a planet’s orbit is constant over equal time intervals?

(\( L = m r v \)), which remains constant for a central force like gravity, ensuring \( \Delta A / \Delta t = L/(2m) \) is constant.

Constant speed
Conservation of angular momentum
Elliptical orbit shape
Conservation of energy
2

Which of the following statements is correct about a satellite in circular orbit?

The gravitational force provides the centripetal force (\( \frac{G M_E m}{r^2} = \frac{m v^2}{r} \)), making option 2 correct.

Its speed increases with altitude
Gravity provides the centripetal force
Its total energy is positive
It has no kinetic energy
2

Which of the following statements is correct about Kepler’s laws?

Kepler’s laws apply to any central force following an inverse-square law (e.g., gravitation), making option 4 correct as they describe planetary motion around the Sun.

They require circular orbits
They apply only to Earth
They depend on planet mass
They apply to inverse-square forces
4

At what depth below Earth’s surface is \( g \) reduced to \( 5.39 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( 5.39 = 9.8 (1 - d/R_E) \).

\( 1 - d/R_E = 0.55 \).

\( d/R_E = 0.45 \).

\( d = 0.45 \times 6.4 \times 10^6 = 2.88 \times 10^6 \, \text{m} \).

2.8 × 10⁶ m
2.9 × 10⁶ m
3.0 × 10⁶ m
3.1 × 10⁶ m
2

Why does the Moon have no atmosphere compared to Earth?

The Moon’s escape speed (2.3 km/s) is much lower than Earth’s (11.2 km/s) due to its smaller mass and radius. Gas molecules with speeds exceeding this low escape speed can easily escape the Moon’s gravitational pull.

It has no magnetic field
Its escape speed is low
It is closer to the Sun
It has no gravitational force
2

A satellite orbits Earth at \( 15 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 15 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{15}} \).

\( v = \sqrt{4.181 \times 10^6} \approx 2.04 \times 10^3 \, \text{m/s} \).

2.0 × 10³ m/s
2.1 × 10³ m/s
2.2 × 10³ m/s
2.3 × 10³ m/s
1

What happens to a satellite’s orbital speed if its altitude increases?

\( v = \sqrt{\frac{G M_E}{r}} \). As altitude increases, \( r \) (distance from Earth’s center) increases, reducing \( v \) since \( v \propto \frac{1}{\sqrt{r}} \). Thus, orbital speed decreases.

It increases
It decreases
It remains constant
It becomes zero
2

What happens to the time period of a satellite if its orbital radius is doubled?

\( T^2 \propto r^3 \) (Kepler’s third law for satellites). If \( r \) is doubled (\( r' = 2r \)), then \( T'^2 = (2r)^3 = 8r^3 \), so \( T' = T \sqrt{8} = T \times 2\sqrt{2} \), meaning the period increases by a factor of \( 2\sqrt{2} \).

It doubles
It increases by \( 2\sqrt{2} \)
It quadruples
It remains the same
2

What is the gravitational potential due to Earth at a distance of \( 1.6 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( U = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{1.6 \times 10^7} \).

\( U = -2.5 \times 10^7 \, \text{J/kg} \).

-2.3 × 10⁷ J/kg
-2.4 × 10⁷ J/kg
-2.5 × 10⁷ J/kg
-2.6 × 10⁷ J/kg
3

A \( 1000 \, \text{kg} \) satellite orbits Earth at \( 1.5 R_E \) from the center. What is its potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 1.5 R_E = 1.5 \times 6.4 \times 10^6 = 9.6 \times 10^6 \, \text{m} \).

\( V = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{9.6 \times 10^6} \).

\( V = -4.17 \times 10^{10} \, \text{J} \).

-4.0 × 10¹⁰ J
-4.2 × 10¹⁰ J
-4.5 × 10¹⁰ J
-4.7 × 10¹⁰ J
2

Why is the gravitational force considered a conservative force?

The work done by gravity is independent of the path taken and depends only on the initial and final positions. This path independence defines a conservative force, allowing a potential energy function to be defined.

It varies with distance
Work done is path-independent
It is always attractive
It follows Newton’s third law
2

A body is launched from Earth at \( 14 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (14)^2 - (11.2)^2 = 196 - 125.44 = 70.56 \).

\( v_f = \sqrt{70.56} \approx 8.4 \, \text{km/s} \).

8.0 km/s
8.2 km/s
8.4 km/s
8.6 km/s
3

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

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

\( v_e = \sqrt{2.09 \times 10^7} \approx 4.57 \times 10^3 \, \text{m/s} \approx 4.6 \, \text{km/s} \).

4.5 km/s
4.6 km/s
4.7 km/s
4.8 km/s
2

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