Gravitation Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A satellite orbits a planet at \( 8 \times 10^7 \, \text{m} \) from its center with a period of 20 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( T = 20 \times 3600 = 72000 \, \text{s} \), \( T^2 = 5.184 \times 10^9 \, \text{s}^2 \).

\( r^3 = (8 \times 10^7)^3 = 5.12 \times 10^{23} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 5.12 \times 10^{23}}{6.67 \times 10^{-11} \times 5.184 \times 10^9} \).

\( M = \frac{2.019 \times 10^{25}}{3.458 \times 10^{-1}} \approx 5.84 \times 10^{25} \, \text{kg} \).

5.7 × 10²⁵ kg
5.8 × 10²⁵ kg
5.9 × 10²⁵ kg
6.0 × 10²⁵ kg
2

Which of the following statements is incorrect about escape speed?

Escape speed depends on planet parameters (option 1 correct), is independent of direction (option 3 correct), and mass (option 4 correct). Option 2 is incorrect as it decreases with altitude.

It depends on the planet’s mass
It increases with altitude
It is independent of launch direction
It is independent of object mass
2

What is the minimum speed to escape from \( 5 R_E \) from Earth’s center? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( v_e = \sqrt{\frac{2 g R_E^2}{5 R_E}} = \sqrt{\frac{2 \times 9.8 \times 6.4 \times 10^6}{5}} \).

\( v_e = \sqrt{2.509 \times 10^7} \approx 5.01 \times 10^3 \, \text{m/s} = 5.0 \, \text{km/s} \).

4.8 km/s
4.9 km/s
5.0 km/s
5.1 km/s
3

A body weighs \( 294 \, \text{N} \) on Earth’s surface. What is its weight at a height \( h = 3 R_E/4 \)? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(h) = \frac{g}{(1 + h/R_E)^2} \).

\( h = 3 R_E/4 \), \( 1 + h/R_E = 1 + 3/4 = 7/4 \).

\( g(h) = \frac{9.8}{(7/4)^2} = 9.8 \times \frac{16}{49} = 3.2 \, \text{m/s}^2 \).

Mass: \( m = 294/9.8 = 30 \, \text{kg} \).

Weight: \( W = 30 \times 3.2 = 96 \, \text{N} \).

94 N
96 N
98 N
100 N
2

At what height above Earth’s surface is \( g \) reduced to \( 5.88 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(h) = \frac{g_0}{(1 + h/R_E)^2} \).

\( 5.88 = \frac{9.8}{(1 + h/R_E)^2} \).

\( (1 + h/R_E)^2 = 1.667 \).

\( 1 + h/R_E = \sqrt{1.667} \approx 1.291 \).

\( h/R_E = 0.291 \).

\( h = 0.291 \times 6.4 \times 10^6 \approx 1.86 \times 10^6 \, \text{m} \).

1.8 × 10⁶ m
1.9 × 10⁶ m
2.0 × 10⁶ m
2.1 × 10⁶ m
2

A body weighs \( 147 \, \text{N} \) on Earth’s surface. What is its weight at a height \( h = R_E/3 \)? (\( g = 9.8 \, \text{m/s}^2 \))

\( g(h) = \frac{g}{(1 + h/R_E)^2} \).

\( h = R_E/3 \), \( 1 + h/R_E = 4/3 \).

\( g(h) = \frac{9.8}{(4/3)^2} = 9.8 \times \frac{9}{16} = 5.51 \, \text{m/s}^2 \).

Mass: \( m = 147/9.8 = 15 \, \text{kg} \).

Weight: \( W = 15 \times 5.51 \approx 82.7 \, \text{N} \).

80 N
82 N
83 N
85 N
3

A satellite orbits Earth at \( 2.5 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 2.5 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2.5}} \).

\( v = \sqrt{2.509 \times 10^7} \approx 5.01 \times 10^3 \, \text{m/s} \).

4.8 × 10³ m/s
5.0 × 10³ m/s
5.2 × 10³ m/s
5.4 × 10³ m/s
2

What is the physical significance of the gravitational constant \( G \) in Newton’s law?

\( G \) is the universal gravitational constant, determining the strength of the gravitational force between two masses. It is a fundamental constant that scales the force in \( F = G \frac{m_1 m_2}{r^2} \).

It measures Earth’s gravity
It determines the force strength
It varies with distance
It depends on the masses
2

What ensures that a planet moves faster at perihelion than at aphelion?

Angular momentum conservation (\( m r v = \text{constant} \)) implies that at perihelion (smaller \( r \)), the speed \( v \) is greater than at aphelion (larger \( r \)), as derived from Kepler’s second law.

Conservation of energy
Conservation of angular momentum
Gravitational force variation
Kepler’s third law
2

A planet orbits the Sun with a period of 5 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

Kepler’s third law: \( \frac{T_p^2}{T_E^2} = \frac{a_p^3}{a_E^3} \).

\( T_E = 1 \, \text{year} \), \( T_p = 5 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{5^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 25 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 25 \times 3.375 \times 10^{33} = 8.4375 \times 10^{34} \).

\( a_p = (8.4375 \times 10^{34})^{1/3} \approx 4.39 \times 10^{11} \, \text{m} \).

4.2 × 10¹¹ m
4.3 × 10¹¹ m
4.4 × 10¹¹ m
4.5 × 10¹¹ m
3

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