Gravitation Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A satellite orbits a planet at \( 2.5 \times 10^7 \, \text{m} \) from its center with a period of 5 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( T = 5 \times 3600 = 18000 \, \text{s} \), \( T^2 = 3.24 \times 10^8 \, \text{s}^2 \).

\( r^3 = (2.5 \times 10^7)^3 = 1.5625 \times 10^{22} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 1.5625 \times 10^{22}}{6.67 \times 10^{-11} \times 3.24 \times 10^8} \).

\( M = \frac{6.158 \times 10^{23}}{2.161 \times 10^{-2}} \approx 2.85 \times 10^{25} \, \text{kg} \).

2.7 × 10²⁵ kg
2.8 × 10²⁵ kg
2.9 × 10²⁵ kg
3.0 × 10²⁵ kg
3

Three masses of \( 18 \, \text{kg} \) each form an equilateral triangle with side \( 12 \, \text{m} \). 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: \( F = G \frac{m^2}{r^2} = 6.67 \times 10^{-11} \frac{18 \times 18}{12^2} = 1.50 \times 10^{-10} \, \text{N} \).

Two forces at 60°: \( F_R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^\circ} \).

\( F_R = \sqrt{(1.50 \times 10^{-10})^2 (1 + 1 + 1)} = 1.50 \times 10^{-10} \sqrt{3} \).

\( F_R \approx 2.60 \times 10^{-10} \, \text{N} \).

2.5 × 10⁻¹⁰ N
2.6 × 10⁻¹⁰ N
2.7 × 10⁻¹⁰ N
2.8 × 10⁻¹⁰ N
2

What is the kinetic energy of a \( 800 \, \text{kg} \) satellite at \( 10 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 800}{2 \times 6.4 \times 10^7} \).

\( K = \frac{3.202 \times 10^{17}}{1.28 \times 10^8} \approx 2.50 \times 10^9 \, \text{J} \).

2.4 × 10⁹ J
2.5 × 10⁹ J
2.6 × 10⁹ J
2.7 × 10⁹ J
2

A satellite near Earth has a period of 90 minutes. What is its period at \( h = 3 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( h = 3 R_E \), \( r = 4 R_E \).

\( T^2 = k (4 R_E)^3 = 64 k R_E^3 \).

\( T = T_0 \sqrt{64} = 90 \times 8 = 720 \, \text{min} \).

700 min
710 min
720 min
730 min
3

A satellite orbits Earth at \( 17 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 17 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{17}} \).

\( v = \sqrt{3.694 \times 10^6} \approx 1.92 \times 10^3 \, \text{m/s} \).

1.9 × 10³ m/s
2.0 × 10³ m/s
2.1 × 10³ m/s
2.2 × 10³ m/s
1

A planet orbits the Sun with a period of 6 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

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

\( T_E = 1 \, \text{year} \), \( T_p = 6 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{6^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 36 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 36 \times 3.375 \times 10^{33} = 1.215 \times 10^{35} \).

\( a_p = (1.215 \times 10^{35})^{1/3} \approx 4.95 \times 10^{11} \, \text{m} \).

4.8 × 10¹¹ m
4.9 × 10¹¹ m
5.0 × 10¹¹ m
5.1 × 10¹¹ m
3

A planet has an orbital period of 4 years around the Sun. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is the planet’s semi-major axis? (\( 1 \, \text{year} = 3.156 \times 10^7 \, \text{s} \))

Kepler’s third law: \( T^2 \propto a^3 \), \( \frac{T_p^2}{T_E^2} = \frac{a_p^3}{a_E^3} \).

\( T_E = 1 \, \text{year} \), \( T_p = 4 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{4^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 16 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 16 \times 3.375 \times 10^{33} = 5.4 \times 10^{34} \).

\( a_p = (5.4 \times 10^{34})^{1/3} \approx 3.78 \times 10^{11} \, \text{m} \).

3.5 × 10¹¹ m
3.8 × 10¹¹ m
4.0 × 10¹¹ m
4.2 × 10¹¹ m
2

Which of Kepler’s laws implies that the gravitational force acting on a planet is a central force?

Kepler’s second law (law of areas) states that the line joining a planet to the Sun sweeps out equal areas in equal times. This is a consequence of the conservation of angular momentum, which holds true for a central force directed along the line joining the two bodies (e.g., the Sun and planet). A central force ensures no torque, preserving angular momentum.

Law of orbits
Law of areas
Law of periods
None of these
2

What role does the Sun play in planetary motion according to Kepler’s laws?

The Sun is at one focus of the elliptical orbits (first law) and provides the central gravitational force driving the equal-area sweep (second law) and period-distance relation (third law).

It rotates the planets
It provides the central force
It balances their mass
It determines their density
2

A body weighs \( 245 \, \text{N} \) on Earth’s surface. What is its weight at a depth \( d = R_E/5 \)? (\( g = 9.8 \, \text{m/s}^2 \))

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

\( d = R_E/5 \), \( g(d) = 9.8 (1 - 1/5) = 9.8 \times 4/5 = 7.84 \, \text{m/s}^2 \).

Mass: \( m = 245/9.8 = 25 \, \text{kg} \).

Weight: \( W = 25 \times 7.84 = 196 \, \text{N} \).

190 N
194 N
196 N
200 N
3

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