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  4. Gravitation Chapter-Wise Test 2

Gravitation Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

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

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

\( r^3 = (7 \times 10^8)^3 = 3.43 \times 10^{26} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 3.43 \times 10^{26}}{6.67 \times 10^{-11} \times 7.465 \times 10^{11}} \).

\( M = \frac{1.353 \times 10^{27}}{4.979 \times 10^1} \approx 2.72 \times 10^{25} \, \text{kg} \).

2.6 × 10²⁵ kg
2.7 × 10²⁵ kg
2.8 × 10²⁵ kg
2.9 × 10²⁵ kg
2

Why is the gravitational force zero at the center of a uniform solid sphere?

At the center (\( r = 0 \)), the gravitational forces from all mass elements of a uniform sphere cancel out due to symmetry, as the net force from opposite directions balances to zero.

Mass is zero at the center
Forces cancel due to symmetry
Distance to the surface is infinite
Density decreases to zero
2

A body is launched from Earth at \( 12 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (12)^2 - (11.2)^2 = 144 - 125.44 = 18.56 \).

\( v_f = \sqrt{18.56} \approx 4.31 \, \text{km/s} \).

4.0 km/s
4.2 km/s
4.3 km/s
4.5 km/s
3

Why does the work done by gravity in moving an object depend only on the vertical displacement near Earth’s surface?

Near Earth’s surface, gravity is approximately constant (\( F = mg \)), and the force acts vertically. Thus, work (\( W = F \cdot \Delta h \)) depends only on vertical displacement (\( \Delta h \)), as horizontal motion is perpendicular to the force.

Gravity varies with height
Gravity is constant and vertical
Horizontal motion cancels out
It depends on the path taken
2

A body is launched from Earth at \( 15.5 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (15.5)^2 - (11.2)^2 = 240.25 - 125.44 = 114.81 \).

\( v_f = \sqrt{114.81} \approx 10.71 \, \text{km/s} \).

10.6 km/s
10.7 km/s
10.8 km/s
10.9 km/s
2

Which of the following statements is incorrect about gravitational force?

It’s attractive (option 1 correct), follows inverse-square (option 2 correct), and is conservative (option 4 correct). Option 3 is incorrect as it is not limited to Earth.

It is always attractive
It follows the inverse-square law
It acts only near Earth
It is a conservative force
3

What is the escape speed from a planet with mass \( 9 \times 10^{23} \, \text{kg} \) and radius \( 2.5 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 9 \times 10^{23}}{2.5 \times 10^6}} \).

\( v_e = \sqrt{4.801 \times 10^7} \approx 6.93 \times 10^3 \, \text{m/s} \approx 6.9 \, \text{km/s} \).

6.7 km/s
6.8 km/s
6.9 km/s
7.0 km/s
3

A satellite near Earth has a period of 89 minutes. What is its period at \( h = 7 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( T_0^2 = k R_E^3 \), \( h = 7 R_E \), \( r = 8 R_E \).

\( T^2 = k (8 R_E)^3 = 512 k R_E^3 \).

\( T = T_0 \sqrt{512} = 89 \times 22.63 \approx 2014 \, \text{min} \).

2000 min
2010 min
2020 min
2030 min
2

What is the kinetic energy of a \( 900 \, \text{kg} \) satellite at \( 11 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 11 R_E = 7.04 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 900}{2 \times 7.04 \times 10^7} \).

\( K = \frac{3.602 \times 10^{17}}{1.408 \times 10^8} \approx 2.56 \times 10^9 \, \text{J} \).

2.5 × 10⁹ J
2.6 × 10⁹ J
2.7 × 10⁹ J
2.8 × 10⁹ J
2

A satellite orbits a planet at \( 3 \times 10^7 \, \text{m} \) from its center with a period of 6 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( T = 6 \times 3600 = 21600 \, \text{s} \), \( T^2 = 4.6656 \times 10^8 \, \text{s}^2 \).

\( r^3 = (3 \times 10^7)^3 = 2.7 \times 10^{22} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 2.7 \times 10^{22}}{6.67 \times 10^{-11} \times 4.6656 \times 10^8} \).

\( M = \frac{1.065 \times 10^{24}}{3.112 \times 10^{-2}} \approx 3.42 \times 10^{25} \, \text{kg} \).

3.3 × 10²⁵ kg
3.4 × 10²⁵ kg
3.5 × 10²⁵ kg
3.6 × 10²⁵ kg
2

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