Gravitation Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What does the variation of \( g \) with altitude indicate about gravitational force?

It increases with distance
It weakens with distance
It remains constant
It depends on mass only
2

Why does the gravitational potential energy depend on the inverse of distance?

Force is constant
Force varies as inverse-square
Mass changes with distance
Energy is zero at the surface
2

What is the gravitational potential at a point \( 9.6 \times 10^6 \, \text{m} \) from Earth’s center due to Earth? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( U = -\frac{4.002 \times 10^{14}}{9.6 \times 10^6} = -4.17 \times 10^7 \, \text{J/kg} \).

-4.0 × 10⁷ J/kg
-4.2 × 10⁷ J/kg
-4.5 × 10⁷ J/kg
-4.7 × 10⁷ J/kg
2

Why does a satellite in a higher orbit have a longer orbital period?

\( T^2 \propto r^3 \) for satellites. A higher orbit (larger \( r \)) increases \( T^2 \) by \( r^3 \), so the period \( T \) increases as the radius grows, reflecting weaker gravitational pull at greater distances.

It moves faster
Gravitational force decreases
Its mass increases
Energy remains constant
2

Why does the gravitational potential decrease (become less negative) as altitude increases?

As altitude increases, \( r \) (distance from Earth’s center) increases, reducing the magnitude of the negative term \( -\frac{G M m}{r} \), making \( V \) less negative.

Force increases
Distance from the center increases
Mass decreases
Energy becomes positive
2

A moon orbits a planet with a period of 12 days and radius \( 9 \times 10^8 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

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

\( T = 12 \times 86400 = 1.0368 \times 10^6 \, \text{s} \).

\( T^2 = 1.075 \times 10^{12} \, \text{s}^2 \).

\( r^3 = (9 \times 10^8)^3 = 7.29 \times 10^{26} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 7.29 \times 10^{26}}{6.67 \times 10^{-11} \times 1.075 \times 10^{12}} \).

\( M = \frac{2.875 \times 10^{27}}{7.171 \times 10^1} \approx 4.01 \times 10^{25} \, \text{kg} \).

3.9 × 10²⁵ kg
4.0 × 10²⁵ kg
4.1 × 10²⁵ kg
4.2 × 10²⁵ kg
2

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

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

\( T = 11 \times 86400 = 9.504 \times 10^5 \, \text{s} \).

\( T^2 = 9.028 \times 10^{11} \, \text{s}^2 \).

\( r^3 = (8 \times 10^8)^3 = 5.12 \times 10^{26} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 5.12 \times 10^{26}}{6.67 \times 10^{-11} \times 9.028 \times 10^{11}} \).

\( M = \frac{2.019 \times 10^{27}}{6.022 \times 10^1} \approx 3.35 \times 10^{25} \, \text{kg} \).

3.2 × 10²⁵ kg
3.3 × 10²⁵ kg
3.4 × 10²⁵ kg
3.5 × 10²⁵ kg
3

A moon orbits a planet with a period of 6 days and radius \( 4 \times 10^8 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

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

\( T = 6 \times 86400 = 5.184 \times 10^5 \, \text{s} \).

\( T^2 = 2.689 \times 10^{11} \, \text{s}^2 \).

\( r^3 = (4 \times 10^8)^3 = 6.4 \times 10^{25} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 6.4 \times 10^{25}}{6.67 \times 10^{-11} \times 2.689 \times 10^{11}} \).

\( M = \frac{2.523 \times 10^{26}}{1.794 \times 10^1} \approx 1.41 \times 10^{25} \, \text{kg} \).

1.3 × 10²⁵ kg
1.4 × 10²⁵ kg
1.5 × 10²⁵ kg
1.6 × 10²⁵ kg
2

What is the significance of the Cavendish experiment in understanding gravity?

It discovered Kepler’s laws
It measured the gravitational constant
It proved Earth’s rotation
It determined escape speed
2

What is the total energy of a \( 100 \, \text{kg} \) satellite orbiting Earth at \( 3 R_E \) from the center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 3 R_E = 3 \times 6.4 \times 10^6 = 1.92 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 100}{2 \times 1.92 \times 10^7} \).

\( E = -\frac{4.002 \times 10^{16}}{3.84 \times 10^7} \approx -1.04 \times 10^9 \, \text{J} \).

-0.9 × 10⁹ J
-1.0 × 10⁹ J
-1.2 × 10⁹ J
-1.4 × 10⁹ J
2

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