Gravitation Chapter-Wise Test 21

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the gravitational potential due to Earth at \( 1.28 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( U = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{1.28 \times 10^7} \).

\( U = -3.125 \times 10^7 \, \text{J/kg} \).

-3.0 × 10⁷ J/kg
-3.1 × 10⁷ J/kg
-3.2 × 10⁷ J/kg
-3.3 × 10⁷ J/kg
2

Which of the following statements is incorrect about gravitational potential energy?

It’s negative (option 1 correct), zero at infinity (option 2 correct), and conservative (option 3 correct). Option 4 is incorrect as it depends on distance, not mass alone.

It is negative for bound objects
It is zero at infinite distance
It is path-independent
It depends only on mass
4

Four \( 9 \, \text{kg} \) masses form a square of side \( 9 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 9 \, \text{m} \), 2 diagonals: \( r = 9\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{9} - 2 \frac{G m^2}{9\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 81 \left(\frac{4}{9} + \frac{2}{9\sqrt{2}}\right) \).

\( V = -5.4027 \times 10^{-9} (0.444 + 0.157) \approx -3.25 \times 10^{-9} \, \text{J} \).

-3.2 × 10⁻⁹ J
-3.3 × 10⁻⁹ J
-3.4 × 10⁻⁹ J
-3.5 × 10⁻⁹ J
1

What is the escape speed from a planet of mass \( 4.8 \times 10^{24} \, \text{kg} \) and radius \( 5 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.8 \times 10^{24}}{5 \times 10^6}} \).

\( v_e = \sqrt{6.403 \times 10^7} \approx 8.0 \times 10^3 \, \text{m/s} = 8.0 \, \text{km/s} \).

7.5 km/s
8.0 km/s
8.5 km/s
9.0 km/s
2

What causes the variation in kinetic energy of a satellite in an elliptical orbit?

In an elliptical orbit, total energy is constant, but kinetic energy (\( KE = \frac{1}{2} m v^2 \)) varies as speed changes with distance from the center (faster closer, slower farther), balancing the changing potential energy.

Constant total energy
Varying distance from the center
Earth’s rotation
Mass loss of the satellite
2

What property of the gravitational force allows Newton’s law to apply to extended objects?

The gravitational force obeys the principle of superposition, meaning the total force on a body is the vector sum of forces from individual point masses. This allows the law to extend to composite objects by summing contributions.

Its conservative nature
Its inverse-square dependence
The principle of superposition
Its dependence on mass only
3

What causes the elliptical shape of planetary orbits according to Newton’s theory?

The inverse-square law (\( F \propto 1/r^2 \)) leads to elliptical orbits as a general solution to the two-body problem, matching Kepler’s first law, unlike a linear force which would produce circular orbits.

Linear gravitational force
Inverse-square gravitational force
Planetary rotation
Solar wind pressure
2

What is the primary reason the gravitational force inside a uniform spherical shell is zero?

Inside a uniform spherical shell, the gravitational forces from all mass elements cancel out due to symmetry, resulting in a net force of zero at any internal point.

The shell has no mass
Forces cancel due to symmetry
The center has no gravity
Mass is concentrated outside
2

Two masses \( 4 \, \text{kg} \) and \( 8 \, \text{kg} \) are \( 8 \, \text{m} \) apart. What is the gravitational potential at a point \( 3 \, \text{m} \) from the \( 4 \, \text{kg} \) mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to \( 8 \, \text{kg} \): \( 8 - 3 = 5 \, \text{m} \).

\( U = -\frac{G m_1}{r_1} - \frac{G m_2}{r_2} \).

\( U = -6.67 \times 10^{-11} \left(\frac{4}{3} + \frac{8}{5}\right) \).

\( U = -6.67 \times 10^{-11} (1.333 + 1.6) = -1.96 \times 10^{-10} \, \text{J/kg} \).

-1.8 × 10⁻¹⁰ J/kg
-1.9 × 10⁻¹⁰ J/kg
-2.0 × 10⁻¹⁰ J/kg
-2.1 × 10⁻¹⁰ J/kg
3

What is the gravitational force on a \( 13 \, \text{kg} \) mass \( 11 \, \text{m} \) from the center of a spherical shell of mass \( 700 \, \text{kg} \) and radius \( 9 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Outside shell: \( F = \frac{G M m}{r^2} \).

\( F = \frac{6.67 \times 10^{-11} \times 700 \times 13}{11^2} \).

\( F = \frac{6.0689 \times 10^{-8}}{121} \approx 5.02 \times 10^{-10} \, \text{N} \).

4.9 × 10⁻¹⁰ N
5.0 × 10⁻¹⁰ N
5.1 × 10⁻¹⁰ N
5.2 × 10⁻¹⁰ N
2

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