Gravitation Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does an object need a minimum speed to escape Earth’s gravity?

Escape speed (\( v_e = \sqrt{\frac{2 G M}{R}} \)) is the minimum speed where kinetic energy equals the gravitational potential energy’s magnitude, ensuring total energy is zero, allowing escape to infinity.

To overcome air resistance
To balance kinetic and potential energy
To match Earth’s rotational speed
To exceed gravitational constant
2

A body is launched from Earth at \( 11.8 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (11.8)^2 - (11.2)^2 = 139.24 - 125.44 = 13.8 \).

\( v_f = \sqrt{13.8} \approx 3.71 \, \text{km/s} \).

3.6 km/s
3.7 km/s
3.8 km/s
3.9 km/s
2

What is the gravitational force on a \( 2 \, \text{kg} \) mass inside a hollow spherical shell of mass \( 1000 \, \text{kg} \) and radius \( 1 \, \text{m} \)?

Inside a hollow spherical shell, the gravitational force is zero due to symmetry (as per the shell theorem).

\( F = 0 \, \text{N} \).

0 N
1.33 × 10⁻⁷ N
2.67 × 10⁻⁷ N
6.67 × 10⁻⁷ N
1

A \( 800 \, \text{kg} \) satellite orbits Earth at \( 12 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 12 R_E = 7.68 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 800}{2 \times 7.68 \times 10^7} \).

\( E = -\frac{3.202 \times 10^{17}}{1.536 \times 10^8} \approx -2.08 \times 10^9 \, \text{J} \).

-2.0 × 10⁹ J
-2.1 × 10⁹ J
-2.2 × 10⁹ J
-2.3 × 10⁹ J
2

How much energy is required to move a \( 500 \, \text{kg} \) satellite from \( 9 R_E \) to \( 18 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

\( \Delta E = -G M_E m \left(\frac{1}{r_2} - \frac{1}{r_1}\right) \).

\( r_1 = 5.76 \times 10^7 \, \text{m} \), \( r_2 = 1.152 \times 10^8 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 500 \left(\frac{1}{1.152 \times 10^8} - \frac{1}{5.76 \times 10^7}\right) \).

\( \Delta E = -2.001 \times 10^{17} (-8.681 \times 10^{-9}) \approx 1.74 \times 10^9 \, \text{J} \).

1.7 × 10⁹ J
1.8 × 10⁹ J
1.9 × 10⁹ J
2.0 × 10⁹ J
1

A body is launched from Earth at \( 13.5 \, \text{km/s} \). What is its speed at infinity? (Escape speed = \( 11.2 \, \text{km/s} \))

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

\( v_f^2 = (13.5)^2 - (11.2)^2 = 182.25 - 125.44 = 56.81 \).

\( v_f = \sqrt{56.81} \approx 7.54 \, \text{km/s} \).

7.4 km/s
7.5 km/s
7.6 km/s
7.7 km/s
2

Why does the gravitational force provide the necessary centripetal force for a circular orbit?

In a circular orbit, the gravitational force (\( \frac{G M_E m}{r^2} \)) equals the centripetal force (\( \frac{m v^2}{r} \)), providing the inward force needed to maintain circular motion.

It is a frictional force
It balances the required inward force
It acts outward
It depends on satellite mass only
2

Why is the gravitational force between two point masses always attractive?

Newton’s universal law of gravitation (\( F = G \frac{m_1 m_2}{r^2} \)) is always positive since masses are positive, and the force vector is directed toward the other mass, making it attractive.

Masses can be negative
Masses are always positive
Distance is squared
It depends on their separation
2

Which of the following statements is correct about a satellite in elliptical orbit?

In an elliptical orbit, speed varies (faster at perihelion, slower at aphelion) due to conservation of angular momentum, making option 3 correct.

Its speed is constant
Its total energy is zero
Its speed varies with distance
Its potential energy is constant
3

What determines the escape speed from a planet?

Planet’s mass and radius
Object’s mass and speed
Altitude and direction
Atmospheric density
1

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