Gravitation Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following statements is correct about the gravitational force inside a hollow spherical shell?

The net gravitational force inside a uniform hollow spherical shell is zero due to symmetry, making option 2 correct.

It increases toward the center
It is zero everywhere inside
It depends on the shell’s thickness
It acts outward from the center
2

What is the significance of the negative sign in the gravitational potential energy formula \( V = -\frac{G M m}{r} \)?

The negative sign indicates that gravitational potential energy is zero at infinite separation (\( r \to \infty \)) and decreases (becomes more negative) as objects approach each other. It reflects the attractive nature of gravity, where work must be done to separate masses.

It shows the force is repulsive
It indicates energy is zero at infinity
It implies energy increases with distance
It accounts for Earth’s rotation
2

A projectile is launched at \( 1 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

\( \frac{1}{2} v_i^2 - \frac{v_e^2}{2} = -\frac{v_e^2}{2} \frac{R_E}{r} \).

\( 0.5 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{62.22} \approx 1.008 \).

\( r = 1.008 \times 6.4 \times 10^6 \approx 6.45 \times 10^6 \, \text{m} \).

6.4 × 10⁶ m
6.5 × 10⁶ m
6.6 × 10⁶ m
6.7 × 10⁶ m
2

Why does the total energy of a satellite in an elliptical orbit remain constant?

Gravity is a conservative force, so the total mechanical energy (\( E = KE + PE \)) is conserved in an elliptical orbit, with kinetic and potential energies trading off but summing to a constant.

Speed remains constant
Gravity is conservative
Orbit is circular
Mass decreases
2

A satellite orbits Earth at \( 11 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 11 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{11}} \).

\( v = \sqrt{5.698 \times 10^6} \approx 2.39 \times 10^3 \, \text{m/s} \).

2.3 × 10³ m/s
2.4 × 10³ m/s
2.5 × 10³ m/s
2.6 × 10³ m/s
2

Four \( 7 \, \text{kg} \) masses form a square of side \( 7 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 7 \, \text{m} \), 2 diagonals: \( r = 7\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{7} - 2 \frac{G m^2}{7\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 49 \left(\frac{4}{7} + \frac{2}{7\sqrt{2}}\right) \).

\( V = -3.2683 \times 10^{-9} (0.571 + 0.202) \approx -2.53 \times 10^{-9} \, \text{J} \).

-2.4 × 10⁻⁹ J
-2.5 × 10⁻⁹ J
-2.6 × 10⁻⁹ J
-2.7 × 10⁻⁹ J
2

A projectile is launched at \( 9 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

\( \frac{1}{2} v_i^2 - \frac{v_e^2}{2} = -\frac{v_e^2}{2} \frac{R_E}{r} \).

\( 40.5 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{22.22} \approx 2.82 \).

\( r = 2.82 \times 6.4 \times 10^6 \approx 1.80 \times 10^7 \, \text{m} \).

1.7 × 10⁷ m
1.8 × 10⁷ m
1.9 × 10⁷ m
2.0 × 10⁷ m
2

A \( 250 \, \text{kg} \) satellite orbits Earth at \( 6 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 6 R_E = 3.84 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 250}{2 \times 3.84 \times 10^7} \).

\( E = -\frac{1.001 \times 10^{17}}{7.68 \times 10^7} \approx -1.30 \times 10^9 \, \text{J} \).

-1.2 × 10⁹ J
-1.3 × 10⁹ J
-1.4 × 10⁹ J
-1.5 × 10⁹ J
2

At what height above Earth’s surface is \( g \) reduced to \( 7.84 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(h) = \frac{g_0}{(1 + h/R_E)^2} \).

\( 7.84 = \frac{9.8}{(1 + h/R_E)^2} \).

\( (1 + h/R_E)^2 = 1.25 \).

\( 1 + h/R_E = \sqrt{1.25} \approx 1.118 \).

\( h/R_E = 0.118 \).

\( h = 0.118 \times 6.4 \times 10^6 \approx 7.55 \times 10^5 \, \text{m} \).

7.4 × 10⁵ m
7.5 × 10⁵ m
7.6 × 10⁵ m
7.7 × 10⁵ m
3

A satellite orbits Earth at \( 19 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 19 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{19}} \).

\( v = \sqrt{3.301 \times 10^6} \approx 1.82 \times 10^3 \, \text{m/s} \).

1.8 × 10³ m/s
1.9 × 10³ m/s
2.0 × 10³ m/s
2.1 × 10³ m/s
1

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