Gravitation Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the gravitational potential due to Earth at \( 1.92 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( U = -2.083 \times 10^7 \, \text{J/kg} \).

-2.0 × 10⁷ J/kg
-2.1 × 10⁷ J/kg
-2.2 × 10⁷ J/kg
-2.3 × 10⁷ J/kg
2

Four \( 4 \, \text{kg} \) masses form a square of side \( 5 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 5 \, \text{m} \), 2 diagonals: \( r = 5\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{5} - 2 \frac{G m^2}{5\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 16 \left(\frac{4}{5} + \frac{2}{5\sqrt{2}}\right) \).

\( V = -1.067 \times 10^{-9} (0.8 + 0.283) \approx -1.15 \times 10^{-9} \, \text{J} \).

-1.1 × 10⁻⁹ J
-1.2 × 10⁻⁹ J
-1.3 × 10⁻⁹ J
-1.4 × 10⁻⁹ J
2

A moon orbits a planet with a period of 9 days and radius \( 6 \times 10^8 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

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

\( T = 9 \times 86400 = 7.776 \times 10^5 \, \text{s} \).

\( T^2 = 6.046 \times 10^{11} \, \text{s}^2 \).

\( r^3 = (6 \times 10^8)^3 = 2.16 \times 10^{26} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 2.16 \times 10^{26}}{6.67 \times 10^{-11} \times 6.046 \times 10^{11}} \).

\( M = \frac{8.51 \times 10^{26}}{4.033 \times 10^1} \approx 2.11 \times 10^{25} \, \text{kg} \).

2.0 × 10²⁵ kg
2.1 × 10²⁵ kg
2.2 × 10²⁵ kg
2.3 × 10²⁵ kg
2

What does the heliocentric model, supported by Kepler’s laws, improve over the geocentric model?

The heliocentric model (Sun at the center) simplifies planetary motion descriptions compared to the geocentric model (Earth at the center), which required complex epicycles. Kepler’s laws align with heliocentric orbits.

It includes Earth’s rotation
It simplifies orbital motion
It accounts for star motion
It predicts gravitational constant
2

A planet orbits the Sun with a period of 7 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

Kepler’s third law: \( \frac{T_p^2}{T_E^2} = \frac{a_p^3}{a_E^3} \).

\( T_E = 1 \, \text{year} \), \( T_p = 7 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{7^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 49 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 49 \times 3.375 \times 10^{33} = 1.65375 \times 10^{35} \).

\( a_p = (1.65375 \times 10^{35})^{1/3} \approx 5.49 \times 10^{11} \, \text{m} \).

5.3 × 10¹¹ m
5.4 × 10¹¹ m
5.5 × 10¹¹ m
5.6 × 10¹¹ m
3

A satellite orbits Earth at \( 7 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 7 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{7}} \).

\( v = \sqrt{8.966 \times 10^6} \approx 2.99 \times 10^3 \, \text{m/s} \).

2.8 × 10³ m/s
2.9 × 10³ m/s
3.0 × 10³ m/s
3.1 × 10³ m/s
3

Four \( 5 \, \text{kg} \) masses form a square of side \( 6 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 6 \, \text{m} \), 2 diagonals: \( r = 6\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{6} - 2 \frac{G m^2}{6\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 25 \left(\frac{4}{6} + \frac{2}{6\sqrt{2}}\right) \).

\( V = -1.6675 \times 10^{-9} (0.667 + 0.236) \approx -1.51 \times 10^{-9} \, \text{J} \).

-1.4 × 10⁻⁹ J
-1.5 × 10⁻⁹ J
-1.6 × 10⁻⁹ J
-1.7 × 10⁻⁹ J
2

What is the gravitational force on a \( 12 \, \text{kg} \) mass \( 10 \, \text{m} \) from the center of a spherical shell of mass \( 600 \, \text{kg} \) and radius \( 8 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 600 \times 12}{10^2} \).

\( F = \frac{4.8036 \times 10^{-8}}{100} = 4.80 \times 10^{-10} \, \text{N} \).

4.7 × 10⁻¹⁰ N
4.8 × 10⁻¹⁰ N
4.9 × 10⁻¹⁰ N
5.0 × 10⁻¹⁰ N
2

What does Kepler’s third law imply about the relationship between orbital period and distance?

Kepler’s third law: the square of the orbital period (\( T^2 \)) is proportional to the cube of the semi-major axis (\( a^3 \)), i.e., \( T^2 \propto a^3 \). This relates time and distance for planetary motion.

\( T \propto a \)
\( T^2 \propto a^3 \)
\( T^2 \propto a^2 \)
\( T \propto a^2 \)
2

A projectile launched at \( 10 \, \text{km/s} \) from Earth reaches a maximum distance of? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( \frac{(10)^2}{2} - \frac{(11.2)^2}{2} = -\frac{(11.2)^2}{2} \frac{R_E}{r} \).

\( 50 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{12.72} \approx 4.93 \).

\( r \approx 4.93 \times 6.4 \times 10^6 \approx 3.15 \times 10^7 \, \text{m} \).

3.0 × 10⁷ m
3.1 × 10⁷ m
3.2 × 10⁷ m
3.3 × 10⁷ m
3

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