Gravitation Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the gravitational potential due to Earth at \( 2.56 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( U = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{2.56 \times 10^7} \).

\( U = -1.5625 \times 10^7 \, \text{J/kg} \).

-1.5 × 10⁷ J/kg
-1.6 × 10⁷ J/kg
-1.7 × 10⁷ J/kg
-1.8 × 10⁷ J/kg
2

Two masses \( 7 \, \text{kg} \) and \( 14 \, \text{kg} \) are \( 14 \, \text{m} \) apart. What is the gravitational potential at a point \( 6 \, \text{m} \) from the \( 7 \, \text{kg} \) mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to \( 14 \, \text{kg} \): \( 14 - 6 = 8 \, \text{m} \).

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

\( U = -6.67 \times 10^{-11} \left(\frac{7}{6} + \frac{14}{8}\right) \).

\( U = -6.67 \times 10^{-11} (1.167 + 1.75) \approx -1.95 \times 10^{-10} \, \text{J/kg} \).

-1.9 × 10⁻¹⁰ J/kg
-2.0 × 10⁻¹⁰ J/kg
-2.1 × 10⁻¹⁰ J/kg
-2.2 × 10⁻¹⁰ J/kg
1

A body weighs \( 98 \, \text{N} \) on Earth’s surface. What is its weight at a height \( h = 2 R_E/3 \)? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( h = 2 R_E/3 \), \( 1 + h/R_E = 1 + 2/3 = 5/3 \).

\( g(h) = \frac{9.8}{(5/3)^2} = 9.8 \times \frac{9}{25} = 3.528 \, \text{m/s}^2 \).

Mass: \( m = 98/9.8 = 10 \, \text{kg} \).

Weight: \( W = 10 \times 3.528 \approx 35.3 \, \text{N} \).

34 N
35 N
36 N
37 N
3

Four \( 3 \, \text{kg} \) masses form a square of side \( 3 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

4 sides: \( r = 3 \, \text{m} \), 2 diagonals: \( r = 3\sqrt{2} \, \text{m} \).

\( V = -4 \frac{G m^2}{3} - 2 \frac{G m^2}{3\sqrt{2}} \).

\( V = -6.67 \times 10^{-11} \times 9 \left(\frac{4}{3} + \frac{2}{3\sqrt{2}}\right) \).

\( V = -6.003 \times 10^{-10} (1.333 + 0.471) \approx -1.08 \times 10^{-9} \, \text{J} \).

-1.0 × 10⁻⁹ J
-1.1 × 10⁻⁹ J
-1.2 × 10⁻⁹ J
-1.3 × 10⁻⁹ J
2

A satellite orbits Earth at \( 13 R_E \) from the center. What is its orbital speed? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( r = 13 R_E \), \( v = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{13}} \).

\( v = \sqrt{4.826 \times 10^6} \approx 2.20 \times 10^3 \, \text{m/s} \).

2.1 × 10³ m/s
2.2 × 10³ m/s
2.3 × 10³ m/s
2.4 × 10³ m/s
2

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

otal energy is negative (option 1 correct), kinetic energy varies in elliptical orbits (option 2 correct), and gravity provides centripetal force (option 3 correct). Option 4 is incorrect as speed decreases with altitude.

Total energy is negative in circular orbit
Kinetic energy varies in elliptical orbit
Gravitational force keeps it in orbit
Speed increases with altitude
4

Why does the acceleration due to gravity decrease both above and below Earth’s surface?

Above the surface, \( g \) decreases as \( g(h) \propto 1/(R_E + h)^2 \). Below the surface, only the mass inside a radius \( r < R_E \) contributes, reducing \( g \) linearly as \( g(d) = g (1 - d/R_E) \). Thus, \( g \) is maximum at the surface and decreases in both directions.

Due to Earth’s rotation
Due to varying mass distribution
Due to distance and mass contribution effects
Due to atmospheric pressure
3

A satellite orbits a planet at \( 5 \times 10^7 \, \text{m} \) from its center with a period of 12 hours. What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( T = 12 \times 3600 = 43200 \, \text{s} \), \( T^2 = 1.866 \times 10^9 \, \text{s}^2 \).

\( r^3 = (5 \times 10^7)^3 = 1.25 \times 10^{23} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 1.25 \times 10^{23}}{6.67 \times 10^{-11} \times 1.866 \times 10^9} \).

\( M = \frac{4.93 \times 10^{24}}{1.245 \times 10^{-1}} \approx 3.96 \times 10^{25} \, \text{kg} \).

3.8 × 10²⁵ kg
3.9 × 10²⁵ kg
4.0 × 10²⁵ kg
4.1 × 10²⁵ kg
3

A satellite near Earth has a period of 83 minutes. What is its period at \( h = 8 R_E \)? (\( R_E = 6.4 \times 10^6 \, \text{m} \))

\( T^2 \propto (R_E + h)^3 \).

\( T_0^2 = k R_E^3 \), \( h = 8 R_E \), \( r = 9 R_E \).

\( T^2 = k (9 R_E)^3 = 729 k R_E^3 \).

\( T = T_0 \sqrt{729} = 83 \times 27 \approx 2241 \, \text{min} \).

2220 min
2230 min
2240 min
2250 min
4

A \( 400 \, \text{kg} \) satellite orbits Earth at \( 5 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 5 R_E = 3.2 \times 10^7 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 400}{2 \times 3.2 \times 10^7} \).

\( E = -\frac{1.601 \times 10^{17}}{6.4 \times 10^7} \approx -2.5 \times 10^9 \, \text{J} \).

-2.3 × 10⁹ J
-2.5 × 10⁹ J
-2.7 × 10⁹ J
-2.9 × 10⁹ J
2

A projectile is launched from Earth’s surface with speed \( 8 \, \text{km/s} \). How far from the center does it reach? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

Energy conservation: \( \frac{1}{2} m v_i^2 - \frac{G M_E m}{R_E} = -\frac{G M_E m}{r} \).

\( v_e^2 = \frac{2 G M_E}{R_E} \), so \( \frac{G M_E}{R_E} = \frac{v_e^2}{2} \).

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

\( \frac{(8)^2}{2} - \frac{(11.2)^2}{2} = -\frac{(11.2)^2}{2} \frac{R_E}{r} \).

\( 32 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{30.72} \approx 2.04 \).

\( r = 2.04 \times 6.4 \times 10^6 \approx 1.31 \times 10^7 \, \text{m} \).

1.2 × 10⁷ m
1.3 × 10⁷ m
1.4 × 10⁷ m
1.5 × 10⁷ m
2

A \( 6 \, \text{kg} \) mass is moved from \( 6 R_E \) to \( 12 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 3.84 \times 10^7 \, \text{m} \), \( r_2 = 7.68 \times 10^7 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6 \left(\frac{1}{7.68 \times 10^7} - \frac{1}{3.84 \times 10^7}\right) \).

\( \Delta V = -2.401 \times 10^{15} (-1.302 \times 10^{-8}) \approx 3.13 \times 10^7 \, \text{J} \).

3.0 × 10⁷ J
3.1 × 10⁷ J
3.2 × 10⁷ J
3.3 × 10⁷ J
2

A projectile is launched at \( 4 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( 8 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{54.72} \approx 1.15 \).

\( r = 1.15 \times 6.4 \times 10^6 \approx 7.36 \times 10^6 \, \text{m} \).

7.2 × 10⁶ m
7.3 × 10⁶ m
7.4 × 10⁶ m
7.5 × 10⁶ m
3

A \( 4 \, \text{kg} \) mass is moved from \( 3 R_E \) to \( 6 R_E \) from Earth’s center. What is the change in potential energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 1.92 \times 10^7 \, \text{m} \), \( r_2 = 3.84 \times 10^7 \, \text{m} \).

\( \Delta V = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 4 \left(\frac{1}{3.84 \times 10^7} - \frac{1}{1.92 \times 10^7}\right) \).

\( \Delta V = -1.601 \times 10^{15} (-2.604 \times 10^{-8}) \approx 4.17 \times 10^7 \, \text{J} \).

4.0 × 10⁷ J
4.1 × 10⁷ J
4.2 × 10⁷ J
4.3 × 10⁷ J
3

What happens to the acceleration due to gravity at Earth’s center?

At Earth’s center (\( r = 0 \)), the gravitational force from all surrounding mass cancels out due to symmetry, reducing \( g \) to zero, as \( g(d) = g (1 - d/R_E) \) becomes zero when \( d = R_E \).

It becomes infinite
It remains constant
It reduces to zero
It doubles
3

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