Gravitation Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A moon orbits a planet with period 8 days and radius \( 3 \times 10^8 \, \text{m} \). What is the planet’s mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2, 1 \, \text{day} = 86400 \, \text{s} \))

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

\( T = 8 \times 86400 = 6.912 \times 10^5 \, \text{s} \).

\( T^2 = 4.776 \times 10^{11} \, \text{s}^2 \).

\( r^3 = (3 \times 10^8)^3 = 2.7 \times 10^{25} \, \text{m}^3 \).

\( M = \frac{4 \times (3.14)^2 \times 2.7 \times 10^{25}}{6.67 \times 10^{-11} \times 4.776 \times 10^{11}} \).

\( M = \frac{1.065 \times 10^{26}}{3.185 \times 10^1} \approx 3.34 \times 10^{24} \, \text{kg} \).

3.2 × 10²⁴ kg
3.3 × 10²⁴ kg
3.4 × 10²⁴ kg
3.5 × 10²⁴ kg
3

What is the gravitational potential due to Earth at \( 5.12 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( U = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{5.12 \times 10^7} \).

\( U = -7.8125 \times 10^6 \, \text{J/kg} \).

-7.8 × 10⁶ J/kg
-7.9 × 10⁶ J/kg
-8.0 × 10⁶ J/kg
-8.1 × 10⁶ J/kg
1

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

\( V = -\frac{G M m}{r} \) is defined with zero at infinity, and work is needed to separate masses, making option 2 correct.

It is zero at Earth’s surface
It is zero at infinite separation
It increases as objects get closer
It is independent of distance
2

What is the kinetic energy of a \( 200 \, \text{kg} \) satellite at \( 4 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 4 R_E = 2.56 \times 10^7 \, \text{m} \).

\( K = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2 \times 2.56 \times 10^7} \).

\( K = \frac{8.004 \times 10^{16}}{5.12 \times 10^7} \approx 1.56 \times 10^9 \, \text{J} \).

1.4 × 10⁹ J
1.5 × 10⁹ J
1.6 × 10⁹ J
1.7 × 10⁹ J
3

How much energy is required to move a \( 300 \, \text{kg} \) satellite from \( 7 R_E \) to \( 14 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 4.48 \times 10^7 \, \text{m} \), \( r_2 = 8.96 \times 10^7 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 300 \left(\frac{1}{8.96 \times 10^7} - \frac{1}{4.48 \times 10^7}\right) \).

\( \Delta E = -1.201 \times 10^{17} (-1.116 \times 10^{-8}) \approx 1.34 \times 10^9 \, \text{J} \).

1.3 × 10⁹ J
1.4 × 10⁹ J
1.5 × 10⁹ J
1.6 × 10⁹ J
1

A planet orbits the Sun with a period of 12 years. If Earth’s orbital radius is \( 1.5 \times 10^{11} \, \text{m} \), what is its semi-major axis?

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

\( T_E = 1 \, \text{year} \), \( T_p = 12 \, \text{years} \), \( a_E = 1.5 \times 10^{11} \, \text{m} \).

\( \frac{12^2}{1^2} = \frac{a_p^3}{(1.5 \times 10^{11})^3} \).

\( 144 = \frac{a_p^3}{3.375 \times 10^{33}} \).

\( a_p^3 = 144 \times 3.375 \times 10^{33} = 4.86 \times 10^{35} \).

\( a_p = (4.86 \times 10^{35})^{1/3} \approx 7.86 \times 10^{11} \, \text{m} \).

7.7 × 10¹¹ m
7.8 × 10¹¹ m
7.9 × 10¹¹ m
8.0 × 10¹¹ m
3

A \( 1000 \, \text{kg} \) satellite orbits Earth at \( 16 R_E \) from the center. What is its total energy? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r = 16 R_E = 1.024 \times 10^8 \, \text{m} \).

\( E = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 1.024 \times 10^8} \).

\( E = -\frac{4.002 \times 10^{17}}{2.048 \times 10^8} \approx -1.95 \times 10^9 \, \text{J} \).

-1.9 × 10⁹ J
-2.0 × 10⁹ J
-2.1 × 10⁹ J
-2.2 × 10⁹ J
1

A projectile is launched at \( 5 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( 12.5 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{50.22} \approx 1.25 \).

\( r = 1.25 \times 6.4 \times 10^6 = 8.0 \times 10^6 \, \text{m} \).

7.8 × 10⁶ m
7.9 × 10⁶ m
8.0 × 10⁶ m
8.1 × 10⁶ m
3

Which of the following statements is correct about Kepler’s third law?

\( T^2 \propto a^3 \), meaning the period squared is proportional to the semi-major axis cubed, making option 2 correct.

Period is proportional to distance
Period squared is proportional to distance cubed
Period is constant for all orbits
Period depends on planet mass
2

What is the escape speed from a planet with mass \( 2.4 \times 10^{24} \, \text{kg} \) and radius \( 4 \times 10^6 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 2.4 \times 10^{24}}{4 \times 10^6}} \).

\( v_e = \sqrt{8.002 \times 10^7} \approx 8.95 \times 10^3 \, \text{m/s} \approx 8.95 \, \text{km/s} \).

8.8 km/s
8.9 km/s
9.0 km/s
9.1 km/s
2

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