Gravitation Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What does the constant \( k \) in Kepler’s third law (\( T^2 = k r^3 \)) depend on for satellites?

\( T^2 = \frac{4\pi^2}{G M_E} r^3 \) for satellites, where \( k = \frac{4\pi^2}{G M_E} \). Thus, \( k \) depends on the mass of the central body (Earth) and universal constants, not the satellite’s mass.

Mass of the satellite
Mass of the central body
Radius of the orbit
Speed of the satellite
2

Which of the following statements is correct about escape speed?

\( v_e = \sqrt{\frac{2 G M}{R}} \) decreases with altitude as \( R + h \) increases, making option 4 correct.

It increases with altitude
It depends on the object’s mass
It is zero at infinity
It decreases with altitude
4

A body weighs \( 392 \, \text{N} \) on Earth’s surface. What is its weight at a height \( h = R_E/6 \)? (\( g = 9.8 \, \text{m/s}^2 \))

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

\( h = R_E/6 \), \( 1 + h/R_E = 1 + 1/6 = 7/6 \).

\( g(h) = \frac{9.8}{(7/6)^2} = 9.8 \times \frac{36}{49} = 7.2 \, \text{m/s}^2 \).

Mass: \( m = 392/9.8 = 40 \, \text{kg} \).

Weight: \( W = 40 \times 7.2 = 288 \, \text{N} \).

286 N
288 N
290 N
292 N
2

How much energy is required to move a \( 400 \, \text{kg} \) satellite from \( 8 R_E \) to \( 16 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( r_1 = 5.12 \times 10^7 \, \text{m} \), \( r_2 = 1.024 \times 10^8 \, \text{m} \).

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 400 \left(\frac{1}{1.024 \times 10^8} - \frac{1}{5.12 \times 10^7}\right) \).

\( \Delta E = -1.601 \times 10^{17} (-9.766 \times 10^{-9}) \approx 1.56 \times 10^9 \, \text{J} \).

1.5 × 10⁹ J
1.6 × 10⁹ J
1.7 × 10⁹ J
1.8 × 10⁹ J
2

At what depth below Earth’s surface is \( g \) reduced to \( 6.86 \, \text{m/s}^2 \)? (\( g_0 = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

\( g(d) = g_0 (1 - d/R_E) \).

\( 6.86 = 9.8 (1 - d/R_E) \).

\( 1 - d/R_E = 0.7 \).

\( d/R_E = 0.3 \).

\( d = 0.3 \times 6.4 \times 10^6 = 1.92 \times 10^6 \, \text{m} \).

1.8 × 10⁶ m
1.9 × 10⁶ m
2.0 × 10⁶ m
2.1 × 10⁶ m
2

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

3 pairs: \( V = -3 \frac{G m^2}{r} \).

\( V = -3 \frac{6.67 \times 10^{-11} \times 5 \times 5}{4} \).

\( V = -3 \times 4.169 \times 10^{-11} = -1.25 \times 10^{-10} \, \text{J} \).

-1.2 × 10⁻¹⁰ J
-1.3 × 10⁻¹⁰ J
-1.4 × 10⁻¹⁰ J
-1.5 × 10⁻¹⁰ J
1

How much energy is required to move a \( 200 \, \text{kg} \) satellite from \( 6 R_E \) to \( 12 R_E \) from Earth’s center? (\( M_E = 6 \times 10^{24} \, \text{kg}, R_E = 6.4 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

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

\( \Delta E = -6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200 \left(\frac{1}{7.68 \times 10^7} - \frac{1}{3.84 \times 10^7}\right) \).

\( \Delta E = -8.004 \times 10^{16} (-1.302 \times 10^{-8}) \approx 1.04 \times 10^9 \, \text{J} \).

1.0 × 10⁹ J
1.1 × 10⁹ J
1.2 × 10⁹ J
1.3 × 10⁹ J
1

Three masses of \( 6 \, \text{kg} \) each form an equilateral triangle with side \( 5 \, \text{m} \). What is the net force on one mass? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Force between two masses: \( F = G \frac{m^2}{r^2} = 6.67 \times 10^{-11} \frac{6 \times 6}{5^2} = 9.61 \times 10^{-11} \, \text{N} \).

Two forces at 60°: \( F_R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^\circ} \).

\( F_R = \sqrt{(9.61 \times 10^{-11})^2 (1 + 1 + 1)} = 9.61 \times 10^{-11} \sqrt{3} \).

\( F_R \approx 1.66 \times 10^{-10} \, \text{N} \).

1.5 × 10⁻¹⁰ N
1.6 × 10⁻¹⁰ N
1.7 × 10⁻¹⁰ N
1.8 × 10⁻¹⁰ N
3

What is the gravitational potential due to Earth at \( 4.48 \times 10^7 \, \text{m} \) from its center? (\( M_E = 6 \times 10^{24} \, \text{kg}, G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( U = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{4.48 \times 10^7} \).

\( U = -8.936 \times 10^6 \, \text{J/kg} \).

-8.9 × 10⁶ J/kg
-9.0 × 10⁶ J/kg
-9.1 × 10⁶ J/kg
-9.2 × 10⁶ J/kg
1

What is the gravitational force on a \( 5 \, \text{kg} \) mass \( 4 \, \text{m} \) from the center of a spherical shell of mass \( 200 \, \text{kg} \) and radius \( 3 \, \text{m} \)? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( F = \frac{6.67 \times 10^{-11} \times 200 \times 5}{4^2} \).

\( F = \frac{6.67 \times 10^{-9}}{16} \approx 4.17 \times 10^{-10} \, \text{N} \).

4.0 × 10⁻¹⁰ N
4.1 × 10⁻¹⁰ N
4.2 × 10⁻¹⁰ N
4.3 × 10⁻¹⁰ N
3

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