Gravitation Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A body weighs \( 343 \, \text{N} \) on Earth’s surface. What is its weight at a height \( h = R_E/5 \)? (\( g = 9.8 \, \text{m/s}^2 \))

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

\( h = R_E/5 \), \( 1 + h/R_E = 1 + 1/5 = 6/5 \).

\( g(h) = \frac{9.8}{(6/5)^2} = 9.8 \times \frac{25}{36} = 6.81 \, \text{m/s}^2 \).

Mass: \( m = 343/9.8 = 35 \, \text{kg} \).

Weight: \( W = 35 \times 6.81 \approx 238 \, \text{N} \).

236 N
238 N
240 N
242 N
2

A projectile is launched at \( 3 \, \text{km/s} \) from Earth’s surface. What is its maximum distance from the center? (Escape speed = \( 11.2 \, \text{km/s}, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( 4.5 - 62.72 = -62.72 \frac{R_E}{r} \).

\( \frac{r}{R_E} = \frac{62.72}{58.22} \approx 1.077 \).

\( r = 1.077 \times 6.4 \times 10^6 \approx 6.89 \times 10^6 \, \text{m} \).

6.8 × 10⁶ m
6.9 × 10⁶ m
7.0 × 10⁶ m
7.1 × 10⁶ m
2

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

\( V = -\frac{G M m}{r} \), which is negative for bound systems and decreases (becomes more negative) as \( r \) decreases, making option 3 correct.

It is positive near Earth
It increases as objects approach
It becomes more negative as objects approach
It is zero at Earth’s surface
3

Three \( 6 \, \text{kg} \) masses form an equilateral triangle of side \( 8 \, \text{m} \). What is the potential energy of the system? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

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

\( V = -3 \frac{6.67 \times 10^{-11} \times 6 \times 6}{8} \).

\( V = -3 \times 3.0015 \times 10^{-11} = -9.0 \times 10^{-11} \, \text{J} \).

-8.8 × 10⁻¹¹ J
-9.0 × 10⁻¹¹ J
-9.2 × 10⁻¹¹ J
-9.4 × 10⁻¹¹ J
2

What determines whether an object can escape Earth’s gravitational influence?

An object escapes if its total mechanical energy (\( E = KE + PE \)) is zero or positive. This occurs when its initial speed equals or exceeds the escape speed (\( v_e = \sqrt{\frac{2 G M}{R}} \)), making energy the key factor.

Its mass
Its total mechanical energy
Its direction of launch
Its height above the surface
2

Three masses of \( 10 \, \text{kg} \) each form an equilateral triangle with side \( 7 \, \text{m} \). 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: \( F = G \frac{m^2}{r^2} = 6.67 \times 10^{-11} \frac{10 \times 10}{7^2} = 1.36 \times 10^{-10} \, \text{N} \).

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

\( F_R = \sqrt{(1.36 \times 10^{-10})^2 (1 + 1 + 1)} = 1.36 \times 10^{-10} \sqrt{3} \).

\( F_R \approx 2.36 \times 10^{-10} \, \text{N} \).

2.2 × 10⁻¹⁰ N
2.3 × 10⁻¹⁰ N
2.4 × 10⁻¹⁰ N
2.5 × 10⁻¹⁰ N
3

Minimum speed to escape from \( 2 R_E \) from Earth’s center is? (\( g = 9.8 \, \text{m/s}^2, R_E = 6.4 \times 10^6 \, \text{m} \))

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

\( v_e = \sqrt{9.8 \times 6.4 \times 10^6} = \sqrt{6.272 \times 10^7} \).

\( v_e \approx 7.92 \times 10^3 \, \text{m/s} \approx 7.9 \, \text{km/s} \).

7.5 km/s
7.7 km/s
7.9 km/s
8.1 km/s
3

Why does the Moon’s orbital period relate to Earth’s mass?

Kepler’s third law (\( T^2 = \frac{4\pi^2}{G M_E} r^3 \)) to satellites like the Moon, where the period \( T \) depends on Earth’s mass \( M_E \), the central body, determining the gravitational force strength.

Moon’s mass affects it
Earth’s mass determines the force
Distance from the Sun varies
Earth’s rotation influences it
2

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

\( V = -\frac{G M m}{r} \), which is negative (option 1 correct), zero at infinity (option 2 correct), and decreases as \( r \) decreases (option 4 correct). Option 3 is incorrect as it is not positive when bound.

It is negative for bound systems
It is zero at infinite separation
It is positive when objects are close
It depends on the inverse of distance
3

Two masses \( 2 \, \text{kg} \) and \( 8 \, \text{kg} \) are separated by \( 4 \, \text{m} \). What is the gravitational potential at the midpoint? (\( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))

Distance to midpoint: \( r = 2 \, \text{m} \).

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

\( U = -6.67 \times 10^{-11} \left(\frac{2}{2} + \frac{8}{2}\right) \).

\( U = -6.67 \times 10^{-11} \times (1 + 4) = -3.335 \times 10^{-10} \, \text{J/kg} \).

-3.0 × 10⁻¹⁰ J/kg
-3.3 × 10⁻¹⁰ J/kg
-3.5 × 10⁻¹⁰ J/kg
-4.0 × 10⁻¹⁰ J/kg
2

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