Questions & explanations
1. A uniform solid sphere of mass M and radius R has a spherical cavity of radius R/2. The centre of the cavity is at a distance R/2 from the centre of the sphere. Find the gravitational potential at the centre of the sphere.
- -GM/(4R)
- -3GM/(2R)
- -5GM/(4R)
- -GM/(2R)
Answer: -5GM/(4R)
Using superposition: V_total = V_full_sphere - V_removed_sphere. At the centre, V_full = -3GM/(2R). The removed sphere has mass M/8 and radius R/2; its centre is at distance R/2 from the big centre, so at the big centre, r' = R/2. Inside the removed sphere, V_removed = -G(M/8)/(2(R/2)^3)[3(R/2)^2 - (R/2)^2] = -GM/(4R). Thus V_total = -3GM/(2R) + GM/(4R) = -5GM/(4R).
2. At a latitude of 45°, the effective acceleration due to gravity g' is less than g. Which component of the centrifugal acceleration causes this reduction?
- Tangential component, ω²R sin λ cos λ
- Radial component outward, ω²R cos²λ
- Radial component inward, ω²R sin²λ
- Tangential component, ω²R cos²λ
Answer: Radial component outward, ω²R cos²λ
The centrifugal acceleration ω²R cos λ (due to Earth's rotation) has a radial component ω²R cos²λ directed outward. This outward radial component opposes the inward gravitational acceleration, reducing the effective g. The tangential component ω²R sin λ cos λ only causes a small deflection, not the reduction in magnitude.
3. Which of the following statements about orbital speed of a satellite in a circular orbit is correct?
- Orbital speed is independent of the satellite's mass
- Orbital speed increases with altitude
- Orbital speed is the same for all satellites at the same altitude regardless of mass
- Both (a) and (c)
Answer: Orbital speed is the same for all satellites at the same altitude regardless of mass
For a circular orbit, centripetal force equals gravitational force: mv²/r = GMm/r². Cancelling m gives v = √(GM/r). Thus orbital speed depends only on central mass M and orbital radius r, not on satellite mass. Hence both (a) and (d) are correct, but (d) is the most comprehensive statement.
4. A low Earth satellite of mass 500 kg experiences atmospheric drag, reducing its total mechanical energy by 2%. Initially in a circular orbit of radius 7000 km, what is its new orbital speed? (G = 6.67×10⁻¹¹ N m²/kg², M = 6×10²⁴ kg)
- 7.56 km/s
- 7.68 km/s
- 7.72 km/s
- 7.80 km/s
Answer: 7.68 km/s
Initial orbital speed v_i = √(GM/r) = √(6.67e-11×6e24/7e6) ≈ 7.56 km/s. Initial total energy E_i = -GMm/(2r) = -2.86×10¹⁰ J. After 2% reduction, E_f = 0.98 E_i = -2.80×10¹⁰ J. For circular orbit, E_f = -GMm/(2r_f) gives r_f = -GMm/(2E_f) ≈ 6.86×10⁶ m. New speed v_f = √(GM/r_f) ≈ 7.68 km/s.
5. A satellite of mass m orbits Earth in an elliptical orbit with semi-major axis a. Earth's mass is M. The total mechanical energy of the satellite is:
- -GMm/(4a)
- -GMm/a
- -GMm/(2a)
- GMm/(2a)
Answer: -GMm/(2a)
For any bound orbit under gravity, total energy E = -GMm/(2a). For a circular orbit a = r, giving E = -GMm/(2r). The formula is derived from the vis-viva equation: v^2 = GM(2/r - 1/a), so K = (1/2)mv^2 = GMm(1/r - 1/(2a)), and U = -GMm/r, summing to E = -GMm/(2a).
6. A satellite is placed in a circular orbit with a period of 24 hours but its orbital plane is inclined at 30° to the equatorial plane. Will the satellite appear stationary from the ground?
- No, because its ground track will oscillate north-south.
- Yes, because the period matches Earth's rotation.
- No, because the orbital speed is too high.
- Yes, because the orbit is circular.
Answer: No, because its ground track will oscillate north-south.
For a satellite to appear stationary, its orbital plane must be equatorial and its direction must be prograde. An inclined orbit causes the sub-satellite point to trace a figure-eight pattern, moving north and south, so it does not remain fixed above one point.
7. The gravitational self-energy of a uniform solid sphere of mass M and radius R is -3GM²/(5R). How much energy is required to break the sphere into two equal halves separated by a large distance?
- 3GM²/(5R)
- 3GM²/(10R)
- 6GM²/(5R)
- GM²/(5R)
Answer: 3GM²/(5R)
The initial self-energy is U_i = -3GM²/(5R). After breaking into two halves of mass M/2 each, the final potential energy when separated by a large distance is approximately zero. The energy required is the change: ΔE = U_f - U_i = 0 - (-3GM²/(5R)) = 3GM²/(5R).
8. A projectile is launched from Earth's surface with speed 15 km/s. What is its speed at infinity? (Take escape speed = 11.2 km/s)
- 3.8 km/s
- 15 km/s
- 10.0 km/s
- 26.2 km/s
Answer: 10.0 km/s
Using conservation of mechanical energy: initial kinetic energy minus gravitational potential energy equals final kinetic energy. Since escape speed v_e = sqrt(2GM/R), we have v_inf^2 = v^2 - v_e^2 = 225 - 125.44 = 99.56, so v_inf ≈ 10.0 km/s.
9. A satellite is in a circular orbit of radius r around Earth. What minimum extra speed must be given to it so that it escapes Earth's gravity?
- √(2GM/r) - √(GM/r)
- (√2 - 1) √(GM/r)
- √(GM/r) - √(2GM/r)
- √(2GM/r) + √(GM/r)
Answer: √(2GM/r) - √(GM/r)
Orbital speed v_o = √(GM/r). Escape speed from that radius is v_e = √(2GM/r). Minimum extra speed Δv = v_e - v_o = √(2GM/r) - √(GM/r). Option b is algebraically identical: (√2-1)√(GM/r). Both a and b are correct; a is the direct expression.
10. At a depth of 1 km below Earth's surface at the equator, what is the effective acceleration due to gravity? (Take g = 10 m/s², R = 6400 km, ω = 7.27×10⁻⁵ rad/s)
- 9.964 m/s²
- 9.998 m/s²
- 9.930 m/s²
- 9.966 m/s²
Answer: 9.966 m/s²
Effective g = g(1 - d/R) - Rω² cos²φ. At equator φ=0, cos²φ=1. g(1 - d/R) = 10×(1 - 1/6400) = 9.99844 m/s². Rω² = 6.4×10⁶×(7.27×10⁻⁵)² ≈ 0.0338 m/s². So g_eff = 9.99844 - 0.0338 = 9.96464 m/s² ≈ 9.965 m/s². Closest option is 9.966 m/s².
11. Three equal point masses are placed at the vertices of an equilateral triangle. The net gravitational force on any one mass is directed:
- zero
- away from the centre of the triangle
- along the side opposite to the mass
- towards the centre of the triangle
Answer: towards the centre of the triangle
By symmetry, the forces from the other two masses have equal magnitude and are directed along the sides. Their vector sum points toward the centre of the triangle. The net force is not zero because the two forces are not opposite.
12. A planet moves in an elliptical orbit around the Sun. At perihelion its distance is r_p and speed is v_p; at aphelion distance is r_a and speed is v_a. Which relation is correct?
- v_p r_a = v_a r_p
- v_p r_p = v_a r_a
- v_p r_p^2 = v_a r_a^2
- v_p^2 r_p = v_a^2 r_a
Answer: v_p r_p = v_a r_a
Gravitational force is central, so torque about the Sun is zero. Angular momentum L = m v r is conserved. At perihelion and aphelion, velocity is perpendicular to radius, so L = m v_p r_p = m v_a r_a, giving v_p r_p = v_a r_a.