Gravitation — Problem Set for Class Discussion
Questions for conceptual discussion and advanced problem-solving
- Every part requires a labelled diagram or sketch where appropriate, and a step-by-step derivation.
- Express final results symbolically before numerical substitution.
- Use G for the gravitational constant. Unless stated otherwise, neglect air resistance, planetary rotation and other bodies.
- Attempt each problem independently, then discuss the reasoning and alternative approaches in class. Difficulty: ★ standard; ★★ moderately hard; ★★★ hard.
Problem 01 — Radial fall into the Sun
Advanced · ★★A probe of mass m is released from rest at Earth's orbital radius r₀ (1 AU). Earth's orbit is circular with period T₀ = 365.25 days. Neglect everything except the Sun's gravity (mass M).
- (a) Using energy conservation, find the speed v(r) of the probe when it is at distance r from the Sun.
- (b) Treat the straight-line fall as half of a very thin ellipse of semi-major axis r₀/2 and use Kepler's third law to show that the time to reach the Sun is T₀/(4√2). Find this time in days.
- (c) Find the time taken to cover the first half of the distance (from r₀ to r₀/2) as a fraction of the total fall time. (Hint: substitute r = r₀cos²θ.)
Reveal final answers — Problem 1
- v(r) = √[2GM(1/r − 1/r₀)]
- t = T₀/(4√2) ≈ 64.6 days
- tfirst half/ttotal = 1/2 + 1/π ≈ 0.8183 (81.8%)
Problem 02 — Ring and rod
Advanced · ★★A uniform ring of mass M and radius R lies in a plane perpendicular to the x-axis with its centre at the origin. A thin uniform rod of mass m and length L lies along the x-axis from x = d to x = d + L.
- (a) Show that the gravitational field of the ring at a point on the axis at distance x from the centre is GMx/(x² + R²)^(3/2), directed towards the centre.
- (b) Find the force of attraction between the ring and the rod.
- (c) Find the force when the rod starts at the ring's centre (d = 0) and is very long (L → ∞), with mass per unit length λ.
Reveal final answers — Problem 2
- gₓ = −GMx/(x² + R²)^(3/2)
- |F| = (GMm/L)[1/√(d² + R²) − 1/√((d + L)² + R²)] (towards ring)
- |F| = GMλ/R
Problem 03 — Sphere with a cavity
Advanced · ★★A uniform solid sphere of radius R has a spherical cavity of radius R/2 whose centre is at distance R/2 from the centre O of the sphere (so the cavity touches the surface and passes through O). Let M be the mass the sphere would have without the cavity.
- (a) Show that the gravitational field everywhere inside the cavity is uniform. Find its magnitude and direction.
- (b) Find the field at the surface point diametrically opposite to the cavity's outermost point (the point on the far side of O, on the line joining the centres).
Reveal final answers — Problem 3
- |g| = GM/(2R²), directed from cavity centre towards O (uniform)
- |g| = 17GM/(18R²), directed towards O
Problem 04 — Planet with non-uniform density
Hard · ★★★A planet of radius R and total mass M has density ρ(r) = ρ₀(1 − r²/R²) for 0 ≤ r ≤ R.
- (a) Express ρ₀ in terms of M and R. Find the mass M(r) enclosed within radius r.
- (b) Find the total gravitational self-energy of the planet in the form −γGM²/R, and find γ. Compare it with the uniform-density value 3/5 and explain the difference.
- (c) Find the radius at which the gravitational field g(r) inside the planet is maximum, and the value of this maximum field in terms of GM/R².
Reveal final answers — Problem 4
- ρ₀ = 15M/(8πR³); M(r) = (M/2)[5(r/R)³ − 3(r/R)⁵]
- U = −5GM²/(7R); γ = 5/7 (greater than 3/5)
- rmax = (√5/3)R; gmax = (5√5/9)GM/R²
Problem 05 — Wire hanging in a borehole
Advanced · ★★A narrow radial shaft is drilled from the surface of a non-rotating uniform planet (mass M, radius R). A uniform flexible wire of mass m, length L (L ≤ R), cross-section A and Young's modulus Y hangs in the shaft from its top end at the surface. Inside the planet, g(r) = g₀r/R with g₀ = GM/R².
- (a) Find the tension T(r) in the wire at distance r from the centre.
- (b) Find the maximum tension. Evaluate it for L = R.
- (c) Find the total elongation of the wire.
Reveal final answers — Problem 5
- T(r) = mg₀[r² − (R − L)²]/(2LR), for R − L ≤ r ≤ R
- Tmax = mg₀(2R − L)/(2R); for L = R: mg₀/2
- ΔL = mg₀L(3R − 2L)/(6RAY); for L = R: mg₀R/(6AY)
Problem 06 — Hohmann transfer
Advanced · ★★A satellite is in a circular orbit of radius r with speed v₀. Its engine fires briefly to place it on an elliptical orbit with perigee r and apogee 4r. At apogee the engine fires again to make the orbit circular.
- (a) Find the speed at perigee just after the first burn and the increase in speed Δv₁ (in terms of v₀).
- (b) Find the speed at apogee before the second burn and the total Δv for both burns.
- (c) Find the time taken to go from perigee to apogee in terms of the period T₀ of the initial orbit.
Reveal final answers — Problem 6
- vp = √(8/5)v₀; Δv₁ = (√(8/5) − 1)v₀ ≈ 0.265v₀
- va = [√(8/5)/4]v₀ ≈ 0.316v₀; Δvtotal = [3√(8/5)/4 − 1/2]v₀ ≈ 0.449v₀
- t = (T₀/2)(5/2)^(3/2) ≈ 1.98T₀
Problem 07 — Unequal binary
Advanced · ★★Two stars of masses M and 2M, separated by a fixed distance D, move in circular orbits about their common centre of mass under mutual gravity.
- (a) Find the angular velocity and the orbital speed of each star.
- (b) Find the total kinetic energy, the total potential energy and the total mechanical energy of the system.
Reveal final answers — Problem 7
- ω = √(3GM/D³); vM = 2√(GM/3D); v2M = √(GM/3D)
- K = GM²/D; U = −2GM²/D; E = −GM²/D
Problem 08 — Braking in orbit
Advanced · ★★A satellite is in a circular orbit of radius r₀ and speed v₀ about a planet of radius Rₚ. A sudden burn reduces its speed to f v₀ (0 < f < 1) with the direction unchanged (still tangential).
- (a) Show that the point of burn becomes the apogee, and find the perigee distance and perigee speed for f = 0.8.
- (b) Find the period of the new orbit for f = 0.8 in terms of the original period T₀.
- (c) If Rₚ = r₀/2, find the largest value of f for which the satellite will hit the planet.
Reveal final answers — Problem 8
- Burn point = apogee; rp = (8/17)r₀; vp = 1.7v₀ (for f = 0.8)
- T = T₀(25/34)^(3/2) ≈ 0.63T₀
- fmax = √(2/3) ≈ 0.8165
Problem 09 — Chord tunnel through a uniform planet
Advanced · ★★A smooth straight tunnel is drilled through a uniform planet (mass M, radius R) along a chord whose closest distance to the centre is d. A particle of mass m is released from rest at one end of the tunnel.
- (a) Show that the component of gravitational force along the tunnel is −(GMm/R³)s, where s is the position measured along the tunnel from its midpoint. Hence show that the motion is simple harmonic and find its period. Does it depend on d?
- (b) Find the maximum speed of the particle and the normal force exerted by the tunnel wall on it. Show that the normal force is constant.
- (c) Find the time taken to travel from one end of the tunnel to the other. Evaluate it for Earth (R = 6400 km, g = 9.8 m/s²), in minutes.
Reveal final answers — Problem 9
- Fs = −(GMm/R³)s; T = 2π√(R³/GM), independent of d
- vmax = √[GM(R² − d²)/R³]; N = GMmd/R³ (constant)
- t = π√(R³/GM) = π√(R/g) ≈ 42.3 min
Advanced practice problems for IIT-JEE 2027 groups. Problem set dated 12 October 2026. For study purposes only. Please do not redistribute commercially. physicsbyrajeevagarwal.com