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ExamsJEE AdvancedPhysics

The mass density of a planet varies with distance from its center as p(r) = p0 * r / R, where R is the radius of the planet and p0 is a constant. A particle of mass m is projected from the surface of the planet with the minimum speed needed to escape the planet's gravitational field. Find the escape speed of the particle.

  1. sqrt(2 * pi * G * p0 * R² / 3)
  2. sqrt(4 * pi * G * p0 * R² / 3)
  3. sqrt(pi * G * p0 * R²)
  4. sqrt(8 * pi * G * p0 * R² / 3)

Correct answer: sqrt(8 * pi * G * p0 * R² / 3)

Solution

With density p(r) = p0 * r / R, integrate to get total mass: M = integral from 0 to R of p0 * (r/R) * 4 * pi * r² dr = 4 * pi * p0 * R³ / (4 * R) * R = pi * p0 * R³. The gravitational potential at the surface is found by integrating the field inward. The gravitational field at radius r inside is g(r) = G * M(r) / r² where M(r) = pi * p0 * r⁴ / R. So g(r) = pi * G * p0 * r² / R. The potential at surface relative to infinity: V_surface = -integral from 0 to R of g(r) dr - integral from R to infinity of G*M/r² dr = -(pi * G * p0 / R)(R³/3) - G * pi * p0 * R³ / R = -pi * G * p0 * R² / 3 - pi * G * p0 * R² = -4 * pi * G * p0 * R² / 3. Escape speed: (1/2) * v² = |V_surface|, so v = sqrt(8 * pi * G * p0 * R² / 3). Wait — re-checking: potential at surface from infinity = -G*M/R - integral₀^R g_inside dr. With M = pi*p0*R³ and g_inside(r) = G*M(r)/r² = pi*G*p0*r²/R. Potential = -G*pi*p0*R² - (pi*G*p0/R)*(R³/3) = -pi*G*p0*R²*(1 + 1/3) = -4*pi*G*p0*R²/3. Escape KE = |V|, giving v_escape = sqrt(8*pi*G*p0*R²/3). This matches option D. The correct answer is sqrt(8 * pi * G * p0 * R² / 3).

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