JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Gravitation

Consider a spherical gaseous cloud of mass density

ρ (r)

in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If

ρ (r)

is constant in time, the particle number density

n (r) = ρ (r) / m

is :
[G is universal gravitational constant]

(A) $\frac{K}{π r ^{2} m ^{2} G}$
(B) $\frac{K}{6 π r ^{2} m ^{2} G}$
(C) $\frac{3 K}{π r ^{2} m ^{2} G}$
(D) $\frac{K}{2 π r ^{2} m ^{2} G}$ ✓ Correct

Answer: (D)

Solution

For a particle of mass m at distance r from center, the gravitational force provides centripetal force. Let M be the total mass within radius r.

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

Since kinetic energy

K = \frac{1}{2} m v^{2}

\frac{GM m}{r} = 2 K

M = \frac{2 K r}{G m}

Differentiating:

d M = \frac{2 K d r}{G m}

But

d M = 4 π r^{2} ρ d r

, so:

4 π r^{2} ρ d r = \frac{2 K d r}{G m}

ρ = \frac{K}{2 π r ^{2} G m}

Therefore:

n = \frac{ρ}{m} = \frac{K}{2 π r ^{2} m ^{2} G}

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Gravitation

Solution

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