Mass density of a planet varies as ρ ( r ) = R ρ 0 r ; where R is radius of planet and r is distance from center of planet :-

Mass of planet: M = ∫ 0 R R ρ 0 r 4 π r 2 d r = π ρ 0 R 3 (A) 2 1 m v e 2 = R GM m ⇒ v e = R 2 GM = 2 π G ρ 0 R 2 (correct) (C) E = R 2 GM = π G ρ 0 R (correct) (D) v 0 = R GM = π G ρ 0 R 2 , so T = v 0 2 π R = 2 π π G ρ 0 R (correct) Answers: (A), (C), (D)

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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationEnergy & PotentialMedium2 minQB

PhysicsMediummultiple choice

Mass density of a planet varies as $ρ (r) = \frac{ρ _{0} r}{R}$ ; where R is radius of planet and r is distance from center of planet :-

Options:(select one or more)

Answer:

A, C, D

Solution:

Mass of planet: $M = \int_{0}^{R} \frac{ρ _{0} r}{R} 4 π r^{2} d r = π ρ_{0} R^{3}$

(A) $\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R} \Rightarrow v_{e} = \frac{2 GM}{R} = 2 π G ρ_{0} R^{2}$ (correct)

(D) $v_{0} = \frac{GM}{R} = π G ρ_{0} R^{2}$ , so $T = \frac{2 π R}{v _{0}} = 2 π \frac{R}{π G ρ _{0}}$ (correct)

Answers: (A), (C), (D)

Stream:JEESubject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: QB

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