A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρr=K/(r^2). Identify the current relation between the radius R of the particle’s orbit and its period T:

Correct answer: (C) TR is a constant. M=∫ρ4πr^2dr =∫0RK/(r^2)4πr^2dr =4πKR Satellite velocity =√GMR=√(4πK) Time period, T=(2πR)/v=(2πR)/(√(4πK)) Now, T/R=(2π)/(√(4πK))= constant

Q85 / 90 · JEE Main 2019 (09 Apr Shift 2) · Physics

A test particle is moving in a circular orbit in the gravitational field produced by a mass density $ρ (r) = \frac{K}{r^{2}} .$ Identify the current relation between the radius $R$ of the particle’s orbit and its period $T :$

A
$T R$ is a constant
B
$\frac{T^{2}}{R^{3}}$ is a constant
Correct answer
C
$\frac{T}{R}$ is a constant
D
$\frac{T}{R^{2}}$ is a constant

Solution

M = \int ρ 4 π r^{2} d r

= \int_{0}^{R} \frac{K}{r^{2}} 4 π r^{2} d r

= 4 π K R

Satellite velocity

= \sqrt{\frac{G M}{R}} = \sqrt{4 π K}

Time period,

T = \frac{2 π R}{v} = \frac{2 π R}{\sqrt{4 π K}}

Now,

\frac{T}{R} = \frac{2 π}{\sqrt{4 π K}} =

constant

Hence, the correct answer is option: 'C'.

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A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρr=Kr2. Identify the current relation between the radius R of the particle’s orbit and its period T:

Solution

Relevant Questions

A test particle is moving in a circular orbit in the gravitational field produced by a mass density $ρ (r) = \frac{K}{r^{2}} .$ Identify the current relation between the radius $R$ of the particle’s orbit and its period $T :$