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

Correct answer: C. (T)/(R) is a constant M = ∫ ρ4π r ^ 2 dr = ∫ _ 0 ^ R K r ^ 2 4π r ^ 2 dr = 4 Satellite velocity = (GM)/(R) = 4 Time period, T = (2 )/(v) = 2 4 Now, (T)/(R) = 2π 4 = constant

JEE Main 20199 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 April
Shift: Evening Shift
Subject: Physics
Chapter: Gravitation

JEE Main 2019 · 9 April · Evening Shift · Physics · Gravitation

A test particle is moving in a circular orbit in the gravitational...

A test particle is moving in a circular orbit in the gravitational field produced by a mass density

ρ\left(r\right) = \dfrac{K}{{r}^{2}} .

Identify the current relation between the radius

R

of the particle’s orbit and its period

T:

Options

A $TR$ is a constant
B $\dfrac{{T}^{2}}{{R}^{3}}$ is a constant
C $\dfrac{T}{R}$ is a constant
D $\dfrac{T}{{R}^{2}}$ is a constant

Correct answer

C. $\dfrac{T}{R}$ is a constant

Step-by-step solution

M = ∫ ρ4\pi{r}^{2}dr

= {∫}_{0}^{R}\dfrac{K}{{r}^{2}}4\pi{r}^{2}dr

= 4\piKR

Satellite velocity

= \sqrt{\dfrac{GM}{R}} = \sqrt{4\piK}

Time period,

T = \dfrac{2\piR}{v} = \dfrac{2\piR}{\sqrt{4\piK}}

Now,

\dfrac{T}{R} = \dfrac{2\pi}{\sqrt{4\piK}} =

constant

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