CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A test particle is moving in circular orbit in the gravitational field produced by a mass density ρ(r)=Kr2. Identify the correct relation between the radius R of the particle's orbit and its period T-

A
TR2 is a constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
TR is a constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
TR is a constant
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
T2R3 is a constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C TR is a constant
Gravitational force experienced by the test particle is given by,

F=GMmr2=Gρ(dV)mr2

=mGR0kr24πr2drr2
=4πkGm[1r]R0
=4πkGmR
Using newton's second law, we have

mv20R=4πkGmR

v0=C(const.)

Time period, T=2πRv0=2πRC

TR=const.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

flag
Suggest Corrections
thumbs-up
36
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Kepler's Law
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon