A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity $\frac{3 π}{G}$ is equal to

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$3 π T^{2} D$

$3 π D^{2} T$

D²T

T ²D

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Using $T = 2 π \sqrt{\frac{R^{3}}{GM}} = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4}{3} {πR}^{3} D}}$
$T^{2} = \frac{4 π^{2} R^{3}}{G \frac{4}{3} π R^{3} D} = \frac{3 π}{D G} \Rightarrow \frac{3 π}{G} = T^{2} D$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!