Question
Physics Question on Keplers Laws
The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The satellite in an orbit at a distance of three times earth radii from 2 jts surface will be
A
83 minutes
B
83×8 minutes
C
664 minutes
D
249 minutes.
Answer
664 minutes
Explanation
Solution
: T2=T1(r1r2)3/2=83(RR+3R)3/2 =83×8=664 minutes.