Answer :

According to Kepler’s Third law, the square of the time period of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit of the satellite. Let T be the time period of revolution and r is the radius of the orbit, then




Since, the revolution of the satellite follows the laws of gravitation the time period of revolution depends on acceleration due to gravity g and the radius of the planet R,




Where k, is a dimensionless constant of proportionality.


Now the dimensional formula for time period [T] = [M0L0T1], for radius of planet [R] = [M0L1T0], for radius of orbit of satellite [r] = [M0L1T0] and for acceleration due to gravity [g] = [M0L1T-2]. Therefore,




Equating the exponents of like terms on both sides, we have





Substituting these values in the expression for T we obtain,




Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

An artificial satPhysics - Exemplar

The volume of a lPhysics - Exemplar

If velocity of liPhysics - Exemplar