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] = [M^{0}L^{0}T^{1}], for radius of planet [R] = [M^{0}L^{1}T^{0}], for radius of orbit of satellite [r] = [M^{0}L^{1}T^{0}] and for acceleration due to gravity [g] = [M^{0}L^{1}T^{-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 :

An artificial satPhysics - Exemplar

The volume of a lPhysics - Exemplar

If velocity of liPhysics - Exemplar