Q. 41

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T =

where k is a dimensionless constant and g is acceleration due to gravity.

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,




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