Q. 41

# An artificial sat

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 :

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