The time period of an earth satellite in circular orbit is independent of

both the mass and radius of the orbit

neither the mass of the satellite nor the radius of its orbit

Gravitation PYQ — JEE Main 2004 · Shift 0 · Q167 with Solution

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Correct answer: C

For a satellite of mass $m$ orbiting Earth (mass $M$ ) in a circular orbit of radius $r$ , the gravitational force provides the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Cancel $m$ :

\frac{GM}{r^2} = \frac{v^2}{r}

So,

v^2 = \frac{GM}{r}

The time period is

T = \frac{2\pi r}{v}

Substitute $v = \sqrt{\frac{GM}{r}}$ :

T = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi \sqrt{\frac{r^3}{GM}}

From

T = 2\pi \sqrt{\frac{r^3}{GM}}

we see that $T$ depends on:

and is independent of the mass of the satellite $m$ .

A: both the mass and radius of the orbit
False, because $T$ depends on the radius $r$ .
B: radius of its orbit
False, because $T \propto r^{3/2}$ .
C: the mass of the satellite
True.
D: neither the mass of the satellite nor the radius of its orbit
False, because $T$ does depend on radius.

The time period is independent of the mass of the satellite.