JEE MAIN - Physics (2004 - No. 52)
The time period of an earth satellite in circular orbit is independent of
both the mass and radius of the orbit
radius of its orbit
the mass of the satellite
neither the mass of the satellite nor the radius of its orbit
Wyjaśnienie
For satellite, gravitational force = centripetal force
$$\therefore$$ $${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$$
$$x=$$ height of satellite from earth surface
$$m=$$ mass of satellite
$$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$$ or $$v = \sqrt {{{GM} \over {R + x}}} $$
We know, $$T = {{2\pi } \over \omega }$$
$$T = {{2\pi \left( {R + x} \right)} \over v} $$ [ as $$\omega = {v \over r}$$ ]
$$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$$
which is independent of mass of satellite
$$\therefore$$ $${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$$
$$x=$$ height of satellite from earth surface
$$m=$$ mass of satellite
$$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$$ or $$v = \sqrt {{{GM} \over {R + x}}} $$
We know, $$T = {{2\pi } \over \omega }$$
$$T = {{2\pi \left( {R + x} \right)} \over v} $$ [ as $$\omega = {v \over r}$$ ]
$$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$$
which is independent of mass of satellite