JEE MAIN - Physics (2004 - No. 49)
A satellite of mass $$m$$ revolves around the earth of radius $$R$$ at a height $$x$$ from its surface. If $$g$$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
$${{g{R^2}} \over {R + x}}$$
$${{gR} \over {R - x}}$$
$${gx}$$
$${\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$$
Explanation
Gravitational force applied on the satellite,
= $${{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
For satellite, gravitational force = centripetal force
$$\therefore$$ $${{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
where $$v$$ is the orbital speed of satellite.
also $$\,\,g = {{GM} \over {{R^2}}}$$ $$ \Rightarrow GM = g{R^2}$$
$$\therefore$$ $${v^2} = {{g{R^2}} \over {R + x}} $$
$$\Rightarrow v = {\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$$
= $${{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
For satellite, gravitational force = centripetal force
$$\therefore$$ $${{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
where $$v$$ is the orbital speed of satellite.
also $$\,\,g = {{GM} \over {{R^2}}}$$ $$ \Rightarrow GM = g{R^2}$$
$$\therefore$$ $${v^2} = {{g{R^2}} \over {R + x}} $$
$$\Rightarrow v = {\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$$
Comments (0)
