JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Time Period of Satellite

It is the time taken by satellite to go once around the earth.

\[\therefore \] \[T=\frac{\text{Circumference}\ \text{of}\ \text{the}\ \text{orbit}}{\text{orbital}\ \text{velocity}}\]

\[\Rightarrow \] \[T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{GM}}\]             [As \[v=\sqrt{\frac{GM}{r}}\]]

\[\Rightarrow \] \[T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \,\,\sqrt{\frac{{{r}^{3}}}{g{{R}^{2}}}}\] [As \[GM=g{{R}^{2}}\]]

\[\Rightarrow \] \[T=2\pi \,\,\sqrt{\frac{{{\left( R+h \right)}^{3}}}{g\,{{R}^{2}}}}\] \[=2\,\pi \,\,\sqrt{\frac{R}{g}}{{\left( 1+\frac{h}{R} \right)}^{3/2}}\][As \[r=R+h\]]

(i) From \[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}\], it is clear that time period is independent of the mass of orbiting body and depends on the mass of central body and radius of the orbit

(ii) \[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}\] \[\Rightarrow \] \[{{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{r}^{3}}\] i.e., \[{{T}^{2}}\propto {{r}^{3}}\]

This is in accordance with Kepler?s third law of planetary motion \[r\] becomes a (semi major axis) if the orbit is elliptic.

(iii) Time period of nearby satellite,

From  \[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{g{{R}^{2}}}}=2\pi \sqrt{\frac{R}{g}}\]                                                                                        [As \[h=0\] and \[GM=g{{R}^{2}}\]]

For earth \[R=6400km\] and \[g=9.8m/{{s}^{2}}\] \[T=84.6\,\text{minute}\ \approx 1.4\ hr\]

(iv) Time period of nearby satellite in terms of density of planet can be given as

\[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\]\[=\frac{2\pi {{\left( {{R}^{3}} \right)}^{1/2}}}{{{\left[ G.\frac{4}{3}\pi {{R}^{3}}\rho  \right]}^{1/2}}}=\sqrt{\frac{3\pi }{G\rho }}\]

(v) If the gravitational force of attraction of the sun on the planet varies as \[F\propto \frac{1}{{{r}^{n}}}\] then the time period varies as \[T\propto {{r}^{\frac{n+1}{2}}}\]

(vi) If there is a satellite in the equatorial plane rotating in the direction of earth?s rotation from west to east, then for an observer, on the earth, angular velocity of satellite will be \[({{\omega }_{S}}-{{\omega }_{E}})\]. The time interval between the two consecutive appearances overhead will be

\[T=\frac{2\pi }{{{\omega }_{s}}-{{\omega }_{E}}}=\frac{{{T}_{S}}{{T}_{E}}}{{{T}_{E}}-{{T}_{S}}}\]                         \[\left[ \text{As}\,\,\,T=\frac{2\pi }{\omega } \right]\]

If \[{{\omega }_{S}}={{\omega }_{E}}\], \[T=\,\infty \] i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.