# JEE Main & Advanced Physics Gravitation Time Period of Satellite

Time Period of Satellite

Category : JEE Main & Advanced

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.

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