Question

For a satellite to be in a circular orbit 780 km above the surface of the earth, what is the period of the orbit (in hours)? ​

Answer 1

I will take 2 hours....

I will take 2 hours....

Answer 2

Given :

Height of the orbit from the surface of the earth (h) = 780 Km

To Find :

Time period of the orbit

Solution :

If the time period of the satellite is T, the distance covered in time T by the satellite = the circumference of the orbit = 2$\pi$r

      ∴ T = $\frac{2\pi r}{v_{0} }$  ( where $V_{0}$ = orbital velocity)

We know, Orbital velocity ( $V_{0}$) = $\sqrt{\frac{GM}{(R+h)} }$ ----------> equation 1

as g = $\frac{GM}{R^{2} }$

(g = acceleration due to gravity; M = mass of earth; r = radius of the earth; G= Gravitational constant)

or, GM = g$R^{2}$

Putting this value of GM in equation 1

     $V_{0}$ = $\sqrt{\frac{gR^{2} }{r} }$  (where r = radius of orbit = R+h)

or,  $V_{0}$ = $R\sqrt{\frac{g}{r} }$

Now, Time period of revolution (T) = $\frac{2\pi r}{v_{0} }$

                                                          =  $2\pi r.\frac{1}{R} \sqrt{\frac{r}{g} }$

                                                          = $\frac{2\pi }{R} \sqrt{\frac{r^{3} }{g} }$

                                                          = $\frac{2 * 3.14}{6.4 * 10^{6} } \sqrt{\frac{(6.4 * 10^{6} +0.78 * 10^{6} )^{3} }{9.8} }$

                                                          = 0.98 x $10^{-6}$ $\sqrt{ \frac{(7.18 *10^{6} )^{3} }{9.8} }$

                                                          = 0.98 x  $10^{-6}$ x $\sqrt{\frac{370.15 * 10^{18} }{9.8}}$

                                                          = 0.98 x 6.146 x $10^{3}$

                                                         = 7126 sec = 2 hours (approx.)

∴ The period of revolution at a height of 780 km from the earth's suface is 2 hours (approximately).