Question

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800 km. The speed of the satellite is (ME = 5.98 × 1024 kg, RE = 6.37 × 106 m, G = 6.67 × 10−11 N·m2/kg2)

Answer 1

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = $0.8 \times 10^{6} m$

    $R_{e} = 6.37 \times 10^{6} m$,    r = R + h = $7.17 \times 10^{6} m$

   $M_{e} = 5.98 \times 10^{24}$ kg,   G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

              $\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

                      v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

               $\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

                            v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

                               = $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

                                = $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$.