Question

A 600 kg satellite is in a circular orbit around Earth at a height above the Earth’s surface equal to the radius of the Earth. What is the satellite’s period of revolution in hours?

Answer 1

Answer:

$T =3.98\ hr$

Explanation:

given,

mass of the satellite = 600 Kg

mass of earth = 5.98 x 10²⁴ Kg

height of the satellite

     r = 2 R_e

R_e is radius of earth = 6.38 x 10⁶ m

Assuming orbital velocity of the satellite

        v = 5.59 x 10³ m/s

$time\ period =\dfrac{2\pi r}{v}$

$time\ period =\dfrac{2\pi \times 2 \times 6.38 \times 10^{6}}{5.59 \times 10^3}$

    T = 14342.3 s

$T = \dfrac{14342.3}{60\times 60}\ hr$

$T =3.98\ hr$