Question

A satellite of mass 230 kg is placed in Earth orbit at a height of 500 km above the surface. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit

Answer 1

Answer:

Orbital period of satellite is 5.83 x 10³ s

Explanation:

The orbital period of satellite revolving around Earth is given by the equation :

$T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }$      .....(1)

Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.

In this problem,

Height of satellite, h = 500 km = 500 x 10³ m

Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).

 $T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }$

T = 5.83 x 10³ s