Question

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit is 1.9 hours. The radius of the planet is 4.38 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?

Answer 1

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

                                            = 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

                             $T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$  second

Squaring the terms and solving it for 'g'

                             g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$   m/s²

Substituting the values in the above equation

                   g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$  

                                    g = 4.84 m/s²      

Therefore, the weight

                                     w = m x g   newton

                                         = 5832 Kg x 4.84 m/s²

                                         = 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N