Question

What is the period of a satellite placed in an orbit 2.78 X 10^10 m above the surface of the earth?

Answer 1

T = 1.5×10^8 s

Step-by-step explanation:

let Fc be the centripetal force and Fg be the gravitational force of attraction between the satelite and earth.

At each point in the orbit, the forces acting on the satelite are Fc and Fg such that:

Fc = Fg

m×v^2/r = G×M×m/(r^2)

v^2 = G×M/(r)

v = \sqrt{G×M/(r)}

v = \sqrt{(6.67408×10^-11)×(5.972×10^24)/(6371×10^3 + 2.78×10^10)}

= 119.72 m/s

then, the satelite covers a displacement of 2×π×r in a period of:

T = 2×π×r/v

= 2×π×(6371×10^3 + 2.78×10^10)/(199.72)

= 1.5×10^8 s

therefore, the time is takes the satelite to orbit Earth is 1.5×10^8 seconds.