Answer to Question #160159 in Field Theory for Keleko

Question

A proposed communication satellite would revolve round the earth in a circular orbit in the equitorial plane, at a height of 35880 km above the earth's surface. Find the period of revolution of the satellite in hours, and comment on the result. (Radius of earth= 6370 km, mass of earth = 5.98 x 10^24 , constant of gravition = 6.66 x 10^-11 Nm^2kg^-2)

Answer 1

Solution,

Given

R_E = 6370km = 6370 * 10³ m

M_{E =} 5.98 * 10²⁴ kg

G = 6.66 * 10 ^-11 N. m² / kg²

h =35880 km = 35880 * 10³ m

Period of revolution of the satellite

$T = 2π \sqrt \frac{(R+h)^3} {GM}$

Put all value in above formula

We got

T = 91,908.78 sec

In hours

T = 91908.78/ 3600

T = 25.53 hours

Comment : it's shows nearly equals to geo-stationary satellite orbit.