Question

A geosynchronous communications satellite in stationary orbit is used to relay radio messages from one point to another around the earth’s curved surface. Apply Newton’s second law to the satellite to find its altitude in meters above the earth’s surface. R_E = 6.38 × 10⁶ m, m_E = 5.97 × 10²⁴ kg, and G = 6.674 × 10⁻¹¹ Nm²/kg².

Answer 1

Let ,

m = mass of the satellite

M_E = mass of the earth = 5.97 × 10²⁴ kg

R_​E = radius of the earth = 6.38×10⁶ m

h = height of the satellite above the surface of earth

r = R_E + h = radius of the orbit

v = orbital velocity of the satellite

As the satellite is revolving in a circular path

mv²/r = GmM_E / r²

v = ( GM_E / r )^1/2

Time period of the satellite

T = 2r / v

T = 2( r³/GM_E)^1/2

(T / 2 )² = r³ / GM_E

GM_E(T/2)² = r³

GM_E(T/2)² = (R_E + h)³

[ GM_E(T/2)²]^1/3 - R_E = h

On putting the numerical values

h = [ 6.674×10^-11×5.97×10²⁴ (24×60×60 /2 )² ]^1/3 - 6.38×10⁶

h = (75.28×10²¹)^1/3 - 6.38×10⁶

h = 42.22×10⁶ - 6.38×10⁶

h = 35.84×10⁶ m