Question

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 555 km above the earth’s surface, while that for satellite B is at a height of 778 km. Find the orbital speed for (a) satellite A and (b) satellite B.

Answer 1

h = height of satellite above earth
v = tangential velocity of satellite , m = mass of satellite

here gravity force is balanced by centripetal force of satellites

centripetal force at height H above the earth surface = mv²/(R+h)

gravity force = mg

so here for proper circular rotation in orbit,

centripetal force = gravity force

mv²/(R+h) = mg

v² = g (R+h)

v =[g (R+h)]^1/2

for satellite A

h = 555 km , R+h =6371+555 = 6926 km, 6926000 m ( radius of earth = 6371 km )

but here also g also change with height, g^' = g(R/R+H)²

g^' = 9.8 ( 6371000/6926000)²

  = 9.8 * 0.8461

= 8.29 m/s²

v =[g (R+h)]^1/2 , but here g= g^' = 8.29 m/s²

v = [ 8.29 * 6926000 ]^1/2

v = 7577.37 m/s , v= 7.577 km/s

this is orbital speed for satellite A

for satellite B,

v =[g (R+h)]^1/2

but here also g also change with height, g^' = g(R/R+H)²

h = 778km , R+h =6371+778 = 7149 km, 7149000 m ( radius of earth = 6371 km )

g^' = 9.8 ( 6371000/7149000)²

  = 9.8 * (0.7941)

g^' = 7.78 m/s²

here, g = g^' = 7.78 m/s²

v =[g (R+h)]^1/2

v = [ 7.78 * 7941000 ]^1/2

v = 7860.08 m/s , v= 7.8 km/s

this is orbital speed for satellite B