Question

Based on the kinetic energy, how much liquid hydrogen (energy density: 106 J/Litre) is at least needed to bring a small 1 kg satellite in an orbit of 400 km. (Use literature to find mE , RE , G.)

Answer 1

Answer:

∴v = 1/212 *G(mE)/(RE)+400

Explanation:

For the satellite to revolve at a distance of 400 Km from surface of earth, its orbital velocity must be such that, the centrifugal force produced is equal to gravitational pull.

i.e F(gravitational)=F(centripetal)

G(mE)(1)/r² = (1)v²/r    (Here is distance from centre of Earth to satellite i.e   r = (RE)+400)

∴v=√(G(mE)/(RE)+400 )

Since the question is based only one kinetic energy,

It is sufficient to provide only K.E through hydrogen fuel(let v litres of fuel is required, hence energy of it is 106v J)

∴106v = 1/2 (1) (√(G(mE)/(RE)+400 ))²

⇒106 v = 1/2 G(mE)/(RE)+400

∴v = 1/212 G(mE)/(RE)+400