a rocket is launched vertically from the surface of the earth with initial speed of 10 km/s . how far above the earth's surface will the rocket go?

Initial speed of the rocket is, u = 10 km/s = 10000 m/s

Total energy of the rocket on the surface of earth is, Ei = PE + KE

=> Ei = -GMm/R + ½ mu2

[m is the mass of the rocket]

Suppose it reaches a height ‘h’ from the surface of the earth. Its total energy at that height is,

Ef = -GMm/(R + h)

[at this height the rocket has no KE]

By law of conservation of energy,

Ei = Ef

=> -GMm/R + ½ mu2 = -GMm/(R + h)

=> ½ u2 = GM/R – GM/(R + h)

=> ½ u2 = GM[R+h-R]/{R(R+h)}

=> ½ u2 = GMh/(R2+Rh)

=> (R2+Rh)u2 = 2GMh

=> R2u2 + Ru2h = 2GMh

=> h = R2u2/(2GM – Ru2)

Here, R = 6400000 m, u = 10000 m/s, G = 6.67 × 10-11 Nm2kg-2, M = 5.97 × 1024 kg

So, h = 2.62 × 107 m

This is the height from the surface of the earth upto which the rocket reaches.

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