Hwo far from the surface of ealth doeas a rocket fired vertically witha speed if 1/4th of escape velocity go
Answers
Velocity of the rocket, v = 5 km/s = 5 × 103 m/s
Mass of the Earth, Me = 6 × 1024 kg
Radius of the Earth, Re = 6.4 × 106 m
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= (1/2)mv2 + (-GMem / Re)
At highest point h,
v = 0
And, Potential energy = -GMem / (Re + h)
Total energy of the rocket = 0 + [ -GMem / (Re + h) ]
= -GMem / (Re + h)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface = Total energy at height h
(1/2)mv2 + (-GMem / Re) = -GMem / (Re + h)
(1/2)v2 = GMe [ (1/Re) – 1 / (Re + h) ]
= GMe[ (Re + h – Re) / Re(Re+ h) ]
(1/2)v2 = gReh / (Re + h)
Where g = GM / Re2 = 9.8 ms-2
∴ v2 (Re + h) = 2gReh
v2Re = h(2gRe – v2)
h = Rev2 / (2gRe – v2)
= 6.4 × 106 × (5 × 103)2 / [ 2 × 9.8 × 6.4 × 106 – (5 × 103)2
h = 1.6 × 106 m
Height achieved by the rocket with respect to the centre of the Earth = Re + h
= 6.4 × 106 + 1.6 × 106 = 8 × 106 m.
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