NCERT Solutions
Class 11 Physics
Gravitation
Q. 7.17
A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.
Velocity of the rocket, v = 5 km/s = 5 × 103 m/s
Mass of the Earth, Me = 6.0 x 1024 kg
Radius of the Earth, Re = 6.4 x 106 m
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= 
mv2 + 
At highest point h, v =0
And Potential energy = - (GMem) (Re +h)
Total energy of the rocket = 0 + (- GMem) (Re +h)
= (- GMem) (Re +h)
By applying law of conservation of energy,
Total energy of the rocket at the Earth’s surface = Total energy at height h
mv2 + = 
v2 = GMe ( 
– 
)
= GMe 
v2 = 
x 
v2 = 
Where g (acceleration due to gravity on earth’s surface)
= 
= 9.8m/s2
Therefore, (v2 Re) = h (2gReh)
h= 
= 
h = 
h = 1.6 x 106 m
Height achieved by the rocket with respect to the centre of the Earth
= (Re + h)
= (6.4 x 106 + 1.6 x 106)
=8.0 x 106 m