1.

Abodyisthrownverticallyupwardswithavelocity-110 km s from earth's surface. Up to which heightwill it go ? Radius of the earth is 6400 km andg 10 m s2Ans. 2.28 x 10 km.

Answer»

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

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

=> Ei= -GMm/R+ ½ mu2

[m is the mass of the body]

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 body 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 × 10-11Nm2kg-2, M = 5.97 × 1024kg

So, h = 2.62 × 107m

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


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