1.

A body is projected from the surface of earth with the velocity three times the escape velocity. Find the velocity beyond the gravitational field of the earth.

Answer»

According to the conservation law of energy, the energy at the surface of the earth is

= -GMem/Re + 1/2m(3ve)2

The energy beyond the gravitational field of the earth, = 1/2mv2 (where v is the velocity beyond the earth's gravity).

Hence, -GMem/Re + 9/2mve2 = 1/2mv2

or, -1/2mve2 + 9/2mve2 = 1/2mv2

or, 8ve2 = v2

v = √2 ve = 2√2 x (11.2) kms-1 = 22.4√2 kms-1.



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