The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Asked by Pragya Singh | 1 year ago |  94

1 Answer

Solution :-

Given,

Escape speed of the projectile on the Earth’s surface, Ve = 11.2 km/s

Speed of projection of the body, v = 3Ve = 3×11.2 = 33.6 km/s

Let v and v’ be the speed of the body at the time of projection and at a point far from the earth.

At the time of projection

Initial kinetic energy of the body =\( \dfrac{1}{2}\)mv²

Initial gravitational potential energy of the body 

\( \dfrac{-GM_em}{R_e}\)

Me is the mass of the Earth

Re is the radius of the Earth

At a larger distance from the earth surface 

KE of the body = \( \dfrac{1}{2}\)mv’²

The gravitational potential energy of the body = 0

According to law of conservation of energy

Total energy at the point of projection = total energy at very far from the earth’s surface.

\( \dfrac{1}{2}\)mv² + (\( \dfrac{-GM_em}{R_e}\)) =\( \dfrac{1}{2}\)mv’²

\( \dfrac{1}{2}\)mv’² = \( \dfrac{1}{2}\)mv²

 – \( \dfrac{-GM_em}{R_e}\)——–(1)

If Ve is the escape velocity , then,

\( \dfrac{1}{2}\) mVe² =\( \dfrac{-GM_em}{R_e}\)———(2)

Substituting  (2) in (1)

\( \dfrac{1}{2}\)mv’² = \( \dfrac{1}{2}\)mv² – \( \dfrac{1}{2}\)mVe²

v’² = v – Ve²

= (3Ve)² – Ve²

= 8Ve²

v’² = 8Ve²

v’ = \( \sqrt{8 \times V_e}\)

v = 2 × 1.414 × 11.2 km/s

= 31.68 km/s

Speed of the body far away from the earth = 31.68 km/s

Answered by Abhisek | 1 year ago

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