Given,

Escape speed of the projectile on the Earth’s surface, V_{e} = 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

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

M_{e} is the mass of the Earth

R_{e} 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 V_{e} is the escape velocity , then,

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

Substituting (2) in (1)

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

v’² = v – V_{e}²

= (3V_{e})² – V_{e}²

= 8V_{e}²

v’² = 8V_{e}²

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

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