Question

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. Question8.18 Gravitation

Answer 1

Given ,
escape speed 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.

Initial K.E of the body = 1/2mv²
Initial Gravitational PE of the body = -GMem/Re
Where, Me , Re are the mass , radius respectively.

At very far from the earth surface .
KE of the body = 1/2mv²
Gravitational PE of the body = 0

A/c to law of conservation of energy .
Total energy at the point of projection = total energy at very far from the earth's surface.
1/2mv² + (-GMem/Re) = 1/2mv'²
1/2 mv'² = 1/2mv² - GMem/Re

If Ve is the escape velocity , then,
1/2 mVe² = GMem/Re

1/2mv'² = 1/2mv² - 1/2mVe²
v'² = v - Ve²
= (3Ve)² - Ve²
= 8Ve²
v'² = 8Ve²
v' = 2√2Ve
v = 2 × 1.414 × 11.2 km/s
= 31.68 km/s

Hence, speed of the body at far away from the earth = 31.68 km/s

Answer 2

Escape velocity of a projectile from the Earth, vesc = 11.2 km/s

Projection velocity of the projectile, vp = 3vesc

Mass of the projectile = m

Velocity of the projectile far away from the Earth = vf

Total energy of the projectile on the Earth = (1/2)mvp2 – (1/2)mvesc2

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = (1/2)mvf2

From the law of conservation of energy, we have

(1/2)mvp2 – (1/2)mvesc2 = (1/2)mvf2

vf = ( vp2 – vesc2 )1/2

= [ (3vesc)2 – vesc2 ]1/2

= √8 vesc

= √8 × 11.2 = 31.68 km/s.