Question

The escape speed of a projectile on a planet's surface is 11.2km/s . If a body is projected at double the speed,find its speed at an infinite distance

Answer 1

Answer:

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

Projection velocity of the projectile, vp= 3v

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)mvp^2– (1/2)mvesc^2

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)mvf^2

From the law of conservation of energy, we have

(1/2)mvp^2– (1/2)mvesc^2= (1/2)mvf^2

vf= ( vp^2– vesc^2 )1/2

= [ (3vesc)^2– vesc^2]^1/2

= √8 vesc

= √8 × 11.2 = 31.68 km/s.

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