InterviewSolution
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InterviewSolution
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20. The escape velocity of a body on the surfaceofthe earth is 11.2 km/s. 1f the earth's massincreases to twice its present value and theradius of the earth becomes half, the escapevelocity becomes:(a) 5.6 km/s(c) 44.8 km/s(b) 11.2 km/s(d) 22.4 km/s |
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Answer» Escape velocity of a projectile from the Earth,vesc= 11.2 km/sProjection velocity of the projectile,vp= 3vescMass of the projectile =mVelocity of the projectile far away from the Earth =vfTotal energy of the projectile on the Earth = (1/2)mvp^2– (1/2)mvesc^2Gravitational 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^2From the law of conservation of energy, we have(1/2)mvp2– (1/2)mvesc^2 = (1/2)mvf2vf= (vp^2–vesc^2)^1/2= [ (3vesc)^2–vesc^2]^1/2= √8vesc= √8×11.2 = 31.68 km/s. |
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