Gravitation Question 117

Question: A projectile is projected with velocity $ [kv_{e}] $ in vertically upward direction from the ground into the space. ($ [v_{e}] $ is escape velocity and $ [k<1)] $ . If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth) [Roorkee 1999; RPET 1999]

Options:

A) $ [\frac{R}{k^{2}+1}] $

B) $ [\frac{R}{k^{2}-1}] $

C) $ [\frac{R}{1-k^{2}}] $

D) $ [\frac{R}{k+1}] $

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Answer:

Correct Answer: C

Solution:

Kinetic energy = Potential energy

$ [\frac{1}{2}m,{{(kv_{e})}^{2}}=\frac{mgh}{1+\frac{h}{R}}] $

Þ $ [\frac{1}{2}mk^{2}2gR=\frac{mgh}{1+\frac{h}{R}}] $

$ [\Rightarrow ,h=\frac{Rk^{2}}{1-k^{2}}] $

Height of Projectile from the earth’s surface = h Height from the centre

$ [r=R+h=R+\frac{Rk^{2}}{1-k^{2}}] $ By solving $ [r=\frac{R}{1-k^{2}}] $



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