A particle of mass m 0 is projected vertically upward from surface of earth with a speed of 4 R 5 GM . The maximum height of the particle from the earth surface is 3 kR . [Mass of the earth is M and R is radius of earth]. What is the value of k ?

Energy conservation: 2 1 m 0 ⋅ 4 R 5 GM − R GM m 0 = − R + h GM m 0 . LHS = − 8 R 3 GM m 0 , so R + h = 3 8 R and h = 3 5 R → k = 5 .

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A particle of mass $m_{0}$ is projected vertically upward from surface of earth with a speed of $\frac{5 GM}{4 R}$ . The maximum height of the particle from the earth surface is $\frac{kR}{3}$ . [Mass of the earth is M and R is radius of earth]. What is the value of k ?

Answer:

Solution:

Energy conservation: $\frac{1}{2} m_{0} \cdot \frac{5 GM}{4 R} - \frac{GM m _{0}}{R} = - \frac{GM m _{0}}{R + h}$ .

LHS $= - \frac{3 GM m _{0}}{8 R}$ , so $R + h = \frac{8 R}{3}$ and $h = \frac{5 R}{3}$ → $k = 5$ .

Stream:JEESubject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: QB

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