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Asked by Prashant DIGHE | 01 Apr, 2020, 10:47: PM
Expert Answer
Total energy of particle at earth's surface = Kinetic energy + Potential energy
If V is projection speed , then Kinetic energy = (1/2)mV2
where m is mass of projectile
Potential energy on Earth's surface = - [ ( G m Me ) / Re ] + Wo
where G is universal gravitational constant, Me is mass of earth, Re is radius of earth and Wo is potential energy at infinity
Hence Total energy of projectile on earth's surface = (1/2)mV2 - [ (G m Me ) / Re ] + Wo ...........................(1)
when projectile reaches the maximum height h = Re / 2 , its kinetic energy becomes zero
Total energy at a height Re/2 above earth surface = - [G m Me ] / [ ( 3/2)Re ] + Wo .......................(2)
By conservaton of energy, we equate eqn.(1) and (2) , then we get after simplification,
V2 = (2/3) G Me / Re ....................(3)
if V = C Ve , where C is constant and Ve is escape speed , then we have
V2 = C2 Ve2 = (2/3) G Me / Re .............................(4)
Escape speed Ve is given as, Ve = [ 2 G Me / Re ]1/2 or Ve2 = 2 G Me / Re .......................(5)
Using eqn.(5) in eqn.(4), we get, C2 = 1/3 or C = 1/√3
Answered by Thiyagarajan K | 02 Apr, 2020, 08:54: AM
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