JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ___________ s.
- (a) √(2Rₑ/g) [ ( 1 + h/Rₑ )^(3/2) - 1 ]
- (b) 1/3√(Rₑ/2g) [ ( 1 + h/Rₑ )^(3/2) - 1 ]
- (c) √(Rₑ/2g) [ ( 1 + h/Rₑ )^(3/2) - 1 ]
- (d) 1/3√(2Rₑ/g) [ ( 1 + h/Rₑ )^(3/2) - 1 ]
Answer and solution
Answer (d) 1/3√(2Rₑ/g) [ ( 1 + h/Rₑ )^(3/2) - 1 ]
Solution
Physics sir · 4 steps
Escape speed means zero total energy, so you know v at every height; then add up dt = dr/v.
- Speed at distance r. A body launched with escape speed has zero total energy all the way up, so 1/2mv² - GMm/r = 0. With GM = gRₑ², the speed at distance r from the centre is v = √(2GM/r) = Rₑ√(2g/r).
- Set up the time. The speed keeps falling, so t = h/v will not work. Write dt = dr/v = (√r dr)/Rₑ√2g and integrate from r = Rₑ to r = Rₑ + h.
t = 1/Rₑ√2g∫_(Rₑ)^(Rₑ+h) r^(1/2) dr = 2/3 ((Rₑ+h)^(3/2)-Rₑ^(3/2))/Rₑ√2g
- Tidy up. Take Rₑ^(3/2) out of the bracket to get t = 2/3√(Rₑ/2g)[(1+h/Rₑ)^(3/2)-1]. Since 2/3√(Rₑ/2g) = 1/3√(4Rₑ/2g) = 1/3√(2Rₑ/g), the prefactor is 1/3√(2Rₑ/g).
- So. t = 1/3√(2Rₑ/g)[(1+h/Rₑ)^(3/2)-1], which is option (d).
The idea to keep. When speed depends on distance, write v as a function of r and use t = ∫ dr/v. A body launched with escape speed has zero total energy, so v = √(2GM/r) at every r.
Answer from the NTA final answer key. Solution by PrepWiser.