A body of mass m is projected with velocity λ ve in vertically upward direction from the surface of the earth into space. It is given that ve is escape velocity and λ < 1. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be: (R: radius of earth) (a) R1+λ² (b) R1-λ² (c) (R)/(1-λ) (d) λ²R1-λ²

Answer: (b). Energy conservation: 1/2mλ²2GM/R = GMm(1/R - 1/r), r from the centre. 1/r = (1-λ²)/(R). r = (R)/(1-λ²). So the answer is (b) R1-λ².

A body of mass m is projected with velocity λ ve in vertically upward… | JEE Main 2022 (27 Jul, Shift 2)

Q40JEE Main 2022 · 27 JUL · SHIFT 2Medium

Escape Velocity

A body of mass $m$ is projected with velocity $\lambda v_e$ in vertically upward direction from the surface of the earth into space. It is given that $v_e$ is escape velocity and $\lambda < 1$ . If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be: (R: radius of earth)

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Answer: (b)

Energy conservation: $\frac{1}{2}m\lambda^2\frac{2GM}{R} = GMm\left(\frac{1}{R} - \frac{1}{r}\right)$ , $r$ from the centre.
$\frac{1}{r} = \frac{1-\lambda^2}{R}$ .
$r = \frac{R}{1-\lambda^2}$ .
So the answer is (b) $\frac{R}{1-\lambda^{2}}$ .

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Escape Velocity question from JEE Main 2022 (27 Jul, Shift 2)

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