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 $\mathrm{v}_{\mathrm{e}}$ is escape velocity and $\lambda<1$. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)

(B) $\frac{R}{1-\lambda^{2}}$

(A) $\frac{\mathrm{R}}{1+\lambda^{2}}$

(C) $\frac{\mathrm{R}}{1-\lambda}$

(D) $\frac{\lambda^{2} R}{1-\lambda^{2}}$

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Gravitation

A body of mass

m

is projected with velocity

λ v_{e}

in vertically upward direction from the surface of the earth into space. It is given that

v_{e}

is escape velocity and

λ < 1

. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)

(A) $\frac{R}{1 + λ ^{2}}$
(B) $\frac{R}{1 - λ ^{2}}$ ✓ Correct
(C) $\frac{R}{1 - λ}$
(D) $\frac{λ ^{2} R}{1 - λ ^{2}}$

Answer: (B)

Solution

Using energy conservation:

- \frac{GM m}{R} + \frac{1}{2} m (λ v_{e})^{2} = - \frac{GM m}{h}

. Since

v_{e}^{2} = \frac{2 GM}{R}

, we have

- \frac{GM m}{R} + \frac{1}{2} m λ^{2} \times \frac{2 GM}{R} = - \frac{GM m}{h}

. Simplifying:

- \frac{1}{R} + \frac{λ ^{2}}{R} = - \frac{1}{h}

, giving

\frac{1}{h} = \frac{1 - λ ^{2}}{R}

, so

h = \frac{R}{1 - λ ^{2}}

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Gravitation

Solution

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