A body of mass m is projected with velocity l | vedclass

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(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$

Vedclass · Accepted Answer

$\frac{R}{1-\lambda^{2}}$

Vedclass · Answer

(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$ from the center of the earth.Total energy at surface = Total energy at height $h$$-\frac{GMm}{R} + \frac{1}{2}m(\lambda v_{e})^{2} = -\frac{GMm}{h} + 0$Since the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,we have $v_{e}^{2} = \frac{2GM}{R}$.Substituting this into the equation:$-\frac{GMm}{R} + \frac{1}{2}m\lambda^{2}(\frac{2GM}{R}) = -\frac{GMm}{h}$$-\frac{GMm}{R} + \frac{\lambda^{2}GMm}{R} = -\frac{GMm}{h}$Dividing both sides by $-GMm$:$\frac{1}{R} - \frac{\lambda^{2}}{R} = \frac{1}{h}$$\frac{1-\lambda^{2}}{R} = \frac{1}{h}$$h = \frac{R}{1-\lambda^{2}}$

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