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(B) The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Given that the projection velocity $v = \frac{1}{3} v_e = \frac{1}{3} \sqrt{\frac{2GM

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$\frac{R}{8}$

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(B) The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Given that the projection velocity $v = \frac{1}{3} v_e = \frac{1}{3} \sqrt{\frac{2GM}{R}}$.Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$:$TE_{	ext{surface}} = TE_{	ext{height } h}$$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0$Substituting $v^2 = \frac{1}{9} 	imes \frac{2GM}{R}$:$-\frac{GMm}{R} + \frac{1}{2}m \left( \frac{2GM}{9R} ight) = -\frac{GMm}{R+h}$$-\frac{GMm}{R} + \frac{GMm}{9R} = -\frac{GMm}{R+h}$$-\frac{8GMm}{9R} = -\frac{GMm}{R+h}$$\frac{8}{9R} = \frac{1}{R+h}$$8(R+h) = 9R$$8R + 8h = 9R$$8h = R$$h = \frac{R}{8}$

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

Similar Questions

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