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(A) Let $v_e$ be the escape velocity,given by $v_e = \sqrt{\frac{2GM}{R}}$.According to the law of conservation of energy,the total energy at the surf

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$800$

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(A) Let $v_e$ be the escape velocity,given by $v_e = \sqrt{\frac{2GM}{R}}$.According to the law of conservation of energy,the total energy at the surface of the Earth equals the total energy at the maximum height $h$.At the surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}m v^2$,where $v = \frac{v_e}{3}$.At maximum height $h$: $E_f = -\frac{GMm}{R+h}$.Equating $E_i = E_f$:$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{v_e}{3}\right)^2 = -\frac{GMm}{R+h}$$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{2GM}{9R}\right) = -\frac{GMm}{R+h}$$-\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} = \frac{6400 \ km}{8} = 800 \ km$.

$A$ body is projected vertically upwards from the surface of the Earth with a velocity equal to one-third of the escape velocity. The maximum height attained by the body will be $...... \ km$. (Take radius of Earth $R = 6400 \ km$ and $g = 10 \ m/s^2$)

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