A particle is fired vertically upwards with a speed of $9.8 k m / s$ . Find the maximum height attained by the particle. Radius of earth = $6400 k m$ and $g$ at the surface = $9.8 {m / s}^{2}$ . Consider only earth’s gravitation.

50100 k m

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20900 k m

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43200 k m

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35200 k m

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Solution

The correct option is B $20900 k m$

$E_{i}$ = PE + KE = $\frac{- G M m}{R} + \frac{m v_{0}^{2}}{2}$
$E_{f}$ = $\frac{- G M m}{R + h} + 0$
From conservation of energy,
$E_{i} = E_{f}$
$\frac{- G M m}{R} + \frac{m v_{0}^{2}}{2} = \frac{- G M m}{R + h}$
$\frac{- g R^{2}}{R} + \frac{(v_{0})^{2}}{2} = \frac{- g R^{2}}{R + h}$
$- g R + \frac{(v_{0})^{2}}{2} = \frac{- g R^{2}}{R + h}$
If we substitute the values,
$g = 9.8 m s^{- 2}, v_{0} = 9.8 m s^{- 1}, R = 6400 k m$
$h = 20900 k m$

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A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle.

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