Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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# P 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. - Physics

Sum

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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#### Solution

The particle attained a maximum height 6400 km, which is equal to the radius of the Earth.

Total energy of the particle on the Earth's surface is given by $E_e = \frac{1}{2} M V^2 + \left( \frac{- gmM}{r} \right) . . . . \left( 1 \right)$

Now, total energy of the particle at the maximum height is given by

$E_3 = \left( \frac{- GMm}{R + h} \right) + 0$

$\Rightarrow E_3 = \left( \frac{- GMm}{2R} \right) . . . \left( 2 \right) \left( \therefore g = R \right)$

From equations (1) and (2), we have:

$- \frac{GMm}{R} + \frac{1}{2}m v^2 = - \frac{GMm}{2R}$

$\Rightarrow \left( \frac{1}{2} \right)m v^2 = GMm\left( - \frac{1}{2R} + \frac{1}{R} \right)$

$\Rightarrow v^2 = \frac{GM}{R}$

$= \frac{6 . 67 \times {10}^{- 11} \times 6 \times {10}^{24}}{6400 \times {10}^3}$

$= \frac{40 . 02 + {10}^{13}}{6 . 4 \times {10}^6}$

$= 6 . 2 \times {10}^7 = 0 . 62 \times {10}^8$

$\therefore v = \sqrt{0 . 62 \times {10}^8}$

$= 0 . 79 \times {10}^4 m/s$

$= 79 km/s$

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 11 Gravitation
Q 37 | Page 227
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