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.

#### 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

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

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

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

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

\[ \Rightarrow v^2 = \frac{GM}{R}\]

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

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

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

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

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

\[ = 79 km/s\]