Question

A particle is fired vertically upward with a speed of 15 km s⁻¹. With what speed will it move in interstellar space. Assume only earth's gravitational field.

Answer 1

Initial velocity of the particle, v = 15 km/s
Let its speed be v' in interstellar space.

Applying the law of conservation of energy, we have:

\[\left( \frac{1}{2} \right)m\left[ v - v '^2 \right] = \int_R^\infty \frac{GMm}{x^2}dx\]

\[ \therefore \left( \frac{1}{2} \right)m\left[ 15 \times {10}^3 - v '^2 \right] = \int_R^\infty \frac{GMm}{x^2}dx\]

\[ \Rightarrow \left( \frac{1}{2} \right)m\left[ \left( 15 \times {10}^3 \right)^2 - v '^2 \right] = GMm\left[ \frac{- 1}{x} \right]\]

\[ \Rightarrow \left( \frac{1}{2} \right)m\left[ \left( 225 \times {10}^5 \right) - v '^2 \right] = \frac{GMm}{R}\]

\[ \Rightarrow 225 \times {10}^5 - v '^2 = \frac{2 \times 6 . 67 \times {10}^{- 11} \times 6 \times {10}^{24}}{6400 \times {10}^3}\]

\[ \Rightarrow v '^2 = 225 \times {10}^6 - \frac{40 . 02}{32} \times {10}^8 \]

\[ = 2 . 25 \times {10}^8 - 1 . 2 \times {10}^8 \]

\[ = {10}^8 \left( 1 . 05 \right)\]

\[\text { Or }\  v' = 1 . 01 \times {10}^4 m/s = 10 km/s\]