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:
$$\begin{gathered} \left(\frac{1}{2}\right)m\left[v-v{\text{'}}^2\right]={\int }_R^{\infty }\frac{GMm}{x^2}dx \\ \therefore \left(\frac{1}{2}\right)m\left[15\times {10}^3-v{\text{'}}^2\right]={\int }_{\mathrm{R}}^{\infty }\frac{\mathrm{G}Mm}{x^2}dx \\ \Rightarrow \left(\frac{1}{2}\right)m\left[{\left(15\times {10}^3\right)}^2-\mathrm{v}{\text{'}}^2\right]=\mathrm{GM}m\left[\frac{-1}{x}\right] \\ \Rightarrow \left(\frac{1}{2}\right)m\left[\left(225\times {10}^5\right)-v{\text{'}}^2\right]=\frac{\mathrm{GM}m}{\mathrm{R}} \\ \Rightarrow 225\times {10}^5-v{\text{'}}^2=\frac{2\times 6.67\times {10}^{-11}\times 6\times {10}^{24}}{6400\times {10}^3} \\ \Rightarrow v{\text{'}}^2=225\times {10}^6-\frac{40.02}{32}\times {10}^8 \\ =2.25\times {10}^8-1.2\times {10}^8 \\ ={10}^8\left(1.05\right) \\ \mathrm{Or}v\text{'}=1.01\times {10}^4\mathrm{m}/\mathrm{s}=10\mathrm{km}/\mathrm{s} \end{gathered}$$