Distance between the centres of two stars is $10 a$. The masses of these stars are $M$ and $16 M$ an

Question

Distance between the centres of two stars is $10 a$. The masses of these stars are $M$ and $16 M$ and their radii a and $2 a$, respectively. A body of mass $m$ is fired straight from the surface of the larger star towards the smaller star. What should be its minimum initial speed to reach the surface of the smaller star?

Professor PC Thomas Classes · Answer

Let $O_{1}$ denote the centre of smaller star $S_{1}$ having radius $(a)$ and mass $(M)$.\ Let $O_{2}$ denote the centre of bigger star $S_{2}$ having radius ( $2 a$ ) and mass ( $16 M$ ).\ Thus $O_{1} O_{2}=$ Distance between two centres or $\quad O_{1} O_{2}=(10 a)$\ Let $A$ denote the point along $O_{1} O_{2}$ where gravitational field strength is zero.\ $	herefore \quad$ Field strength due to $S_{1}=$ Field strength due to $S_{2}$\ or $\frac{G M}{r_{1}^{2}}=\frac{G(16 M)}{r_{2}^{2}}$ or $r_{2}=4 r_{1}$\  But $r_{1}+r_{2}=10 a$ $$\begin{equation*} 	herefore \quad r_{1}+4 r_{1}=10 a 	ag{i} \end{equation*}$$ or $r_{1}=2 a$\ and $r_{2}=8 a$\ When the body of mass $m$ is fired straight from the surface of $S_{2}$ towards $S_{1}$ along $\mathrm{O}_{2} \mathrm{O}_{1}$, it travels under the gravitational forces exerted by $S_{2}$ and $S_{1}$.\ From $O_{2}$ to $A$, it is attracted by $S_{2}$.\ At $A$, there is a region of no attraction.\ From $A$ to $O_{1}$, it is attracted by $S_{1}$.\ Let $v$ denote velocity with which the body of mass $m$ is fired so that it just crosses $A$ along $A O_{1}$.\ The energy is conserved.\ $	herefore \frac{1}{2} m v^{2}=\left[-\frac{G m M}{r_{1}}-\frac{16 G m M}{r_{2}}ight]$ $$ -\left[-\frac{G m M}{(10 a-2 a)}-\frac{16 G m M}{2 a}ight] $$ or $\frac{1}{2} m v^{2}=-G m M\left[\frac{1}{2 a}+\frac{16}{8 a}+\left(\frac{-1}{8 a}ight)ight.$ $\left.-\left(\frac{16}{2 a}ight)ight]$\ or $\frac{1}{2} m v^{2}=\frac{45 G m M}{8 a} \Rightarrow v^{2}=\frac{45 G M}{4 a}$\ or $\quad v=\frac{3}{2} \sqrt{\frac{5 G M}{a}}$.\