Question

A particle of mass $m$ is subjected to an attractive central force of magnitude $k/r^2$, $k$ being a constant. If at the instant when the particle is at an extreme position in its closed orbit, at a distance $a$ from the centre of force, its speed is $\sqrt{\left(k/2ma\right)}$, if the distance of other extreme position is $b$. Find $a/b$

Answer 1

$F=-K/r^2$ (negative sign is for attractive force)
Potential energy $U=\int Fdr=\int \frac{K}{r_2}dr=-\frac{K}{r}$
Conservation of energy gives (let the other extreme position $r=b$)
$K_1+U_1=K_2+U_2$
$\frac{1}{2}m{v}_1^2-\frac{K}{a}=\frac{1}{2}m{v}_2^2-\frac{K}{b}$
where $v_1=\sqrt{\frac{K}{2ma}}$
Conservation of angular momentum gives
$m{v}_{1}a=m{v}_{2}b$
$v_2=\frac{a}{b}{v}_1=\frac{a}{b}\sqrt{\frac{K}{2ma}}$
Therefore, from Eq, $\left(i\right)\Rightarrow \frac{1}{2}m\frac{K}{2ma}-\frac{K}{a}=\frac{1}{2}m{\left(\frac{a}{b}\right)}^2\frac{K}{2ma}-\frac{K}{b}$
$-\frac{3K}{4a}=\frac{aK}{4b^2}-\frac{K}{b}\Rightarrow b^2-\frac{4a}{3}b+\frac{a^3}{3}=0$
Hence, $b=a/3\Rightarrow a/b=3$