A satellite is orbiting the earth in a circular orbit of radius $r$ with velocity $v_0$. A small impulse is given to the satellite such that its velocity increases to $v=\alpha v_0$, where $\alpha >1$. What is the eccentricity $e$ of the new elliptical orbit?

Solution

1. The initial orbital velocity is $v_0=\sqrt{\frac{GM}{r}}$, and the new velocity after the impulse is $v=\alpha v_0$.
2. The specific angular momentum is $$\begin{array}{rl}h& =rv\\ & =r(\alpha v_0)\\ & =\alpha \sqrt{GMr}\end{array}$$, and the specific energy is $$\begin{array}{rl}ϵ& =\frac{1}{2}{v}^2-\frac{GM}{r}\\ & =\frac{GM}{r}(\frac{{\alpha }^2}{2}-1)\end{array}$$.
3. Using the eccentricity formula $e=\sqrt{1+\frac{2ϵh^2}{(GM{)}^2}}$, substitute the expressions for $ϵ$ and $h$ to get $e=\sqrt{1+\frac{2(GM/r)({\alpha }^2/2-1)({\alpha }^{2}GMr)}{(GM{)}^2}}$.
4. Simplifying the expression yields $$\begin{array}{rl}e& =\sqrt{1+({\alpha }^2-2){\alpha }^2}\\ & =\sqrt{{\alpha }^4-2{\alpha }^2+1}\\ & =\sqrt{({\alpha }^2-1{)}^2}\\ & ={\alpha }^2-1\end{array}$$.
5. Hence the answer is (A).