A planet is in a circular orbit about a massive star. The star undergoes a spherically symmetric explosion in which one percent of its mass is suddenly ejected to a distance well beyond that of the planet’s orbit. Find the eccentricity of the new orbit of the planet assuming that the planet is unaffected by the explosion.

Before explosion orbital velocity of the planet is,  ${\upsilon }_0={\left(\frac{GM}{r_0}\right)}^{1/2}$. The planet’s position and velocity immediately after explosion are not affected. The process is sudden, infinitesimal in time; therefore it would require infinite velocity and acceleration to change position and velocity of planet. At the instant of explosion planet is moving tangentially, it is nearest point of the new orbit. In order to find farthest distance we have to apply conservation of angular momentum and energy.
$m{v}_0{r}_0=m{v}_1{r}_1$
 $v_1=\frac{v_0{r}_0}{r_1}$                  ……………(1)
$\frac{1}{2}m{\upsilon }_0^2-\frac{GMm}{r_0}=\frac{1}{2}m{\upsilon }^2-\frac{GM\text{'}m}{r}$

$\frac{GMm}{2r_0}-\frac{GMm}{r_0}=\frac{1}{2}m{\left(\frac{{\upsilon }_0{r}_0}{r_1}\right)}^2-\frac{GM\text{'}m}{r}$

$\frac{GMm}{2r_0}-\frac{GM\text{'}m}{r_0}=\frac{GMm{r}_0}{2r_1^2}-\frac{GM\text{'}m}{r} \dots \dots \dots \dots ..\left(2\right)$

Now we have to solve the above expression. Consider   $\frac{M\text{'}}{M}=a$    And  $\frac{r_1}{r_0}=b$
Equation (2) reduces to  $1-2a=\frac{1}{b^2}-\frac{2a}{b}$
Or  $b^2(1-2a)+2ab-1=0$
Or  $(b-1)[b(1-2a)+1]=0$
Since  $b\ne 1,$ we must have
$b=\frac{1}{2a-1}$