(D)

$v _0=\sqrt{\frac{ GM }{ R }} \Rightarrow GM = v _0^2 R$

By conservation of angular momentum

$I _1 \beta v _0= r _2 \alpha v _0 \Rightarrow \beta=\frac{ r _2}{ r _1} \alpha$

By conservation of mechanical energy

$\frac{1}{2} \beta^2 v _0^2-\frac{ GM }{ I _1}=\frac{1}{2} \alpha^2 v _0^2-\frac{ GM }{ I _2}$

$\Rightarrow GM \left[\frac{ I _1- I _2}{ I _1 I _2}\right]=\frac{1}{2} v _0^2\left(\alpha^2-\beta^2\right)$

$\Rightarrow GM \left(\frac{ I _1- I _2}{ I _1 I _2}\right)=\frac{1}{2} v _0^2 \alpha^2\left[1-\frac{ I _2^2}{ I _1^2}\right]$

$\Rightarrow GM \left(\frac{ r _1- r _2}{ r _1 r _2}\right)=\frac{1}{2} v _0^2 \alpha^2\left[\frac{\left( r _1+ r _2\right)\left( r _1- r _2\right)}{ r _1^2}\right]$

$\Rightarrow \frac{ GM }{ r _2}=\frac{1}{2} v _0^2 \alpha^2 \frac{\left( r _1+ r _2\right)}{ r _1}$

$\Rightarrow \frac{ r _1+ r _2}{2}=\frac{ GMr _1}{ r _2 v _0^2 \alpha^2}$

We know for elliptical orbit

$T ^2=\frac{4 \pi^2}{ GM }\left(\frac{ r _1+ I _2}{2}\right)^3$

$\Rightarrow T ^2=\frac{4 \pi^2}{ GM } \frac{ G ^3 M ^3 \alpha^3}{\beta^3 v _0^6 \alpha^6}$

$=\frac{4 \pi^2 v _0^2 R _0^2}{\beta^3 \alpha^3 v _0^6}$

$\Rightarrow T =\frac{2 \pi R }{ v _0}(\alpha \beta)^{-\frac{3}{2}}$