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The speed of a satellite in a circular orbit of radius $R_0$ around the earth is $v_0$. Another satellite is in an elliptic orbit around the earth. If the minimum and maximum speeds of the second satellite are $\alpha v_0$ and $\beta v_0$ respectively, then its time period is
$\frac{2 \pi R_0}{v_0}\left(\frac{\alpha^2+\beta^2}{2}\right)^{\frac{3}{2}}$
$\frac{2 \pi R _0}{ v _0}\left(\frac{\alpha+\beta}{2}\right)^{\frac{3}{2}}$
$\frac{2 \pi R_0}{v_0}(\alpha \beta)^{\frac{3}{2}}$
$\frac{2 \pi R_0}{v_0}(\alpha \beta)^{-\frac{3}{2}}$
Solution
(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
$T ^2=\frac{4 \pi^2}{ GM }\left(\frac{ r _1+ I _2}{2}\right)^3$
$\Rightarrow T ^2=\frac{4 \pi^2}{ GM } \frac{ G ^3 M ^3 \alpha^3}{\beta^3 v _0^6 \alpha^6}$
$=\frac{4 \pi^2 v _0^2 R _0^2}{\beta^3 \alpha^3 v _0^6}$
$\Rightarrow T =\frac{2 \pi R }{ v _0}(\alpha \beta)^{-\frac{3}{2}}$
