Answer :
The period of a satellite that is orbiting the earth is given by,
[tex]T=2\pi\sqrt[]{\frac{r^3}{GM}}[/tex]Where G is the gravitational constant, M is the mass of the earth, and r is the radius of the satellite.
Let us assume that the period of the satellite is decreased to half its value.
Then the new period is given by,
[tex]\begin{gathered} T_n=\frac{T}{2} \\ =\frac{2\pi}{2}\sqrt[]{\frac{r^3}{GM}} \end{gathered}[/tex]On simplifying the above equation,
[tex]\begin{gathered} T_n=2\pi\sqrt[]{\frac{r^3}{4GM}} \\ =2\pi\sqrt[]{\frac{(\frac{r}{\sqrt[3]{4}})^3}{GM}} \\ =2\pi\sqrt[]{\frac{r^3_n}{GM}} \end{gathered}[/tex]Where r_n is the decreased radius of the orbit of the satellite.
The value of r_n is,
[tex]\begin{gathered} r_n=\frac{r}{\sqrt[3]{4}} \\ =0.63r \end{gathered}[/tex]Thus for the period of the satellite to be reduced to one-half of its initial value its radius should be reduced to 0.63 times its initial velue.
To calculate the percentage decrease of the radius,
[tex]\begin{gathered} P=\frac{r-r_n}{r}\times100 \\ =\frac{r-0.63r}{r}\times100 \\ =(1-0.63)100 \\ =37\% \end{gathered}[/tex]Therefore to reduce the orbital period of a satellite by a factor of one-half, its radius should be decreased by 37%.