Semi-minor axis of the orbit of satellite

3.92 × 10⁷ m

Eccentricity of the orbit e = $\left(1+\frac{2 E L^2}{G^2 M^2 m^3}\right)$

= 1 + $\frac{2 \times\left(-1.5 \times 10^{10}\right) \times\left(3.6 \times 10^{14}\right)^2}{\left(6.67 \times 10^{-11}\right)^2 \times\left(5.97 \times 10^{24}\right)^2 \times(5000)^3}$ = 0.804

Hence, semi–minor axis b = $a \sqrt{1-e^2}=6.6 \times 10^7 \sqrt{1-(0.804)^2}$ = 3.92 × 10⁷ m