A geostationary satellite orbits around the earth in a circular...

By Kepler's law of planetary motion,

$T^{2} ∝ r^{3}$, hence ${T}_{1}^{2} ∝ {r}_{1}^{3}$

Similarly ${T}_{2}^{2} ∝ {r}_{2}^{3}$

$∴ {T}_{1}^{2} ∝ \left(36000\right)3$

${T}_{2}^{2} ∝ \left(6400 + h\right)3$

Therefore,~

\left({T}_{2}\right)2 = {T}_{1}^{2}\left(\dfrac{6400 + h}{36000}\right)3$</p><p>$> {T}_{1}^{2}\left[\dfrac{6400}{3600}\right]3$</p><p>$> \left(24\right)2\left[\dfrac{8}{45}\right]3$</p><p>$∴ {T}_{2} > \dfrac{24 \times 8 \times \sqrt{8}}{45 \times \sqrt{45}}$</p><p>$> 1 . 8h$</p><p>So,${T}_{2} ≅ 2h