# If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the center of the planet will become

$(a)\;2^{\frac{1}{3}}\;times \quad (b)\;2^{\frac{2}{3}}\;times \quad (c)\;2^{\frac{3}{2}}\;times \quad (d)\;4\;times$

$T^2 \; \alpha \;r^3$
$T \; \alpha \; r^{3/2}$
$\large\frac{2 \pi}{w}$$\; \alpha \; r^{3/2}$
$w \; \alpha \; r^{-3/2}$
If w is halved . distance will become
$w_2 =\large\frac{1}{2} w_1$
$\large\frac{w_2}{w_1}=\bigg(\large\frac{r_2}{r_1}\bigg)^{-3/2}$
$\large\frac{1}{2}=\bigg(\frac{r_2}{r_1}\bigg)^{-3/2}$
$r_2 =r_1 \bigg(\frac{1}{2}\bigg)^{-2/3}$
$r_2=r_1 \;2^{2/3}$
it become $2^{2/3}$ times
Hence b is the correct answer.

edited Feb 18, 2014 by meena.p