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The radius of planet is R. A satellite revolves around it in a circular orbit of radius r with angular speed 'w'. The acceleration due to gravity on the planet's surface will be

$(a)\;\frac{r^3w}{R} \quad (b)\;\frac{r^3w^2}{R^2} \quad (c)\;\frac{r^2w^3}{R^2} \quad (d)\;\frac{r^2w^2}{R}$
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Let M be mass of planet and m mass of satellite
$mrw^2=\large\frac{GMm}{r^2}$
$GM=r^3w^2$
But $g= \large\frac{GM}{R^2}$
$\therefore g= \large\frac{r^3w^2}{R^2}$
Hence d is the correct answer.

answered Aug 29, 2013 by
edited Feb 18, 2014 by meena.p

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