Question

What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring

Answer 1

Answer:

The orbital velocity $v = 16.4 \ km/s$

The period is  $T = 14.8 \ hours$

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the planet , this is mathematically represented as

       $\frac{GM_s * m }{r^2 } = m w^2 r$

=>    $w = \sqrt{ \frac{GM}{r^3} }$

Here G is the gravitational constant with value  $G = 6.67*10^{-11}$

        $M_s$  is the mass of with value  $M_s =5.683*10^{26} \ kg$

        r is the is distance from the center of the  to the  outer edge of the  A ring

i.e r = R  + D  

Here R  is the radius of the planet   with value  $R = 60300 \ km$

         D  is the distance from the  equator to the outer edge of the  A ring with value  $D = 80000 \ kg$

So  

       $r =80000 + 60300$

=>    $r =140300 \ km = 1.4*10^{8} \ m$

So

    =>    $w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }$

    =>    $w = 1.175*10^{-4} \ rad/s$

Generally the orbital velocity is mathematically represented as

       $v = w * r$

=>     $v = 1.175*10^{-4} * 1.4*10^{8}$

=>     $v = 1.64*10^{4} \ m /s = 16.4 \ km/s$

Generally the period is mathematically represented as

     $T = \frac{2 \pi }{w }$

=> $T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }$

=> $T = 53473 \ second = 14.8 \ hours$

Answer:

The orbital velocity $v = 16.4 \ km/s$

The period is  $T = 14.8 \ hours$

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the , this is mathematically represented as

       $\frac{GM_s * m }{r^2 } = m w^2 r$