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Serhud [2]
3 years ago
13

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

Physics
1 answer:
mote1985 [20]3 years ago
3 0

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

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