Answer:
The orbital velocity [tex]v = 16.4 \ km/s[/tex]
The period is [tex]T = 14.8 \ hours[/tex]
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
[tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]
=> [tex]w = \sqrt{ \frac{GM}{r^3} }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67*10^{-11}[/tex]
[tex]M_s[/tex] is the mass of with value [tex]M_s =5.683*10^{26} \ kg[/tex]
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 [tex]R = 60300 \ km[/tex]
D is the distance from the equator to the outer edge of the A ring with value [tex]D = 80000 \ kg[/tex]
So
[tex]r =80000 + 60300[/tex]
=> [tex]r =140300 \ km = 1.4*10^{8} \ m[/tex]
So
=> [tex]w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }[/tex]
=> [tex]w = 1.175*10^{-4} \ rad/s[/tex]
Generally the orbital velocity is mathematically represented as
[tex]v = w * r[/tex]
=> [tex]v = 1.175*10^{-4} * 1.4*10^{8}[/tex]
=> [tex]v = 1.64*10^{4} \ m /s = 16.4 \ km/s[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{2 \pi }{w }[/tex]
=> [tex]T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }[/tex]
=> [tex]T = 53473 \ second = 14.8 \ hours[/tex]
Answer:
The orbital velocity [tex]v = 16.4 \ km/s[/tex]
The period is [tex]T = 14.8 \ hours[/tex]
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as
[tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]
