Asked by ne
Find the orbital speed of an ice cube in the rings of Saturn, if the mass of Saturn is 5.67 × 1026 kg and the rings
have an average radius of 100,000 km.
have an average radius of 100,000 km.
Answers
Answered by
FredR
Newton's law of gravitation:
F = G m1 m2 /r^2
let m2 = mass of ice cube and
s = G m1/r^2
so,
F = s m2
rearranging,
s = m2/F
let V = orbital speed
centripetal acceleration = V^2/r
For an object to remain in orbit s must equal the centripetal acceleration so,
s = V^2/r
F = G m1 m2 /r^2
let m2 = mass of ice cube and
s = G m1/r^2
so,
F = s m2
rearranging,
s = m2/F
let V = orbital speed
centripetal acceleration = V^2/r
For an object to remain in orbit s must equal the centripetal acceleration so,
s = V^2/r
Answered by
drwls
Taking up where FredR left off,
V^2/r = G m/r^2
V^2 = G m/r
where m is the mass of Saturn.
G = 6.674*10^-11 N*m^2/kg^2
Solve for V
V^2/r = G m/r^2
V^2 = G m/r
where m is the mass of Saturn.
G = 6.674*10^-11 N*m^2/kg^2
Solve for V
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