Physics: Hard Rotation Question?

Question

Jupiter's four large moons - Io, Europa, Ganymede, and Callisto - were discovered by Galileo in 1610. Jupiter also has dozens of smaller moons. Jupiter's rocky, and volcanically - active moon Io is about the size of Earth's moon. Io has a radius of about 1.82 x 10^6 m, and the mean distance between Io and Jupiter is 4.22 x 10^8m.

1) If Io's orbit were circular, how many days would it take for Io to complete one full revolution around Jupiter?

Answer 1

For Io to stay in orbit needs a centripetal force to balance gravitational.
Fg=Fc
GMm/R^2=mV^2/R

V=sqrt( GM/R)

M- mass of Jupiter = 1.899 E+27 kilograms
G - gravitational constant = 6.673 E-11 m3 / kg1 s^2
Rej- Jupiter's equatorial Radius =71.492 E+6 m

R= 1.82 x 10^6 + 71.492 E+6 m = 73.312 E+6 m

V= sqrt( 6.673 E-11 x 1.899 E+27 / 73.312 E+6)
V=41,575 m/s

T= 2piR/V= 2 pi 73.312 E+6/ 41,575
T=11080 sec
T=0.128 Earth days

Answer 2

Not enough info for a numeric answer. Io's radius is irrelevant.
From Kepler's 3rd law,
T = (2pi*r^1.5) / sqrt(Gm)
Now you can look up G (the universal grav. constant) and the mass m of Jupiter and plug them in.
EDIT: 1st ans. plugged them into a different set of equations so I did the same with Kepler's. I get T = 1.53E5 s = 42.5 h. This agrees pretty well with the ref. Ans. 1 used the wrong value for r (the sum of both bodies' radii?); the correct value is given in the question.

Answer 3

ok, first, don't be discouraged! that isn't any longer a stressful subject as quickly as you realize the thank you to innovations-set it. enable me to offer you the achieved working: First, you will in all probability prefer to place in writing down down the equation of around action centripetal acceleration = speed^2 / radius Now, in case you view the section station end-on, you will in basic terms see a circle. This circle is rotating. as a fashion to offer an effect resembling gravity, we require acceleration via employing gravity = speed^2 / radius g = v^2 / r we can paintings out the radius, with the aid of actuality we've the diameter. so which you will in all probability prefer to paintings out v = sqrt(rg), remembering to maintain on with the surprising gadgets, and you gets the rotation speed mandatory. in spite of the indisputable actuality that, from the form you have phrased the question i think you prefer the angular rotation speed. in basic terms use v = r * w the area w is the angular rotation speed. Then in basic terms convert it to revolutions consistent with day, and you have your answer!! i desire that helps - it particularly is a splash common subject as quickly as you realize the thank you to innovations-set it.