how many days would it take for a certain planet to make a full revolution? Wat would be it's orbital speed?

Question

Jupitar's rocky, 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 mean distance between Io and Jupitar is 4.22 x 10^8 m. 

a. If Io's orbit were circular, how many days would it take for Io to complete one full revolution around Jupitar?
b. If Io's orbit were circular, wat would its orbital speed be?

so I have Io's radius to be 1.82 x 10^6
mean dist between Io and Jupitar is 4.22 x 10 ^ 8m. 
Radius of Jupitar is 2.0 x 10^8m. 
mass of jupitar is 1.9 x 10^7kg.
mass of earth's moon =(?) mass of Io = 7.35 x 10^22. 

Now the answers are a) 2.25 days and b) 1.6 x 10^4 m/s. 

I THINK I should use the following formulas: T = (2pir)/(Orbital velocity) and Orbital velocity^2 = (GM)/r.

I think these are the right formulas but when I think I am putting in the wrong numbers. Because I am not getting my answers. All help will be appreciated! I am trying not to cheat on my hw but I am really confused.

DH · Accepted Answer

===>>>Your mass of Jupiter is wrong Jupiter's mass = 1.90x10^27kg <<====

Now use GMm/r^2 = m*v^2/r So GM/r = v^2 But v = (2*pi*r)/T So

GM/r = 4*pi^2*r^2/T^2 So T = sqrt([(4*pi^2/GM)*r^3]

= sqrt[(4*pi^2/(6.67x10^-11*1.9x10^27))*(4.22x10^8)^3] = 1.53x10^5s = 1.53x10^5/86400s/day

= 1.77days

Then v = 2*pi*r/T = 2*pi*4.22x10^8/1.53x10^5 = 1.73x10^4m/s

? · Answer

None of those concepts, considering the fact which you're using the incorrect equation. the in basic terms right equation is F = ma (no longer V) = GMm/r^2. because of the fact m = mass of the planet is on the two factors of the equation, they cancel out so the equation turns right into a (acceleration, no longer speed) = GM/r^2. because of the fact planetary orbits are elliptical, the speed of a planet isn't consistent. you will desire to start up with Kepler's 2d regulation of planetary action, that a planet sweeps out equivalent aspects of an ellipse in equivalent quantities of time to make your suggestions up the speed of the planet. All you desire for that's the gap of the planet from the solar at 3 or extra diverse circumstances, alongside with a( for acceleration, no longer the semi-significant axis of the ellipse) a = GM/r'2. the different mistake you made grew to become into taking the sq. root of GM. Taking the sq. root of r would not make any experience in any respect. Edit: Caleb is sturdy. You needto use kepler's 0.33 regulation too.

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how many days would it take for a certain planet to make a full revolution? Wat would be it's orbital speed?

2 Answers