Jupiter is up to 58 moons! (more likely)

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  • #1
Phobos
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Main Question or Discussion Point

egad!

http://www.cnn.com/2003/TECH/space/04/05/jupiter.moons.ap/index.html [Broken]
 
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  • #2
drag
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Great ! There's a moon for everyone !
(After you buy it from Dennis Hope... )
 
  • #3
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This shows how miserable our planet is, with only one moon :frown:
Probably some of them have been captured by Jupi from the asteroid belt.
 
  • #4
russ_watters
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The moons are tiny, perhaps just a mile or so across, and orbit Jupiter at a distance of tens of millions of miles.
So at what point do you consider it not a moon and just-a-hunk-of-rock-near-Jupiter?
 
  • #5
Phobos
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Originally posted by russ_watters
So at what point do you consider it not a moon and just-a-hunk-of-rock-near-Jupiter?
Somewhere between that small size and the smaller sizes of stuff that make up planetary rings. I suppose small stuff around a planet is either (1) a moon (2) part of a ring system or (3) negligable.

Yet another astronomical gray area. (giant planet/brown dwarf, comet/asteroid, planet/KBO, etc.)
 
  • #6
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Isn't it very difficult to get time on big telescopes these days? :wink:
 
  • #7
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I think a "moon" should have enough gravitation to maintain (hypothetically) a binary system with an equivalent body.
 
  • #8
Nicool003
I think these are very small to be considered moons... But hey we are up to 58! If they say it's official let's move on to numbers 59,60, and 61 shall we?!:wink:
 
  • #9
BoulderHead
Originally posted by Loren Booda
I think a "moon" should have enough gravitation to maintain (hypothetically) a binary system with an equivalent body.
I'd be happy with just enough gravity to keep me from jumping off into outer space.
How big would that be anyway?
 
  • #10
FZ+
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Let's see....
Escape velocity = [squ] (2 * G * mass / distance from centre of mass)

Consider an adequate escape velocity to be, say, 2 m/s...

average density of asteroids = mass/volume = 2120 kg/m^3(based on ceres)

therefore: (assuming a cube shaped asteroid )
mass/distance = density * distance ^2

2 = [squ] (2*6.67*10^-11 * 2120 * d^2)
d = [squ] (2 * 10^11 / 14140)
d = 3.76 * 10^3 m

So the minimum size of cube shaped asteroid from which you cannot jump off is one of side 3.7 kilometers.
 
  • #11
Nicool003
Darn FZ looks like you have to do that all over again... You see we were talking Miles :wink:
 
  • #12
BoulderHead
Originally posted by FZ+
Let's see....
Escape velocity = [squ] (2 * G * mass / distance from centre of mass)

Consider an adequate escape velocity to be, say, 2 m/s...

average density of asteroids = mass/volume = 2120 kg/m^3(based on ceres)

therefore: (assuming a cube shaped asteroid )
mass/distance = density * distance ^2

2 = [squ] (2*6.67*10^-11 * 2120 * d^2)
d = [squ] (2 * 10^11 / 14140)
d = 3.76 * 10^3 m

So the minimum size of cube shaped asteroid from which you cannot jump off is one of side 3.7 kilometers.
All right, thanks!

My last question involves a statistical analysis to determine the likelihood that if we each lived alone on our own little moons we'd still end up taking aim with our squirrel guns and firing pot shots at our neighbors.
 
  • #13
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BoulderHead
My last question involves a statistical analysis to determine the likelihood that if we each lived alone on our own little moons we'd still end up taking aim with our squirrel guns and firing pot shots at our neighbors.
Reminiscent of the fantastic movie featuring Baron von Munchausen.
 
  • #14
BoulderHead
  • #15
Janus
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Originally posted by BoulderHead
I'd be happy with just enough gravity to keep me from jumping off into outer space.
How big would that be anyway?
Going about it slighty diferently than has already been done:

Vesc = [squ](2GM/d)

M = VD (volume x density)

Assuming a spherical (or at least roughly so) body,

V = 4[pi]r³/3

and M = 4[pi]r³D/3

then

Vesc = [squ](8G[pi]r³D/3d)

At the surface, d =r, so

Vesc = [squ](8G[pi]r³D/3r) = [squ](8G[pi]r²D/3)

= r[squ](8G[pi]D/3)

solve for r

r = Vesc/[squ](8G[pi]D/3)

For the density assumed, r = 1.837 km or 1.14 miles
 
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  • #16
BoulderHead
I very much appreciate everyone's time!

I'd like to say a big 'Thank you' to FZ+ and Janus for working this question out for me.:smile:

I know this thread was about all those moons and gadzooks, 58 is a bunch of moons, but now I'm wondering something;

1) How high might an average person jump, and how rough would the impact be upon landing?

2) If you poured a gallon of water out onto the soil would anything out of the ordinary happen?

3) I'm assuming that the core of this little moon could not be molten and that it should be easy enough, judging from the depths attained here on Earth, to drill a hole completely through the center and out the other side. What would happen if someone jumped down this hole? I mean, would you shoot out the other side into the atmosphere (well, not that there really is any atmosphere worthy of note on this tiny hunk of material) before falling back down through it again, and would you eventually come to rest at the center (just 'floating')?

[edited]
On question 3 I'm thinking you wouldn't make it much further on the far side than you were from the surface where you jumped, that is, like a pendulum you lose a little 'ground' each time you pass the center.
 
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