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luckyfrank
2011-Jan-05, 09:49 AM
So we know that jupitar's gravity pulls and streches the moon's closest to it causing gravitational tidal heating and in some cases fractures the surface of it's moons, my question is if an astronaut were to set foot on one of these moons what would jupitar's immense gravity feel like for him

Jens
2011-Jan-05, 10:22 AM
Interesting question. In a sense, a first answer would be that since the planet is of course free-falling, the astronaut would not feel the gravity. But then I suppose the issue is, where is the astronaut standing on the planet? If the astronaut is standing on one of the poles (assuming the planet is in line with Jupiter), then he would feel nothing. If he were on the equator, he would feel lighter and heavier depending on whether he is on the near side or the far side as the planet rotates. I don't know how much difference it would make, though. It would depend on how far from Jupiter the planet is, not really the strength of the gravitational field alone. After all, the sun is much, much more massive than Jupiter, and yet we don't feel any difference as the earth spins, because the difference is so tiny at this distance.

Also, the title should read "its close moons," without an apostrophe. . .

StupendousMan
2011-Jan-05, 12:46 PM
I grabbed numbers from The Nine Planets, http://nineplanets.org/.

A person of mass 100 kg standing on the surface of Io would have a weight of 179 Newtons, if we consider Io's gravitational pull alone.

If the person were on the side of Io facing Jupiter, the gravitational pull of Jupiter would be about 72 Newtons,
pulling him up and away from the surface. He would feel much lighter.

If the person were on the side of Io facing AWAY from Jupiter, the gravitational pull of Jupiter would about 71
Newtons, pulling him down and into the surface. He would feel much heavier.

baric
2011-Jan-05, 05:07 PM
I grabbed numbers from The Nine Planets, http://nineplanets.org/.

A person of mass 100 kg standing on the surface of Io would have a weight of 179 Newtons, if we consider Io's gravitational pull alone.

If the person were on the side of Io facing Jupiter, the gravitational pull of Jupiter would be about 72 Newtons,
pulling him up and away from the surface. He would feel much lighter.

If the person were on the side of Io facing AWAY from Jupiter, the gravitational pull of Jupiter would about 71
Newtons, pulling him down and into the surface. He would feel much heavier.

Interesting! I wonder if this means that the level volcanic activity on Io fluctuates according to its rotation with respect to Jupiter? Or perhaps it is already tidally locked because of this pull.

antoniseb
2011-Jan-05, 06:39 PM
A person of mass 100 kg standing on the surface of Io would have a weight of 179 Newtons, if we consider Io's gravitational pull alone. If the person were on the side of Io facing Jupiter, the gravitational pull of Jupiter would be about 72 Newtons, pulling him up and away from the surface. ...

My first reaction was "that can't be right" ... though I was aware that Io is not THAT far from the Roche limit. So I calculated it, and got more or less the same numbers. I got a downward force of 182 Newtons at the poles, but I could have made some excess simplifying assumptions. This is cool to think about.

NEOWatcher
2011-Jan-05, 06:52 PM
Also, the title should read "its close moons," without an apostrophe. . .
Interesting that you caught that, and not the spelling of the planet's name.

baric
2011-Jan-05, 07:41 PM
My first reaction was "that can't be right" ... though I was aware that Io is not THAT far from the Roche limit. So I calculated it, and got more or less the same numbers. I got a downward force of 182 Newtons at the poles, but I could have made some excess simplifying assumptions. This is cool to think about.

Ya, native Ionians could use a slingshot to launch probes to Jupiter :P

IsaacKuo
2011-Jan-05, 07:59 PM
The answer is most emphatically NO.

Io has an orbital period of 1.769 days and orbital radius of 421700km, meaning that centripetal acceleration is r*(2pi/period)^2 = 0.71m/s/s.

The tidal force is easiest to calculate if the person is standing on the great circle perpendicular to Jupiter. This includes the poles. The person sees Jupiter on the horizon--half of Jupiter is above the horizon; half is below. This is where the tidal force is strongest downward.

When standing at one of these locations, the person has the same distance from Jupiter as Io's centerpoint. His circular motion thus has the same radius, so the centripetal acceleration is the same--0.71m/s/s. The only difference is a slight difference in direction. The difference between those two vectors is the tidal force experienced.

Io's radius is 1821.3km, so the difference in direction is 1821.3/421700 = 0.00432 radians. This implies a tidal force of only 0.00432*0.71m/s/s = 0.0031m/s/s, or 0.31 milligees. That is not perceptible. Contrast this with Io's own surface gravity of 1.8m/s/s, and the tidal force is swamped!

What about the locations on the bulge nearest and furthest from Jupiter? These two locations, on the equator, represent the two points where the tidal force is strongest in the upward direction. Gravitational acceleration is inversely proportional to the square of the distance, so we end up with accelerations of:

0.71m/s/s*421700^2/(421700+1821.3)^2 = 0.7039066m/s/s
0.71m/s/s*421700^2/(421700-1821.3)^2 = 0.7161728m/s/s

Subtracting the centripetal acceleration of 0.71m/s/s, we get a tidal force of 0.0061m/s/s, or 0.62 milligees.

The bottom line is that the tidal force is not humanly perceptible. It's totally swamped by Io's own gravity of 1.8m/s/s.

Ilya
2011-Jan-05, 08:27 PM
Also, this 0.62 milligees is in the direction away from Io's center in both sub-Jovian and anti-Jovian points. In BOTH places you are lighter (by a very very small amount) than at Io's poles.

That may be the hardest part for people to grasp about the tides:

http://curious.astro.cornell.edu/question.php?number=143

StupendousMan must have gotten his numbers by imagining a sphere with mass of Jupiter and size of Io's orbit, and calculatins its surface gravity. That's not how tides work.

Hornblower
2011-Jan-05, 08:27 PM
I grabbed numbers from The Nine Planets, http://nineplanets.org/.

A person of mass 100 kg standing on the surface of Io would have a weight of 179 Newtons, if we consider Io's gravitational pull alone.

If the person were on the side of Io facing Jupiter, the gravitational pull of Jupiter would be about 72 Newtons,
pulling him up and away from the surface. He would feel much lighter.

If the person were on the side of Io facing AWAY from Jupiter, the gravitational pull of Jupiter would about 71
Newtons, pulling him down and into the surface. He would feel much heavier.
My bold. No, he would seem lighter than at a polar location, just as with the near side. Jupiter's gravitational action is strongest on the near side, intermediate at the center, and weakest on the far side. It "pulls" an object on the near side away from Io, and it pulls Io away from an object on the far side. That is why there are tidal bulges on both sides.

As Isaac just pointed out, it is not the absolute value of Jupiter's gravitational action that counts, it is the gradient across Io that determines the amount of tidal stress.

publius
2011-Jan-05, 09:37 PM
I grabbed numbers from The Nine Planets, http://nineplanets.org/.

A person of mass 100 kg standing on the surface of Io would have a weight of 179 Newtons, if we consider Io's gravitational pull alone.

If the person were on the side of Io facing Jupiter, the gravitational pull of Jupiter would be about 72 Newtons,
pulling him up and away from the surface. He would feel much lighter.

If the person were on the side of Io facing AWAY from Jupiter, the gravitational pull of Jupiter would about 71
Newtons, pulling him down and into the surface. He would feel much heavier.

Be careful. :) Your calculations would be correct *if Io were be held stationary against Jupiter's field*. But it's not, as Io is in free fall. It's an orbital free fall, but locally the same as a simple radial free fall. The only effects can be that of the *tides* in Jupiter's gravity.

This is just our old friend, the Equivalence Principle, at work here. Consider a rigid (non-rotating to keep things simple) gravitating sphere in free fall in a uniform gravitational field. The magnitude of surface gravity of the sphere is 'g_s' and the external uniform field is g_e. Consider observers standing on the surface of the sphere on opposite ends of diameter aligned with g_e, the uniform external field.

The sphere is free falling at g_e. The "force you feel" is of course the deviation from the geodesic. If your acceleration is 'a', then the deviation from the geodesic is simply a - g, where g is the geodesic. On the surface of the earth, using an earth fixed coordinate system, the geodesic is -g, and a is 0, since we're stationary with the earth. So we have a - g = 0 - (-g) = +g, that is 1g *up*. We're being pushed up against our geodesic and feel
1g of acceleration.

Now, do this at the ends of that diameter for the sphere in free fall. On the side facing the direction of g_e, the total geodesic g = g_e - g_s, since they point in opposite directions. But since the sphere is rigid and free falling, 'a' is simply g_e. So the deviation form geodesic is
a - g = g_e - (g_e - g_s) = g_s. On the opposite side, g = g_e + g_s, since they point in the same direction and a - g = -g_s.

Note g_e cancels, and if you do it vector form, it cancels everywhere on the surface of the sphere (and inside) since we always have g_e - g_e in the expression. We feel nothing from the external field g_e, since we're everywhere free-falling in it. We feel only the local gravity of the sphere, which is the only thing we're resisting, the only contribution to the total geodesic we're deviating from.

The uniform external field can be completely transformed away. Free falling in it is the same as the sphere alone in free space.

For Io, the only thing that matters is Jupiter's tides, the difference in g_e at different points on the sphere. Tidal deformations make things get complicated, but you can consider the simple case of a rigid sphere free-falling in a tidal field. The sphere will free fall at the value of g_e at the center of the sphere (center of mass for a general rigid body).

-Richard

Sauna
2011-Jan-05, 11:40 PM
My bold. No, he would seem lighter than at a polar location, just as with the near side. Jupiter's gravitational action is strongest on the near side, intermediate at the center, and weakest on the far side. It "pulls" an object on the near side away from Io, and it pulls Io away from an object on the far side. That is why there are tidal bulges on both sides.

As Isaac just pointed out, it is not the absolute value of Jupiter's gravitational action that counts, it is the gradient across Io that determines the amount of tidal stress.

Actually, since the question is, "does a human feel it?", the gradient across a human being should be considered, NOT Io, and it will be even smaller.

Sauna
2011-Jan-05, 11:41 PM
My bold. No, he would seem lighter than at a polar location, just as with the near side. Jupiter's gravitational action is strongest on the near side, intermediate at the center, and weakest on the far side. It "pulls" an object on the near side away from Io, and it pulls Io away from an object on the far side. That is why there are tidal bulges on both sides.

As Isaac just pointed out, it is not the absolute value of Jupiter's gravitational action that counts, it is the gradient across Io that determines the amount of tidal stress.

Actually, since the question is, "does a human feel it?", the gradient across a human being should be considered, NOT Io, and it will be even smaller.

WayneFrancis
2011-Jan-06, 01:16 AM
So we know that jupitar's gravity pulls and streches the moon's closest to it causing gravitational tidal heating and in some cases fractures the surface of it's moons, my question is if an astronaut were to set foot on one of these moons what would jupitar's immense gravity feel like for him

Depends on what effect you are talking about. If the person was on the side facing Jupiter would they feel "lighter" then if they where on the opposite side? I'll go away and do the maths for that but I suspect that the difference would be on the order of mangitude of being on a mountain or being at sea level....Went away and did the maths and got roughly the same answers as you see here. I did it for an orbiting and non orbiting body and you do get 2 very different answers. I didn't think that the none orbiting body's difference was going to be that big. Interesting as IO is bigger then our moon, slightly, and I didn't think the effect was that big but it is VERY close to Jupiter. But if you are asking if they'd feel a tidal force as the planet does then no. The difference over that 2m is not noticeable by a person at those locations.

StupendousMan
2011-Jan-08, 05:28 PM
Be careful. :) Your calculations would be correct *if Io were be held stationary against Jupiter's field*. But it's not, as Io is in free fall. It's an orbital free fall, but locally the same as a simple radial free fall. The only effects can be that of the *tides* in Jupiter's gravity.

-Richard

Darn you, equivalence principle!

I _knew_ that those values couldn't be right. Grrr.

whatdoctor
2011-Jan-21, 05:19 AM
I think you are missing the centrifugal effect here - which counteracts the gravity - basically saying it in freefall. The difference is the 1 Newton - not a lot really (1.4%). i don't think you would notice.