Gravitational force on Io

[eris]

I have recently read that the gravitational force between Earth and Moon is something like 2 x 10^20, and we get tides.

What is the force on Io which creates all the activity?

[bystander]

I have recently read that the gravitational force between Earth and Moon is something like 2 x 10^20, and we get tides.

What is the force on Io which creates all the activity?

Io has an eccentric orbit around Jupiter, due to orbital resonances with Europa and Ganymede. This eccentricity, along with Jupiter's immense gravitational pull causes a large difference (100 m) in the tidal bulge on Io between the points of its least and greatest distance from Jupiter. The stress on Io causes it to heat up internally and that heat is released as volcanic activity. This is known as tidal heating.

See http://en.wikipedia.org/wiki/Io_(moon)#Tidal_heating

[Chris Peterson]

I have recently read that the gravitational force between Earth and Moon is something like 2 x 10^20, and we get tides.

What is the force on Io which creates all the activity?

Tides are not determined just by the gravitational force, but by the gradient of that force. In other words, we see a lunar tide on Earth because one side of the planet is significantly closer to the Moon than the other side, and therefore each side experiences a different gravitational force. The force of gravity between the Earth and Moon is 1.98e20 N, and the force of gravity between the Earth and Sun is 3.52e22 N. However, in calculating the tidal force, you need to multiply this by 2r/R, where r is the radius of the body you are calculating the force on, and R is the distance between the two bodies. In the case of the Earth-Moon, you get a tidal force of 6.58e18 N, and in the case of the Earth-Sun, 2.98e18 N. So we see that even though the force of gravity between the Earth and Sun is much larger than between the Earth and Moon, the tidal force on the Earth is still greater from the Moon than the Sun.

Considering Jupiter-Io, the gravitational force is 6.35e22 N, similar to the gravitational force between the Earth and Sun. But the tidal force between the two is a whopping 5.48e20 N, 83 times higher than the tidal force exerted on the Earth by the Moon.

But there is one more thing to consider. Io is nearly tidally locked to Jupiter, meaning it always keeps the same face towards the planet. If the tidal locking were perfect, there would be no heating of Io's interior by the tide, since the bulges would always be in the same location. But Io isn't orbiting Jupiter in isolation; Europa and Ganymede in particular influence its orbit. These other moons have prevented Io's orbit from becoming perfectly circular. Because the orbit is eccentric, the tidal forces raise and lower tidal bulges just because the distance from Jupiter is changing- even without rotation of the satellite itself with respect to Jupiter. Also, the eccentric orbit prevents perfect tidal locking, so Io experiences libration just like the Moon does- from Jupiter's view, the satellite appears to wobble a bit. This also causes the tidal bulges to move around, dissipating energy in Io and producing internal heating.

Hope that's not more detail than you were looking for... I'm a physicist, so I actually enjoy doing these kinds of calculations over my morning coffee <g>.

[eris]

bystander, Chris, thanks you both for those answers.

Now, if 2r/R is the calculation for the tidal force, then the force on the moon is about a quarter of the force on Earth, isn't it? Because the r of the moon is about a quarter of the r of Earth. If I understand your comment about tidal locking correctly, the moon would be tidally locked to the Earth, and so there would be 'no change' to the moon caused by the tidal force, were it not for the effects of triple libration.

Can you give me the one coffee answers to these two questions, please? [I don't think I could cope with the two coffee answer yet] I'll get a nice hot cup of tea to read your answers.

What effect does the tidal force of the earth have on the moon?
How does the tidal force of the sun on the moon affect the moon?

I have questions about the forces that determine where natural satellites situate themselves, but they will have to wait.

[Chris Peterson]

Now, if 2r/R is the calculation for the tidal force, then the force on the moon is about a quarter of the force on Earth, isn't it? Because the r of the moon is about a quarter of the r of Earth.

That's right. Because the Moon is smaller than the Earth, it has a smaller force gradient across it (a point source experiences no tidal force at all, no matter how large a gravitational field it is in).

If I understand your comment about tidal locking correctly, the moon would be tidally locked to the Earth, and so there would be 'no change' to the moon caused by the tidal force, were it not for the effects of triple libration.

Also correct. The Earth raises tidal bulges on the Moon, but they don't travel much because the Moon's rotation is nearly locked to its orbit. Like Io, however, the Moon is in an eccentric orbit, so the magnitude of tidal bulges increases and decreases with the Moon's distance from Earth.

What effect does the tidal force of the earth have on the moon?

As you note, the tidal bulges raised on the Moon by the Earth do move around a little because of libration, and also in magnitude because of an eccentric orbit. That means that lunar material is being stretched and relaxed, which means that energy is being dissipated. So the Moon is experiencing some tidal heating from Earth tides. It isn't generally considered significant- certainly not enough to result in volcanic or tectonic activity. But it is considered in studies of lunar evolution, since in the distant past the Moon was much closer to the Earth, and tidal heating was probably significant.

How does the tidal force of the sun on the moon affect the moon?

Short answer: in a complex way. There will be a pair of tidal bulges, which are not fixed. So there is some tidal heating produced. Also, because the Moon orbits the Earth, its distance from the Sun is always changing. The solar tide is slightly altering the Moon's orbit around Earth. In fact, most three-body systems are not perfectly stable. Given enough time, the bodies will enter into new orbits, or a body might even be ejected from the system entirely. I believe that numerical simulations of the Earth-Moon-Sun system suggest that the system, while not truly stable, is not likely to experience any kind of radical adjustment over the lifetime of the Sun.

[eris]

Thanks Chris. I don't want to impose upon you further, BUT I want to understand why the Earth is where it is, and the day and year are as long as they are.

Can you tell me where to look to read about the elements of celestial mechanics that determine the orbit and speed of circulation and spin of a satellite, such as the Earth?

[Chris Peterson]

Thanks Chris. I don't want to impose upon you further, BUT I want to understand why the Earth is where it is, and the day and year are as long as they are.

Can you tell me where to look to read about the elements of celestial mechanics that determine the orbit and speed of circulation and spin of a satellite, such as the Earth? :?:

Celestial mechanics allows you to specify orbits, and to understand how the masses and velocities of objects determine their orbits. But it can't tell you why the Earth is where it is, or how the day and year got to be what they are. Those things fall in the category of "just happened"- a consequence of the way the Solar System formed. The year is the length it is because of the distance of the Earth from the Sun (which has no special significance), and the day is what it is because of the angular momentum of the material that formed the Earth (also, no special significance), the Earth's subsequent collision history, and the effects of the Moon.