# Thread: Tidal Locking

1. Oops!!! I was off by a factor of a thousand! Make that "350-400 thousand miles (560-640 thousand Km).

2. big oops.

:surprised

3. [Stuff] happens!

4. Originally Posted by Kaptain K
Oops!!! I was off by a factor of a thousand! Make that "350-400 thousand miles (560-640 thousand Km).
which is only about half again as far as it is right now, so it's still in our neighborhood.

5. Yep! And still firmly "in the grasp".

6. At that distance, the tidal effects from the Moon will be less than half of the effects from the Sun.

7. Sounds right.

After that the tidal lobe cause by the sun will act as a periodic influence on the orbit parameters of the Moon instead of a secular one. The semi-major axis should stabilize while the other numbers bounce around instead.

8. Join Date
Dec 2004
Posts
16
Thanks for yuor responses. .

9. Established Member
Join Date
Dec 2001
Posts
199
If we had a sun with a single planet in orbit around it (with the planet spinning on its axis faster than the sun was spinning its axis), and that sun was completely elastic to tidal forces caused by the orbiting planet whereas the planet was not completely elastic to tidal forces caused by the sun, this should eventually lead to the same side of the planet always facing the sun *without* an increase in the orbital distance of the planet from the sun.

Is that correct?

(edit: added "with the planet spinning on its axis faster than the sun was spinning its axis")

10. well, you can see that Mercury already exhibits something similart to that.

wouldn't happen on the earth however, our moon has stronger effects than does the sun.

11. Originally Posted by DALeffler
If we had a sun with a single planet in orbit around it (with the planet spinning on its axis faster than the sun was spinning its axis), and that sun was completely elastic to tidal forces caused by the orbiting planet whereas the planet was not completely elastic to tidal forces caused by the sun, this should eventually lead to the same side of the planet always facing the sun *without* an increase in the orbital distance of the planet from the sun.
I don't think this is correct. As I see it, if the Sun responds completely elastically, then its rotation is irrelevant, its bulges should line right up with the planet and never cause any torques. The planet is not elastic, so its bulges get out of line. This causes an orbital torque as well as a torque against the planet's spin. As a closed system, these torques compensate and there is no change in total angular momentum. If the planet's spin is reduced (which happens if the spin is faster than the orbit), then you'll get more orbital angular momentum, and the separation increases. The opposite holds if the planets spin is slower than its orbit. As I see it, the Sun can spin all it likes, it won't matter if its elastic. But you are right that eventually the same side will always face the Sun, and it will be tidally locked. This all assumes a circular orbit-- since Mercury is in an elliptical orbit, it actually flips the side that faces the Sun each time it reaches perihelion.

12. Established Member
Join Date
Dec 2001
Posts
199
I don't think this is correct. As I see it, if the Sun responds completely elastically, then its rotation is irrelevant, its bulges should line right up with the planet and never cause any torques.

I see this: If the sun is completely elastic, rotation is irrelevant.

This causes an orbital torque as well as a torque against the planet's spin.

This I don't see: If the tidal bulges on the sun caused by the planet are not swept forward in the direction of the suns rotation due to the suns rotation and inelasticity, how is there a torque delivered to the orbital motion of the planet?

The planet loses rotational speed around its axis due to its internal friction caused by the suns tidal forces. Is there any angular momentum left to conserve?

13. Torque on the planet is delivered through the bulge in the planet, interacting with the sun's point gravity. There is no need to have a bulge on the Sun. Since the system is closed, the sum of the orbital angular momentum, the rotational angular momentum of the planet, and the rotational angular momentum of the Sun, is a constant number. Since nothing is happening to the rotation of the Sun, the sum of orbital plus planet angular momentum is fixed. If the planet rotation goes away (which it would not completely, it would eventually equal its orbital period), this has to be picked up by the orbit, causing it to spiral outward since that what happens to an orbit that picks up angular momentum. It would not spiral outward indefinitely however-- there is likely not enough energy in the planet rotation to allow it to escape orbit. Eventually it would reach a tidally locked orbit.

14. Man, I see things completely different.

First of all, the sun isn't completely elastic. It is not a liquid and it suffers from internal friction, just like any other body.

As most of us know, the earth's high tides do NOT line up with the moon. internal friction actually causes the high tides to be in front of the lunar line.

The sun works similarly.

15. My response was to DALeffler's previous posting. I should have embedded a quote to clarify this. The hypothetical issue was about what would happen even in the absence of a solar bulge, but you are right cross that in reality there is a bulge and there is a slowing of the solar rotation as a result (negligible in practice, of course).

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•