Earth's velocity is 30 km/s.
What would be Earth's velocity without the moon ? What's the formula?
velocity with respect to what?
Our revolution speed around the sun? Unchanged.
Would Earth spin much faster without its orbiting moon ?
The pull of the moon, puts the brakes on Earth's daily rotation, right ?
Is there a way to calculate how fast the Earth would be spinning on its axis, If we didn't have the moon ?
Probably not. Three-body problems are very difficult to solve analytically, so this wouldn't be a simple calculation.
Someone correct me if I'm wrong, but as I understand it, the gravitational force of the moon causes the earth to rotate about an axis perpendicular to the center of mass between the earth and the moon. So I suppose that without a moon, the earth would rotate about its geographic axis instead.
deleted: missed the change in subject from the OP...
Don't know if there is a way to calculate it because the sun also has an effect on this.
Angular momentum is what physicists call for the Earth-Moon system zero-sum game. Currently, the Moon's distance from the Earth is increasing by about 3 centimeters per year. Since the Earth's angular momentum is decreasing, the Moon's angular momentum must increase to keep the overall angular momentum of the Earth-Moon system the same.
Days are increasing in length.
If the moon were just to disappear, then what would happen to the angular momentum ? Do you suppose that Earth would slow down ?
If the moon were to suddenly disappear the length of the day would not change.
However, the angular momentum stored in the Earth-Moon pair would change the Earths orbit somewhat.
Depending on just when the moon disappeared the year might get longer or shorter or the orbit might become more elliptical.
This could be computed, with some degree of accuracy, if you care to learn orbital mechanics.
if L means angular momentum.
L(no moon) = L(earth+moon) + L(moon) + L(moon's orbit)
If we toss in L=Iw, hence we can figure out how fast Earth would be spinning.
The Barycentre Eart-Moon system will change, the new barycentre will be at the centre of mass of the Earth.
If the moon "just disappears", why would it leave behind its angular momentum? I know that angular momentum is conserved, but so are a lot of other things whose conservation would be violated by the moon "just disappearing.
Can't change, Orbit speed is nothing to do with the mass of the object.
Check the space shuttle or the International Space Station, - the astronauts float around inside it at same speed. Identical orbital speed is same reason they experience "weightlessness"
Check Keple's laws (which say the same as Newton's inverse square law)
This is the basis of general relativity too. The object's mass is not descernable from its motion in space.
Or in simpler terms, Galileo would have been wrong when he said that all ojects fall to earth at same speed regardless of their mass.
Aren't you people over complicating things ?
Finally, the only thing that really changes is the position of the centre of mass. That WOULD affect the orbit because the mass has 'moved' to a different position (carries the same speed). But to be more exact, the orbit speed only changes if the orbit position of centre of mass becomes higher or lower (ie closer in or further out from the Sun)
So it actually depends entirely on WHEN the Moon disappears. And nothing else.
Firstly the earth does not have a velocity, it has a speed. velocity requires a direction.
secondly, the only thing that will affect us is less light during the night and max/min tide heights.
the net force of the earth onto the moon is zero. the moon moves in a circular motion, if you take the infinately many directions of the the force of the moon, it should add to zero.
our earth actually moves around the sun in a sine/cos motion due to the moons gravitational force. but on average the speed of earth is unchanged
Oops, sorry didn't realise a couple of other people (no time) had already said that it only depends on the exact moment the moon disappears. If the timing is right, it could have no effect whatsoever on the orbit time of the earth (length of a year). This is quite an interesting question in my opinion.
You just put the average distance (semi-major axis) of the Earth from the Sun into Kepler's 3rd law T=k.r^3/2
Probably the simplest way to get the exact solution.
Mmm, if the moon has the ability to slow down the earth's rotation due to its mass then by the same token the earth must have a similar effect on the moon's rotation.
Now, the moon does rotate, but at a rate that makes it appear stationary to an observer on the earth. Therefore by the thoughts above the earth ought to influence the lunar rotation and slow it down, this seems common sense to me, but no-one has ever seen the dark side except till the space age. The moon's rotation is obviously static, but the earth's is changing.
Conclusion - it can not be 'lunar' tidal forces that are retarding the earth 's rotation.
Three body problem? nah bruv, common sense problem more like.
earth to moon = 384 403 km
earth to sun = 150 000 000 km
Earth mass 5.9742 × 10^24 kilograms
Moon 7.36 × 10^22 kilograms
centre of mass is .'. = 384,403 * M/(M+E) from Earth
= 4678 km from centre of the Earth
Worst case is when the Earth is furthest from sun (moon closest) and r -> r + 4678 km (lol)
so where r = normally (on average) 150 000 000 km,
from Kepler #3,
T' = T * (r'/r)^3/2
= 365 * (150,004,678/150,000,000) ^3/2
(assumed a typical year is exactly 365 days)
so maximum difference is about 0.01707510616413248734767419157 of a day, per year.
= 24 minutes 35 seconds (not much)
minimum = 0 seconds
(p.s. I don't think tidal effects make any difference whatsoever, though I'm not absolutely certain of that.)
cheers, Yellow Taxi
edit: corrected earth to moon distance
Why not?. Where does the tide get the energy everyday? if not from the Moon's motion. In fact, the mutual forces between Moon and Earth are not conservative any more because of the water layer on earth. It's something soft, not elastic.
It is the difference between rotational period and orbital period that causes the braking effect.
Yes, the Earth has an effect on the Moon's rotation. This effect is the very reason the moon rotates at the same rate as it orbits. If the moon were rotating faster than it orbited the tidal interaction with the Earth would slow its rotation down, if it rotated slower, the same interaction, would speed the Moon's rotation up. In both cases the end result will be the moon ending up in the situation it is in now.
In the case of the Earth, the slowing of Earth's rotation will stop once its rotation matches the orbit of the Moon and it continually presents one side to the Moon.
I don't see what difference the orbital speed or rotation/spin of anything makes to this. Only a change in the centre of mass is going to affect the trajectory.
The way the moon and earth interact is like they are just one & the same planet moving in a perfect ellipse around the Sun (practically a circle actually). Centre of mass is always right on the ellipse, simple as that.
Add tidal effects if you must, but that don't make any difference to the centre of mass. Not even if the Earth and Moon stopped spinning & turning around each other altogether, the centre of mass would still be in exactly the same place, period.
It's like trying to argue that the orbital speed of a planet around the Sun can be affected by its spin rate on its own axis. It can't.
That would be true if objects were point masses, but they aren't.
When a cylinder or any long object is orbiting, there's a stable mode where one side of the object will always point to the center of mass of the object it's oribting. NASA twice attempted to duplicated this effect with a space tether and a very long wire, but in both cases the spool of wire jammed before it could be released to any appreciable length. Here's one link about this:
Although the moon isn't a cylinder, it's distribution of mass isn't perfectly spherical, so the reason that it's rate of rotation matches it's rate of orbit is the same as that for a cylinder or space tether.
So the real question is what was the moons original orbit speed and what slowed it down to the point it would stabilize into a orbital rate of rotation. This would require some component of drag.
I don't see how tidal effects on the earth could affect the moon, but tidal effect on the moon could slow it down. Maybe the moons interior isn't solid. Maybe it's small atmosphere, some of which is due to captured solar winds provided enough drag over time to slow it down.
Regarding tidal effect on the earth, the moons "response" is to increase distance from the earth. This is because the earths center of mass is slightly moved towards the moon due to tidal motion, and the moons orbit is relative to the slightly "orbiting" center of mass of the earth. I reiterate, I don't see how this could affect the rate of rotation of the moon.
Tidal stresses affect not only bodies of water but, the crust moves as well.
If the moon rotates relative to the Earth, not the case now.
Then it will experience tidal stresses, even without an ocean.
The stresses cause friction, turning the angular momentum into heat.
An extreme example is shown in the Roche Limit.
Agreed, I was just pointing out that tidal stresses on the earth aren't going to affect the rate of rotation of the moon.
No it's true for any object, not just point masses. You don't need to know the distribution of mass on a planet to use Newton's Law of gravity do you, - You only locate it's centre of mass and treat it just like a point mass. Newtons gravity law would be incredibly complicated otherwise wouldn't it
And any physics problem is vastly simplified by analysing the motion of the centre of mass of the system.
In the absence of a gravitational field, the c-o-m of any system moves in a perfect straight line. When the space shuttle blasts off from Houston, the centre of mass of Earth+SpaceShuttle stays in exactly the same place, doesn't move.
If you put the solution I found back into newton's law, you find the orbital velocity could only change by a maximum of +/- 0.0015 %
You could use conservation of angular momentum if you want, or whatever, but I expect you get the same result. You could even add in the effect of the angular speed of the Earth just when the moon disappears, but the difference would be incredibly small. Only the position of centre of mass and the instantaneous orbital velocity is required to get the answer. The solution is actually very simple and I think you people are making it sound very difficult. It's not.
Well, the moon does still oscillate back and forth relative to it's tidal-locked position. There is a complicated interaction of effects there, but part of it is an oscillation due to the force component responsible for tidal locking.
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