Earth's Velocity without the Moon: Formula & Answers

[Jean-Louis]

Earth's velocity is 30 km/s.

What would be Earth's velocity without the moon ? What's the formula?

 

[tim_lou]

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?

 

[russ_watters]

Our revolution speed around the sun? Unchanged.

 

[Jean-Louis]

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 ?

 

[arunma]

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.

 

[russ_watters]

deleted: missed the change in subject from the OP...

 

[NoTime]

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 ?

Yes
Yes
Don't know if there is a way to calculate it because the sun also has an effect on this.

http://en.wikipedia.org/wiki/Tidal_acceleration][/PLAIN]
620 million years ago: the day was 21.9±0.4 hours

More
http://www.astronomynotes.com/gravappl/s10.htm

 

[Jean-Louis]

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 ?

 

[NoTime]

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.

 

[Jean-Louis]

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.

 

[Janus]

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.

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.

 

[YellowTaxi]

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.

 

[phlik]

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
simplest explination.

 

[YellowTaxi]

Oops, sorry didn't realize 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.

 

[Ian]

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.

 

[YellowTaxi]

Roughly,

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
=365.01707510616413248734767419157 days
(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
good question

edit: corrected Earth to moon distance

 

[pixel01]

..
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.

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.

 

[Janus]

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.

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.

 

[YellowTaxi]

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.

 

[russ_watters]

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.

That would be true if objects were point masses, but they aren't.

 

[rcgldr]

the moon does rotate, but at a rate that makes it appear stationary to an observer on the earth. ... tidal

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:

http://science.nasa.gov/headlines/y2000/ast09jun_1.htm

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 Earth's 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.

 

[NoTime]

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.

 

[rcgldr]

Tidal stresses affect not only bodies of water but, the crust moves as well.

Agreed, I was just pointing out that tidal stresses on the Earth aren't going to affect the rate of rotation of the moon.

 

[YellowTaxi]

That would be true if objects were point masses, but they aren't.

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.

 

[russ_watters]

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.

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.

http://www.phy6.org/stargaze/Smoon4.htm

 

[russ_watters]

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

There is more to Newton's gravity than just that.

And any physics problem is vastly simplified by analysing the motion of the centre of mass of the system.

True, but in this case, the simplification simplifies-away the effect that we're discussing.

The solution is actually very simple and I think you people are making it sound very difficult. It's not.

No, it is more complicated than you realize. See all of the above posts about tidal locking. If what you were saying were true, there'd be no such thing as tidal locking. Heck, there'd be no such thing as tides if we could only consider the objects as point masses! To calculate tidal forces you, at the very least, need to consider a dumbell-shaped object with two point masses.

2/3 of the way down this page is a diagram showing the vector resultant fore doesn't pass through the CoM:

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html

Here's a discussion of where the torque comes from:

Resulting torque: Since the bulges are now displaced from the A-B axis, A's gravitational pull on the mass in them exerts a torque on B. The torque on the A-facing bulge acts to bring B's rotation in line with its orbital period, while the "back" bulge which faces away from A acts in the opposite sense. However, the bulge on the A-facing side is closer to A than the back bulge by a distance of approximately B's diameter, and so experiences a slightly stronger gravitational force and torque. The net resulting torque from both bulges, then, is always in the sense which acts to synchronise B's rotation with its orbital period, leading inevitably to tidal locking. [/qutoe]
http://www.answers.com/topic/tidal-locking

 

[D H]

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

This simplification is perfectly valid if and only if the object in question has a spherical mass distribution. For any other mass distribution, this simplification is not quite valid. Whether the results change much depends on the problem. The behavior of the Apollo subsatellites exemplify how a non-spherical mass distribution can drastically change the results. See http://science.nasa.gov/headlines/y2006/06nov_loworbit.htm" .

 

[YellowTaxi]

That can't be true DH,
the Earth plus the Moon is definitely not a sphere, but their centre of mass can still be treated as a point mass moving on a perfect ellipse around the Sun.
If the Sun wasn't there, the c-of-m would move in a perfect straight line.

And like I already said, the wobble of the Earth by the moon is actually very small compared to the velocity of the orbit around the Sun, so I think it's safe to ignore that.

Or if you really want to complicate things just add in the tidal effect of the Earth on the plasma of the Sun, and all the perturbation effects of the other planets in the solar system ;-)

 

[D H]

That can't be true DH,
the Earth plus the Moon is definitely not a sphere, but their centre of mass can still be treated as a point mass moving on a perfect ellipse around the Sun.

The Earth-Moon center of mass can be approximated as moving in an ellipse. Even if the Sun, Earth, and Moon were the only objects in the universe, this approximation would not be exact.

If the gravity could be decoupled as you imply the N-body problem would be soluble in terms of elementary functions.

 

[YellowTaxi]

The Earth-Moon center of mass can be approximated as moving in an ellipse. Even if the Sun, Earth, and Moon were the only objects in the universe, this approximation would not be exact.

I don't believe that. Can you proove it?


And in any case, since you're at it, why not add in all the subtle corrections due to the general relativity theory.

 

[D H]

The result is well-known (google "three body problem"). It is incumbent upon you to prove otherwise.

 

[YellowTaxi]

"The result is well-known. It is incumbent upon you to prove otherwise."

That doesn't make sense. I asked you to answer what you meant and now you saying I should answer my own question.!
Leads me to assume that you can't answer my simple question yourself.
It only suggests to me you don't understand what you're talking about.

I didn't say can NASA scientists or anybody else debate with me whether the trajectory of the Earth-Moon system is essentially a perfect ellipse, (it probably isn't due to gen rel effects etc). I said can you prove it to me (I'm thick) without resorting to gen rel or pasting links to obscure websites that aren't your own straightforward explanation. ;-)

And would it really add anything to getting closer to a good, straightforward answer to the initial problem posted by the OP. I don't think so...

For example, prove to me that the tidal effect of the Earth's pull on the plasma of the Sun makes a significant difference to the problem under discussion and I'll eat all my words. I'm sure there would be some effect caused by the change in the mass of the Earth-Moon system and its position, but I doubt it would be worth the effort computing. ;-) maybe just a second per year or less - whatever. It's boring.


Frankly this has become an absurd discussion. And I still think Kepler's 3rd Law is all you really need to get a good enough answer to the OP.:zzz:

 

[D H]

Replacing the Earth and Moon positions/masses with the Earth/Moon barycenter/total mass is not (completely) valid. It is approximately valid; whether that approximation is good enough depends on the problem to be solved.

The issue at hand is to compare the acceleration of the Earth-Moon barycenter due to the gravitational attraction of the Moon, Earth, and Sun to each other versus versus the acceleration of a combined Earth-Moon (located at the Earth-Moon barycenter) toward the Sun.

Nomenclature:
<br /> \begin{array}{lll}<br /> M_x&amp;&amp;\text{x=e,m,s: Earth, Moon, Sun mass} \\[12pt]<br /> \vec r_x&amp;&amp;\text{x=e,m,s: Earth, Moon, Sun position (inertial coords)} \\[12pt]<br /> \vec r_b&amp;&amp;\text{Earth-Moon barycenter position} \\[12pt]<br /> \vec r_{p\to q}&amp;&amp;\text{Vector from point p to point q}<br /> \end{array}

If the universe comprised two point masses, the Sun and the Earth/Moon, the acceleration of the Earth/Moon toward to Sun would be
<br /> \ddot{\vec r}_b = -\,\frac{GM_s}{||\vec r_{s\to b}||^3}\vec r_{s\to b}<br /> \end{array}<br />

Now suppose the Earth and Moon are distinct point masses. The vector from the sun to each of these point masses is

<br /> \begin{array}{rl}<br /> \vec r_{s\to e}&amp;=\vec r_{s\to b} - \frac{M_m}{M_e+M_m}\vec r_{e\to m} \\[12pt]<br /> \vec r_{s\to m}&amp;=\vec r_{s\to b} + \frac{M_e}{M_e+M_m}\vec r_{e\to m} \end{array}<br />

The gravitational acceleration of the Earth is
<br /> \begin{array}{rl}<br /> \ddot{\vec r_e} &amp;=<br /> -\,\frac{G M_s}{||\vec r_{s\to e}||^3}\vec r_{s\to e}<br /> + \frac{G M_m}{||\vec r_{e\to m}||^3}\vec r_{e\to m}<br /> \\[12pt]<br /> &amp;=<br /> -\,\frac{G M_s}{||\vec r_{s\to e}||^3}<br /> (\vec r_{s\to b}-\frac{M_m}{M_e+M_m}\vec r_{e\to m})<br /> + \frac{G M_m}{||\vec r_{e\to m}||^3}\vec r_{e\to m}<br /> \\[12pt]<br /> &amp;=<br /> -\,\frac{G M_s}{||\vec r_{s\to e}||^3}\vec r_{s\to b}<br /> +\left(<br /> \frac{G M_s}{||\vec r_{s\to e}||^3}\frac{M_m}{M_e+M_m}+\frac{G M_m}{||\vec r_{e\to m}||^3}<br /> \right)<br /> \vec r_{e\to m}<br /> \end{array}<br />

The gravitational acceleration of the Moon is similarly

<br /> \ddot{\vec r}_m<br /> =<br /> -\,\frac{G M_s}{||\vec r_{s\to m}||^3}\vec r_{s\to b}<br /> -\left(<br /> \frac{G M_s}{||\vec r_{s\to m}||^3} \frac{M_e}{M_e+M_m}+\frac{G M_e}{||\vec r_{e\to m}||^3}<br /> \right)<br /> \vec r_{e\to m}<br />

The second derivative of the center of mass location is
<br /> \begin{array}{rl}<br /> \ddot{\vec r_b}&amp;=<br /> \frac1{M_e+M_m}(M_e \ddot{\vec r}_e + M_m \ddot{\vec r}_m)<br /> \\[12pt]<br /> &amp;=<br /> -\,GM_s<br /> \left(<br /> \frac{M_e}{M_e+M_m}\,<br /> \frac 1 {||\vec r_{s\to e}||^3} +<br /> \frac{M_m}{M_e+M_m}\,<br /> \frac 1 {||\vec r_{s\to m}||^3}<br /> \right)<br /> \vec r_{s\to b} +<br /> GM_s<br /> \frac{M_e M_m}{(M_e+M_m)^2}<br /> \left(\frac 1 {||\vec r_{s\to e}||^3} -<br /> \frac 1 {||\vec r_{s\to m}||^3}\right)<br /> \vec r_{e\to m}<br /> \end{array}<br />

which is obviously not the same as the simple combined Earth/Moon acceleration.

 

[YellowTaxi]

Now, how exactly does that affect the time it takes the Earth to orbit the Sun if the Moon suddenly disappears ?

- Don't tell me, it doesn't, right ?

As far as I can tell, if the orbit is near as dammit to circular - which our Earth orbit IS ,
then it IS safe to use the Keplerian mechanics, and anything more complicated won't be worth the effort computing.

Thanks for making the effort all the same.

 

[tim_lou]

DH is correct, the center of mass concept is only valid for a constant external force (the usual 9.8 acceleration on the surface of the earth, or zero external force).

as DH explained in the specific example of the three body system, the point mass assumption fails for the Earth moon system as the gravitational field caused by the sun varies as position changes.

in general, let R denote the center of mass,
M\ddot{\vec{R}}=\sum_i m \ddot{\vec{r}_i}=\sum_i \left(\vec{F}^\text{ext}(\vec{r}_i)+\sum_j\vec{F}_{ij}\right)

where F_ij denote the force exerted on i by j (F_ii = 0 vector by convention).
the double sum
\sum_{i,j}\vec{F}_{ij} goes to zero by Newton's third law (equal and opposite force), hence,
M\ddot{\vec{R}}=\sum_i \vec{F}^{\text{ext}}(\vec{r}_i)

but in general,
\sum_i \vec{F}^{\text{ext}}(\vec{r}_i) \neq n\cdot \vec{F}^{\text{ext}}(\vec{R})

unless there are some symmetries one can invoke or F^ext is a constant vector (zero or some other values like g)

 

[YellowTaxi]

DH is correct..

Probably true tim_lou, but I don't think the effect would make a noticeable difference because we're dealing with basically a circular orbit, and I think the complexity only applies to elliptic orbits with extreme excentricity.

It's like arguing that Galileo was wrong because he ignored the effects of air viscosity when he dropped two masses off that tower

Pedantic

 

[tim_lou]

Re-reading the original question, the question is not precise enough to be answered, because the mechanism in which the moon disappeared is not specified, and/or some conditions (total angular momentum conserved?) are not specified.

And even if the moon ceases to exist, the Earth can stay in the sum's orbit with an arbitrary amount of angular momentum associated with spin. (the orbital angular momentum and spin angular momentum are decoupled in this situation).

 

[YellowTaxi]

what ?
So how much was I out by (1 second a year or maybe less)

Maybe we forgot to include the extra drag induced by new blades of grass growing near lake Baikal.

I'll go back and recalculate ...

 

[Janus]

Roughly,

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
=365.01707510616413248734767419157 days
(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
good question

edit: corrected Earth to moon distance

You can't just plug 150,004,678,000 in for r' in the Kepler
formula. In this case r' is the semimajor axis of the new orbit and will be larger than the value you used.

Assume, for the sake of argument, that the original Earth-Moon orbit is circular. Now when the Earth is furthest from the Sun it is not only 4,678 km further from the Sun than the Earth-Moon barycenter, but it is also moving faster as it is also orbiting the barycenter at 12.5 m/sec. This means that it is traveling faster than it would for a circular orbit at that distance.

So if the Moon were to disappear at this moment, the Earth would enter a new elliptical orbit with its present distance at perhelion. This puts the new value of r' for this orbit at a further distance out.

Using 149,600,000km for r, the new r' works out to 149,735,000km

Plugging these values into the Kepler equation yields a difference in orbital periods of about 11.82 hrs; about 30 times the answer you got.