# Earth-Moon system

russ_watters
Mentor
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]

Last edited:
D H
Staff Emeritus
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" [Broken].

Last edited by a moderator:
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 ;-)

Last edited:
D H
Staff Emeritus
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.

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.

Last edited:
D H
Staff Emeritus
The result is well-known (google "three body problem"). It is incumbent upon you to prove otherwise.

"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:

Last edited:
D H
Staff Emeritus
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:
$$\begin{array}{lll} M_x&&\text{x=e,m,s: Earth, Moon, Sun mass} \\[12pt] \vec r_x&&\text{x=e,m,s: Earth, Moon, Sun position (inertial coords)} \\[12pt] \vec r_b&&\text{Earth-Moon barycenter position} \\[12pt] \vec r_{p\to q}&&\text{Vector from point p to point q} \end{array}$$

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

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

$$\begin{array}{rl} \vec r_{s\to e}&=\vec r_{s\to b} - \frac{M_m}{M_e+M_m}\vec r_{e\to m} \\[12pt] \vec r_{s\to m}&=\vec r_{s\to b} + \frac{M_e}{M_e+M_m}\vec r_{e\to m} \end{array}$$

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

The gravitational acceleration of the Moon is similarly

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

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

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

Last edited:

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.

Last edited:
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)

Last edited:
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

Last edited:
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).

Last edited:
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 ...

Last edited:
Janus
Staff Emeritus
Gold Member
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.