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Gravitational Forces Acting on the Moon

Date: 08/27/2005 at 05:38:13
From: Veet
Subject: Quadratic equation: WHY IS THE MOON ORBITING EARTH

I'm looking for the point between the sun and the Earth where their 
gravitational attraction is numerically equal.

  x = distance from Earth to the balance point
  y = distance from sun to the balance point

At that balance point the following equation applies:

  m1/x^2 = m2/y^2

If d = distance from Earth to the sun then

  y^2 = (d - x)^2 

and the equation may be written as 

  m1/x^2 = m2/(d - x)^2

This simple quadratic equation has two solutions (one on each side 
of the Earth).  I got following data from the Internet:

   d = Mean Earth orbit 149,000,000 Km
  m1 = Earth mass 5.979 *10^24Kg
  m2 = Sun mass 1.99 *10^30 Kg

Solving this equation with the above data, I get the balance point
distance between Earth and sun to be 257,758 Km from the Earth, and on 
the opposite side of the Earth the distance is -258,653 Km.  I also 
calculated the distance perpendicular to the Earth and it is 258,205 
Km which is approximately half way between the two above distances.

What I find most difficult about this is that I also found the 
following number for the mean moon orbit: 384,000 Km.  If this is 
true, then why is the moon orbiting around Earth rather than being
pulled toward and orbiting around the sun?

Date: 08/27/2005 at 20:43:18
From: Doctor Rick
Subject: Re: Quadratic equation: WHY IS THE MOON ORBITING EARTH

Hi, Veet.

You got me for a few minutes!

I did your calculation to confirm that you got the answer correct, 
and you did (at least pretty close; using your numbers I got 257,822 
km).  So, as I understand it, your point is that when the moon is 
closer to the sun than the balance point, the sun is pulling harder 
on the moon than the earth is, so you'd expect the sun to pull the 
moon away from the earth.  At the very least, the sun would 
significantly distort the path of the moon from an elliptical orbit.

Here is what you are forgetting: yes, the sun is pulling the moon 
toward itself--but it is also pulling the earth toward itself!  The 
moon's orbit is indeed greatly distorted from an elliptical orbit: in 
fact the moon follows a path that wobbles around an elliptical path 
around the sun.  In effect, the moon does orbit the sun.  But the 
elliptical path around the sun, around which the moon's orbit 
wobbles, is in fact the earth's orbit!  In short, the moon orbits the 
earth, which in turn is orbiting the sun.

If you want to determine the extent to which the sun perturbs the 
orbit of the moon around the earth, you need to consider not the ratio 
of the forces on the moon by the sun and the earth, but the difference 
in the forces of the sun on the moon and the earth--essentially a 
"tidal" force.  As the moon pulls the water in the seas toward itself 
(and also toward a point opposite the moon, so that there are two high 
tides a day), in the same way the sun pulls the moon away from the 
earth at the points in the moon's orbit that are nearest to and 
farthest from the sun.  But this tidal force (the difference between 
the sun's pull on the moon and the earth) will be much less than the 
gravitational force of the earth on the moon; therefore the 
perturbation of the moon's orbit is much less than your calculation 

I suggest that you do this calculation and tell me what you get.  I 
should clarify a bit: we aren't really calculating the difference in 
*forces*; of course the force the sun exerts on the earth will be 
much greater than the force exerted on the moon, because the mass of 
the earth is much greater.  Instead, compute the *acceleration* 
resulting from those forces.  The difference in the accelerations is 
the relative acceleration of the moon away from the earth; multiply 
this relative acceleration by the mass of the earth gives you an 
effective force, the "tidal force".  This is what you should compare 
with the gravitational force of the earth on the moon.

- Doctor Rick, The Math Forum 

Date: 08/28/2005 at 01:47:18
From: Veet
Subject: Quadratic equation: WHY IS THE MOON ORBITING EARTH

Aloha Dr. Rick,
Thank you for your answer.  I now understand that Earth is “falling"
towards the sun just as the moon is and that only the *difference*
between gravitational accelerations counts.  I can see that even if
sun’s acceleration was a million times stronger, it would make no
difference as long as it was the same on Earth and on the moon. 

What really interests me is how far would the moon need to be from 
the Earth in order to become a planet in its own right?  Or 
alternatively, how close would Mars or Venus need to come to Earth 
in order to become a twin planet rotating around the common center 
of gravity?

I spent the whole day figuring out the calculations you suggested.  My
wife thinks that you must be a doctor of Psychiatry and I a mentally
ill person trying to figure out what it would take to lose our moon.

Either way, a shrink or Doctor Math, I would very much appreciate if 
you could check my thinking and my results:

  Units	 m, Kg, s

Assuming a circular orbit of moon with center in Earth (actual center 
is some 4700 Km from Earth’s center, that is inside Earth).

  m1	Sun’s mass 				1.991 10^30 
  m2	Earth’s mass 				5.979 10^24
   d	Distance from sun to Earth 		149000000000
   r	Avg. distance from Earth to moon 	384000000
   G	Gravitational constant			6.67259 10^-11 

1. Sun's acceleration at Earth: G m1/d^2 = 0.00598402

2. Sun's acceleration at moon when moon is closest to the sun: 
     G (m1/(d - r)^2) = 0.00601499

3. Sun's acceleration at moon when it is farthest from the sun:
     G(m1/(d + r)^2)=0.0059533

4. EAM = Earth acceleration at moon: G m2/r^2 = 0.00270558

5. DSA = the difference between sun's acceleration at moon and Earth
     (when the moon is closest to the sun:  -0.0000309635)
     (when the moon is farthest from sun:    0.000030725 )

Both Earth and moon are falling towards the sun.  At moon’s closest 
point to the sun it is accelerating towards sun just a little faster 
than Earth (about 0.51 percent faster than Earth)

6. MARE = (Moon’s Acceleration Relative to Earth) = EAM - DSA =
     G(m1/(d - r)^2 - m1/d^2 - m2/r^2)

7. When the moon is between Earth and sun, 
     MARE = 0.00270558 - 0.0000309635 = 0.00267462
   When the moon is on the other side of Earth 
     MARE = 0.00270558 + 0.000030725 = 0.00273631
   When moon and Earth are at the same distance from the sun
     MARE = EAM = 0.00270558. 

Now the question: How far from Earth should the moon be in order to 
become a planet?

That would happen if the difference between the moon's acceleration
towards sun and Earth's acceleration towards sun is larger than the
moon’s acceleration towards Earth.  That is, when MARE <0.  The moon's
relatively weakest acceleration to Earth is when it is between the sun
and Earth.  Playing with numbers I found that if the moon were 4.42
times farther away, the sun would steal it from us.

I tried to solve this equation looking for MARE = 0:

  m1/(d - x)^2 - m1/d^2 - m2/x^2 = 0

I stuck it into an old version of “Mathematica” I have on my hard 
drive and got:

  -8.48195*10^8-1.48607*10^9 i,

The only result that makes sense is 1.69645*10^9.  I don’t understand
the other results.  How can there be more than one answer to this
problem in our physical world?

Please tell me if my thinking is correct, that if the the moon were
4.42 times farther away it would become a planet.

If my thinking is correct, I might attempt to find out how far would 
a planet the size of Earth need to be in order for us to be a double 

I suppose the situation would be quite different in such a case as we 
would be rotating around a common center of gravity half way between 
Earth and that planet.

Thanks again, Doctor.  It is great to have someone I can ask my 
mentally ill questions!


Date: 08/29/2005 at 22:04:40
From: Doctor Rick
Subject: Re: Quadratic equation: WHY IS THE MOON ORBITING EARTH

Hi, Veet.

I'm a bit surprised you haven't come up with the word "lunatic" to 
describe yourself!  (Get it? Luna is the moon.)  ;-)

I can explain the four solutions that Mathematica gave you: the 
equation you provided is equivalent to a quartic (fourth-power) 
equation, and quartic equations have four solutions.  In this case, 
two of the solutions are complex as opposed to real numbers, so they 
are not physical solutions (distances must be real numbers).  The last 
corresponds to the moon being on the opposite side of the sun from 
the earth; it is physical in the sense that if the moon were there, 
the difference in accelerations due to the sun would equal the 
acceleration due to the earth, but it is not relevant because in that 
configuration the moon would clearly not be in orbit around the 
earth.  This leaves the one solution you identified.

To tell the truth, I was surprised that the moon would only have to 
be 4.4 times farther from earth to meet your condition.  Thinking 
more, though, I realized that the point you are calculating is 
essentially one of the Lagrange points of the sun-earth system (L1, 
in particular).  See for instance,

  Microwave Anisotropy Probe: The Lagrange Points 

Following the link to SOHO, you'll read that L1 is about 1.5 * 10^6 
km sunward of earth.  That's the order of magnitude of your figure, so 
I guess your work is fine.

- Doctor Rick, The Math Forum 

Date: 08/30/2005 at 03:29:51
From: Veet
Subject: Quadratic equation: WHY IS THE MOON ORBITING EARTH

Dear Doc--

“Lunatic”!  Ha-ha-ha!  Still laughing, that’s a good one.  Thank you
so much for taking the time to help me go deeper into my lunacy.

Thanks for explaining the second real solution.  I did not notice 
before that 2.98*10^11 is exactly twice the Earth orbit.  Of course 
the difference between the moon’s and Earth’s acceleration towards the 
sun is zero and moon’s acceleration towards Earth also (practically) 
zero.  I understand the solution. 

I have not studied the complex, just surprised that there should be 
solutions that have no physical meaning whatsoever.  I wonder what 
kind of imaginary reality they represent. ;-)

I just Googled “La Grange”: “JOSEPH LA GRANGE (1736-1813 C.E.).  First
introduced the method of Lagrange Multipliers.” 

His L1 is only 12% off from my result.  Perhaps he was using different
numbers for the mass of Sun and Earth and for the distance between
them.  I will need more time to understand the website that you 
recommended.  It looks very interesting.

Thanks again.

Associated Topics:
College Physics

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