Gravitational Forces Acting on the MoonDate: 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 indicates. 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 http://mathforum.org/dr.math/ 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, -8.48195*10^8+1.48607*10^9i 1.69645*10^9 2.98*10^11 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 planet. 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! Veet 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 http://www.physics.montana.edu/faculty/cornish/lagrange.html 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 http://mathforum.org/dr.math/ 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. Veet |
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