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In an ideal situation, we can use a simple formula to determine the velocity at which a body revolves around a massive body at a particular distance. I was wondering if there is an intuitive way to understand the motion of a body influenced by two other massive bodies (say the moon being influenced by the earth and the Sun). Is there any analogy to the motion of a wheel that can be considered as a combination as translational and rotational motion (and both these motions can be mathematically treated separately)? Or do we need a simulator to visualize the actual path?

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I think this page, this page, this video and this page will help you –  Gigi Butbaia May 8 at 12:25
    
For sufficiently distance objects you should be able to use the center of mass of the 2-body system. –  Brandon Enright May 8 at 14:51

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I was wondering if there is an intuitive way to understand the motion of a body influenced by two other massive bodies (say the moon being influenced by the earth and the Sun .

No, intuition is not of real use in a three body gravitational problem, more so in many body.

In 1887, mathematicians Ernst Bruns [4] and Henri Poincaré showed that there is no general analytical solution for the three-body problem given by algebraic expressions and integrals. The motion of three bodies is generally non-repeating, except in special cases.[5]

Actually the motions of the planets are studied as possible dynamic chaos.

Planetaria solve the many body problem with numerical approximations.

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Because there's a clear hierarchy of masses (the moon is much lighter than the earth, is much lighter than the sun), there are a few useful approximations we can make. Here are some, roughly from big effects to little effects.

  • The earth-moon system (whose barycenter is within the earth) orbits the sun on an ellipse, passing through perihelion in January. The plane of this ellipse is the ecliptic.

  • The moon orbits the earth on an ellipse tilted at about 5º from the ecliptic.

  • The plane of the moon-earth orbit changes direction with respect to the ecliptic ("precession of nodes"). This is pretty fast: in 2010 we had eclipses in June-ish and December-ish, while in 2014 we have eclipses in April-ish and October-ish. A shift in the nodes by two months in four years gives roughly 24 years for the plane of the moon's orbit to wobble all the way back to its starting point.

  • The direction of the moon's perigee changes from year to year. Also there's a pretty big difference in perigee differences month-to-month.

  • The moon undergoes libration. The rate of the Moon's spin around its axis is quite stable, while the moon's revolution around the earth is faster near perigee than near apogee. This means that the moon appears to "turn" slightly in its orbit, not quite presenting always the same face.

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First of all, you have to be familiar with uniform circular motion (UCM) and the law of universal gravitation. $$\Sigma F=ma$$$$G\frac{m_1m_2}{r^2}=m_1\frac{v^2}{r}$$where $G$ is the gravitational constant, or $6.67\times10^{-11}$, $r$ is the distance between the centers of the two bodies, v is the velocity, and $m_1$ and $m_2$ are the masses of the two bodies ($m_1$ is orbiting $m_2$). By the rules of UCM, $$v=\frac{2\pi r}{T}$$where $T$ is the time it takes for one orbit. By substituting $v$ into the equation, you will get $$G\frac{m_1m_2}{r^2}=m_1\frac{4\pi^2r}{T^2}$$Rearrange this to get $$\frac{T^2}{r^3}=\frac{4\pi^2}{Gm_2}$$Just use this equation to solve for $T$, then plug it into $$v=\frac{2\pi r}{T}$$to find the value of $v$. Hope this helped! (from Giancoli Physics 6th edition)

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