How does the Sun's gravity affect the shape of the orbit of the Moon around the Earth?

Question

If we look at the Earth-Moon system as a two-body Keplerian system, we would expect Earth and the Moon to orbit their center of mass (barycenter) in perfect ellipses, with the barycenter at one of the foci of each body's ellipse, and with the bodies moving faster when closer to the barycenter and slower when farther from the barycenter. If the orbital eccentricity was zero, the distance between either body and the barycenter would be constant, as would the orbital speed.

However, such a two-body model does not tell the whole story of the Moon's motion, as the gravity of the Sun distorts the orbits of Earth and the Moon about their barycenter. Wikipedia calls this effect variation. If the undisturbed orbits were perfect circles (with zero eccentricity), the disturbed orbits would take the form of ellipses, but with the barycenter at the center of the ellipse rather than at a focus of the ellipse. As with an eccentric orbit, the bodies would move faster when closer to the barycenter and slower when farther from the barycenter.

I want to know how to calculate the shape of the disturbed orbits, assuming zero eccentricity in the undisturbed orbits. Specifically, how much closer and farther do the bodies get to the barycenter, compared to the constant distance they would have in an undisturbed two-body zero-eccentricity system? And how much does their speed vary from the constant speed in an undisturbed two-body zero-eccentricity system?

For the purpose of simplification, let's assume that the three bodies are point masses and the orbits are in the same plane with zero eccentricity. For my Keplerian prediction, I'm using the following information:

• MassSun = 1.989 × 10³⁰ kg
• MassEarth = 5.9742 × 10²⁴ kg
• MassMoon = 7.34767309 × 10²² kg
• TimeEarthOrbit = 31,557,600 s
  • 365.25 days of 86,400 seconds each
• TimeMoonOrbit = 2,427,507.6923 s
  • TimeEarthOrbit / 13, making one solar year consist of exactly 12 lunar months, for simplification purposes
• G = 6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻²
• DistanceSunSubsystem = cbrt((TimeEarthOrbit² × G × (MassSun + MassEarth + MassMoon)) / (4π²))
  • Distance between the center of the Sun and the center of mass of the Earth-Moon subsystem
• DistanceEarthMoon = cbrt((TimeMoonOrbit² × G × (MassEarth + MassMoon)) / (4π²))
  • Distance between the center of the Earth and the center of the Moon
  • With the numbers I'm using this works out to 392,029,361.3123 meters, well within the real range of 356,500,000 to 406,700,000 meters and only slightly larger than the real time-averaged value of 384,399,000 meters
• BarycenterDistanceSun = DistanceSunSubsystem × (MassPlanet + MassEarth) / (MassSun + MassEarth + MassMoon)
  • Distance between the Sun-Earth-Moon barycenter and the Sun
• BarycenterDistanceSubsystem = DistanceSunSubsystem × MassSun / (MassSun + MassEarth + MassMoon)
  • Distance between the Sun-Earth-Moon barycenter and the Earth-Moon barycenter
• BarycenterDistanceEarth = DistanceEarthMoon × MassMoon / (MassEarth + MassMoon)
  • Distance between the Earth-Moon barycenter and the center of the Earth
• BarycenterDistanceMoon = DistanceEarthMoon × MassEarth / (MassEarth + MassMoon)
  • Distance between the Earth-Moon barycenter and the center of the Moon
• VelocitySun = BarycenterDistanceSun × 2π / TimeEarthOrbit
  • Velocity of the Sun in its orbit around the Sun-Earth-Moon barycenter
• VelocityEarth = BarycenterDistanceSubsystem × 2π / TimeEarthOrbit + BarycenterDistanceEarth × 2π / TimeMoonOrbit
  • Velocity of the Earth at the exact moment of new-moon
• VelocityMoon = BarycenterDistanceSubsystem × 2π / TimeEarthOrbit − BarycenterDistanceMoon × 2π / TimeMoonOrbit
  • Velocity of the Moon at the exact moment of new-moon

Answer 1

You could simply use the Conservation of Energy for the Moon at diametrically opposite points. Two equations at the two points and the two unknowns speed and distance from sun.

You may safely assume that speed at diametrically opposite points are same since you already assuming the final orbit to be ellipsoidal with earth at the center.

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For example consider the situation where the moon is in between earth and sun, together with moon away from the sun (in both cases all three objects are in a line) -

For the former,
Total energy of Moon (calculate from some circular model) = KE at $v$ velocity + PE due to Earth at $r$ + PE due to sun at $(r_e-r)$

For the latter,
Total energy of Moon (calculate from some circular model) = KE at $v$ velocity + PE due to Earth at $r$ + PE due to sun at $(r_e+r)$

were $r_e$ is distance of Earth from Sun. Here you have two equations and two unknowns ($v, r$) which can be solved.