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B Effect of the Sun's Gravity on the Moon

  1. Jul 6, 2017 #1
    Is the gravitational pull of Sun on moon greater than the gravitational pull that Earth has over moon?

  2. jcsd
  3. Jul 6, 2017 #2


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  4. Jul 6, 2017 #3
    That seems a rather counterintuitive statement. Doesn't the Moon rotating around the Earth imply that that interaction is the dominant force?
  5. Jul 6, 2017 #4


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    It is a true statement.

    If we ignore the Sun and consider the Earth and moon in isolation, the moon clearly orbits the Earth. Both are pulled by the Sun with roughly the same force, so this is a reasonable way of looking at the situation.

    But if you plot the path of the moon around the sun, the path is always concave toward the Sun and is not always concave toward the Earth.

    Or you could do it the hard way. Look up the mass of the Sun, the mass of the Earth, the sun to earth distance and the sun to moon distance. Do the inverse square calculation and see which is larger.

    Edit: I just looked up the relevant figures. The sun wins by a factor of roughly 333,000 on mass. The Earth wins by a factor of roughly 152,000 on r2. Net is a 2 to 1 win for the Sun over the Earth. That matches what I Googled up earlier and what my teachers told me half a century ago.
    Last edited: Jul 6, 2017
  6. Jul 6, 2017 #5
    Then why is moon not ripped away from Earth...
  7. Jul 6, 2017 #6


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    The Sun pulls on the Earth too.
  8. Jul 6, 2017 #7


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    Both the Earth and the Moon are accelerated towards the Sun at approximately the same rate. They are in "free fall" around the Sun. In such a case, the Earth's gravity keeps the Moon orbiting around it while both of them orbit the Sun.
  9. Jul 6, 2017 #8
    I can definitely say "today I learned", heh. But yeah, I guess it makes sense. The difference of the sun's attraction to the moon, being in front of the Earth vs behind it is negligible, so for all intents and purposes the two are entirely oblivious of the sun and thus just circle around each other.
  10. Jul 6, 2017 #9


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    As already pointed out both the Earth and Moon are being pulled on by the Sun by pretty much the same amount. I say "pretty much" because at Full moon, the Moon is slightly further from the Sun than the Earth is and during the new moon is is slightly closer. This produces a small difference in the gravitational accleration on them caused by the Sun. This has the tendency to "stretch" the Moon's orbit along the Earth-Sun line. This is much like the effect the sun has on Earth's ocean tides, So we can say that the Sun has a tidal effect on the Moon's orbit. Unlike straight gravitational pull, tidal forces fall off by the cube of distance rather than by the square of the distance. (this is why, even though the Moon's gravitational pull on the Earth is much weaker than the Sun's, its tidal effect is larger. It is 1/400 the distance from the Earth than the Sun is.)
    Now the larger the orbit around the Earth the stronger the Sun tidal effect on it is, and the weaker the Earth's grip on the orbiting object. At some point, the Sun wins this tug of war, an the object will drift off into an independent orbit around the Sun. This boundary of this critical distance is known as the Hill sphere. The Moon is nestled quite safely at a distance that is ~1/2 that of the radius of Earth's Hill sphere.
  11. Jul 6, 2017 #10
    ##m_{earth} \approx 5.97237 \cdot 10^{24} \cdot kg##
    ##m_{sun} \approx 1,98855 \cdot 10^{30} \cdot kg##
    ##m_{moon} \approx 7.342\cdot 10^{22} \cdot kg##
    ##G \approx 6.67384 \cdot 10^{-11} \cdot kg##
    ##r_{moon-sun} \approx 149598023000 \cdot m##
    ##r_{earth-moon} \approx 384399000 \cdot m##

    ##F_{moon-sun}=\frac{G \cdot m_{moon} \cdot m_{sun}}{r_{moon-sun}^2} \approx 4.35386519 \cdot 10^{20} \cdot N##
    ##F_{earth-moon}=\frac{G \cdot m_{earth} \cdot m_{moon}}{r_{earth-moon}^2} \approx 1.98048874 \cdot 10^{20} \cdot N##
  12. Jul 7, 2017 #11
    Thanks everybody...
  13. Jul 7, 2017 #12
    Ok. I have deleted my post...I want to be cautious..
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