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As the earth rotates around the Sun our distance from the center of the galaxy is varying - what sort of variations does this cause in the force of gravity here on earth?

[Edit] Given that Earth orbits around the Sun at a radius of 150e+6 km while the Sun orbits within the Milky Way at a distance somewhere between 2.283e+17 km and 2.685e+17 km of the center (with a motion that weaves up and down in relation to the galaxies flattened plane).

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  • $\begingroup$ This doesn't really need the GR tag IMO . . . all the answers below are based on Newtonian arguments anyway. $\endgroup$
    – m4r35n357
    Jan 19, 2023 at 9:45

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None that we could measure.

The space station orbits the earth. Astronauts are weightless inside. If the space station was far from Earth orbiting the galaxy on its own, it would be the same.

On Earth, there is an effect from the gravity of the Moon. One side of the Earth is closer to the moon. Gravity from the moon is a little stronger on the near side. This creates tides.

The galaxy does the same, but the galaxy is so far away that the effect is unmeasurably small.

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    $\begingroup$ How small is unmeasurably? $\endgroup$
    – user263399
    Jan 18, 2023 at 16:55
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    $\begingroup$ I think that the distribution of mass largely being on a disk and the solar system zipping in and out of this disk might introduce measurable effects. Enough to correlate with mass extinctions due to the distubrances in the Oort cloud. $\endgroup$ Jan 18, 2023 at 18:23
  • $\begingroup$ @user263399 - Unmeasurably for me is when you have to start using scientific notation. The sun is 333,000 times the mass of the Earth. Ok, that's feasible. The current value of Sagittarius A is ~4.154 million solar masses, which is ~1.38e12 times the mass of the Earth, which is a number that's absolute gobbledygook to me. What it speaks to is the distances you stated, to the power of 17.... completely unfathomable. Even if we could measure it, it would be meaningless. $\endgroup$
    – Mazura
    Jan 19, 2023 at 2:20
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Easy to estimate: the gravitational force goes like $1/(\mathrm{distance})^2$. With the distances of Earth to Sun and Sun to the Galactic center respectively,

\begin{equation} \Delta F \sim \left( \frac{150\times 10^6 \mathrm{km}}{2\times 10^{17}\mathrm{km}} \right)^2 \sim 10^{-20} , \end{equation}

i.e. by nothing.

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  • $\begingroup$ While 1/(distance)^2 applies to Earth around the Sun, surely something else applies to the Sun around the Galactic center since we are 'inside' and apparently the gravitational force inside a 'solid' sphere is linear (what it is for a disk instead of sphere is currently beyond me). $\endgroup$
    – user263399
    Jan 19, 2023 at 0:07
  • $\begingroup$ true, good point. so, add an order of magnitude or three to that ratio. still: utterly negligible. $\endgroup$
    – rfl
    Jan 19, 2023 at 0:19
  • $\begingroup$ (the actual discrepancy is more like a factor two or so, judging by the rotation curve of the Milky Way as it's decomposed into the various contributions, in particular that of the disk and that of the dark matter halo) $\endgroup$
    – rfl
    Jan 19, 2023 at 0:20
  • $\begingroup$ While negligible from a practical standpoint from a theoretical view I'm curious how much that is written off as 'randomness' could be accounted for by taking into account the relative motions between the Earth:Sun:Milky Way $\endgroup$
    – user263399
    Jan 19, 2023 at 0:51
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    $\begingroup$ @user263399 look deeply into Earth:Sun:Jupiter before you start worrying too much about the rest of the galaxy. There's enough there to last you years, if not decades.... $\endgroup$
    – nitsua60
    Jan 19, 2023 at 1:41

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