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As I understand, "dark matter" is what they call a theoretical substance which is only known by it's influence upon velocity curve of the galaxy.

If indeed the gravity of "dark matter" is so strong that it keeps our solar system in orbit, then it's force effects on Earth should be comparable with that of our Moon.

However, we can observe the moon tides, and combined Moon-solar tides, but we never observe the tides resulting from pull towards the center of the galaxy. Why would that be?

Could you please share your thoughts?

enter image description here

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    $\begingroup$ It is also much more uniform across the globe that than the gravity of the moon (or the sun for that matter). Tides are an effect of non-uniformity of gravitational influence. $\endgroup$ – dmckee Mar 11 '14 at 3:45
  • $\begingroup$ Rotation of Earth should provide for non-uniformity, am I wrong? $\endgroup$ – Division by Zero Mar 11 '14 at 3:54
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    $\begingroup$ @LexPodgorny Rotation causes a continuous reorientation of our surroundings. But if our neighborhood of dark matter is the same in every direction, reorientation does nothing. $\endgroup$ – Blackbody Blacklight Mar 11 '14 at 5:18
  • $\begingroup$ If it is the same in every direction then their gravitational forces would cancel each other out and the winner would be the centrifugal force, which is not the case. Am I just not getting something or this is one of the cases where Newtonian laws go out of the window? Perhaps you can provide a link to an article that clarifies that? $\endgroup$ – Division by Zero Mar 11 '14 at 5:31
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    $\begingroup$ Our tides are a composite of all the gravitational forces earth feels. The moon is the closest and contributes most. The sun is far away with a large mass and contributes less than 50% . The galactic center ( center of mass of dark + shiny) is light years away and the effect on the tides will unmeasurable, dark halo or not. If there is dark matter around the solar system we will only feel the one within our orbit and it will be added up to the sun ( cms of solar system, close to the sun) and will be indistinguishable $\endgroup$ – anna v Mar 11 '14 at 5:45
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To put a little mass into the discussion, the gravitational field due to a compact body of mass $M$ and distance $R$ is $$ g = G\frac{M}{R^2} \,.$$

Tides arise because another body of size $r$ (measured from the subject body's center of mass) experiences a different gravitational attraction on either side of the body. That difference is

$$ \begin{align} \delta &= g_\text{surface} - g_\text{CoM} \\ &\approx \left. \frac{\partial g}{\partial R} \right|_\text{center} r \\ &= -2G\frac{M}{R^3} r \,. \end{align} $$ Notice that this result (good only to first order in $r/R$) is linear in the mass of the thing causing the tides and in the size of the subject body, but goes by the inverse cube of the distance to the thing causing the tides.

So not only is the galactic dark matter distribution fairly even, but most of it is at huge distance from the Earth which strongly suppresses the contribution from both the dark matter core of the galaxy and the Baryonic matter core as well.

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The solar-system effects of dark matter are almost zero, if not exactly zero. While dark matter is the majority of matter in the galaxy, it is also very diffuse when compared to things like the sun and the Earth.

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  • $\begingroup$ Correct me if I am wrong, but if dark matter gravitation did not affect solar system, the solar system would be among the first ones to depart the proximity of our galaxy as it is basically on the fringe of it. $\endgroup$ – Division by Zero Mar 11 '14 at 4:02
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    $\begingroup$ @LexPodgorny: sure, the dark matter content of the galaxy keeps us orbiting the galaxy. But we are basically free falling through an empty space in the region we're in (the dark matter pulling uis into the galaxy is far from us), and the equivalence tells us that this is not distinguishable from just moving in empty space with no gravity. And it has no effect on the gravitational forces BETWEEN bodies in the solar system. $\endgroup$ – Jerry Schirmer Mar 11 '14 at 4:17
  • $\begingroup$ I understand that it has little affect on interaction of the bodies WITHIN the solar system, but the Earth rotation should continuously expose different part of the ocean to the center of the galaxy and thus towards it's center of gravity. That is of course if we still agree that the effect of dark matter is directional and it's resulting force vector connects centers of Earth and the galaxy. Being new to this topic and I may be missing something. Perhaps, you could suggest an article that disproves my supposition? $\endgroup$ – Division by Zero Mar 11 '14 at 5:38
  • $\begingroup$ Lex: You should consider whether your reasoning also applies to the visible matter of the galaxy. Why isn't there a specifically galactic component to Earth tides, resulting from the visible matter throughout the galaxy? $\endgroup$ – Mitchell Porter Mar 11 '14 at 7:11
  • $\begingroup$ @MitchellPorter Quoting NASA: "It turns out that roughly 68% of the universe is dark energy. Dark matter makes up about 27%. The rest - everything on Earth, everything ever observed with all of our instruments, all normal matter - adds up to less than 5% of the universe." [science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy] $\endgroup$ – Division by Zero Mar 4 '17 at 2:45
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Dark matter is supposed to exist in a halo with radial symmetry about the galactic center. The gravitational effect is therefore indistinguishable from any other constituent of the galaxy. Things closer than us to the center pull us inward; things farther than us from the center have no effect.

In the same way, we cannot gravitationally observe the composition of the interior of the earth, from a vantage point on the surface. Lowering a probe deep into the planet would allow such structure to be mapped. This is analogous to observing the revolution of stars around other galaxies.

We infer that dark matter exists because the pull of gravity upon stars relative to their distance from galactic center is too much. But either way, in the neighborhood of an individual star, the pull toward the galactic center is roughly uniform.

If we could measure the tidal force of the galaxy, i.e. the difference in acceleration on opposite sides of the solar system, then dark matter might be apparent. Essentially this would be measuring the local density of dark matter presumed to uniformly fill the solar system. But it would be a small effect, intuitively on the order of the galactic orbital acceleration times the ratio of the measured distance, to the distance to the galactic center, squared.

It seems doable, with sufficiently precise probes launched into diverging orbits. However, gravitational effects are what we do know about dark matter so it doesn't seem a very useful experiment. Perhaps a very large space-based gravitational wave observatory would need to correct for the effect, though (as an offset from ordinary tides resulting from difference in galactic orbital distance).

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The distribution of dark matter in the Milky Way is (probably) flatter than the distribution of baryonic matter (see figure 1 in the linked paper - the Solar System is at about 8 kparsecs). While there would be some tidal effects they would be less than the tides generated by the visible stars, dust clouds etc, and these in turn would be less than in a setup like the Solar System where most of the mass is concentrated at the centre.

Actually calculating the tidal force would be hard, but you can immediately see that it's small by looking at galactic rotation curves. The tangential velocity is independant of radius over a large range, so there would be little or no tidal force until you got outside this range. I found this rotation curve for the Milky Way:

Milky Way

here. There's no reference for this so treate it with care, but it does make my point.

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