<p>The height of the tidal bulge raised on the Sun by the Earth is about $10^{-13}$ times the radius of the Sun, which is about a 20th of a millimetre.</p>

<p>(This presumes that the Sun is made of a frictionless inertialess fluid: it is a measure of the <em>driving force</em> behind the tides).</p>

CC BY-SA 3.0
2016-08-04T14:35:18.803Z
This answer needs explanation, and maths.
Nico
CC BY-SA 3.0
2016-08-04T13:02:30.303Z
2016-08-04T13:02:30.303Z
Martin Kochanski
<p>The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$a_{\rm tidal} = 2\frac{Gm R}{r^3},$$
where $R$ is the radius of the Sun.</p>

<p>You can work out that Jupiter has the largest tidal effect on the Sun, <em>very closely followed</em> by Venus, and both produce tidal forces that are only a factor of $\sim 2.2$ times greater than that produced by the Earth on the Sun.</p>

<p>The distortion of the Sun could be modelled as following the consequent equipotentials (well inside the Roche lobe) in a binary system, but this is not analytically solvable and I could not locate any straightforward approximations. One way to judge the size of the effect would be to compare the tidal acceleration with the centrifugal acceleration at the Sun's surface $R\omega_{\rm Sun}^2$.</p>

<p>So $a_{\rm tidal}$ for Jupiter is $3.7\times10^{-10}$ m/s$^2$, whereas the centrifugal acceleration (at the solar equator) is 7 orders of magnitude larger at 0.006 m/s$^2$. Given that the Sun is <a href="https://physics.stackexchange.com/questions/208344/why-is-the-sun-almost-perfectly-spherical/208348#208348">almost spherical</a> to 1 part in 100,000, because the centrifugal acceleration is only 20 millionths of the Sun's surface gravity, we can safely assume that tides due to planets that are 7 orders of magnitude weaker do not distort the shape of the Sun to any degree worth worrying about.</p>

<p>I was surprised to find that the potential influence of planetary tides on solar activity is still an area of recent research. Suggestions that planetary tides might be responsible for sunspots or the sunspot cycle on the Sun go back a long way. <a href="http://adsabs.harvard.edu/abs/1996Ap%26SS.243..393G" rel="nofollow noreferrer">Grandpierre (1996)</a> suggested that planetary tides influence flows in the Sun and could be responsible for the 11-year solar cycle. A NASA report by <a href="http://ntrs.nasa.gov/search.jsp?R=20070025111" rel="nofollow noreferrer">Hung (2007)</a>, although not peer-reviewed, does suggest a link between solar activity and planetary positions. The claim is made that major solar flares happen "underneath" the locations of the major tide-inducing planets (Jupiter, Venus, Earth, Mercury [in that order]). </p>

<p>I very quickly found a peer-reviewed paper in "Solar Physics" by <a href="http://link.springer.com/article/10.1007/s11207-013-0288-6" rel="nofollow noreferrer">Seker (2013)</a>, which discusses all this and reviews the evidence on sunspots. They conclude that the appearance of new sunspots is <em>not</em> correlated at all with the positions of the planets. However, they do raise the caveat that because the magnetic fields producing the spots are produced at the tachocline between the radiative core and convective envelope of the Sun and take several days to buoyantly rise. Thus the longitude at which the spots appear and where they are generated, with respect to the positions of the planets, can change significantly.</p>

<p>I guess the jury is still out, but this is quite "fringe" material. The solar cycle is certainly thought to be a product of the dynamo mechanism that produces the magnetic field. However, the location of sunspots definitely cannot be predicted with such models at the moment. Interest in this topic has been revived recently by claims of correlations between magnetic activity and the orbital positions of hot Jupiter-type exoplanets and even the suggestion that tidal effects or interactions between stellar and planetary magnetic fields might be responsible for increased magnetic activity (see <a href="http://arxiv.org/abs/1408.6049" rel="nofollow noreferrer">Lanza 2014 for a recent review</a>). For these close-in exoplanets of course the tides are many orders of magnitude larger than in the solar system (see equation above).</p>

CC BY-SA 3.0
2016-08-04T09:51:39.283Z
Any idea of the ballpark magnitude for the tidal effect of Jupiter on the *shape* of the sun? It's probably small and difficult to measure, but possibly a slight deviation from an axisymmetric oblate shape might break some degeneracies or otherwise be detectable with solar seismography maybe? Or possibly the orbit of Solar Probe Plus (see [here](http://astronomy.stackexchange.com/q/16707/7982) and [here](http://space.stackexchange.com/q/17562/12102))?
uhoh
CC BY-SA 3.0
2016-08-04T12:02:58.003Z
@uhoh utterly negligible - see edit.
ProfRob
CC BY-SA 3.0
2016-08-04T12:40:08.710Z
Thanks! I remember something like 25 years ago the [Eöt-Wash Group](https://www.washington.edu/research/pathbreakers/1990c.html) was stacking a few lead blocks around a torsional pendulum to try to cancel the *gravitational quadruple moment of the hill* next to the Nuclear Physics building because (as Eric Adelberger enthusiastically explained to me) it was overwhelming the experiment. To them the words "...to any degree worth worrying about." would have some extra zeros as well. Anyway, I'll try to formulate a question based on "potentially measurable"-like language.
uhoh
CC BY-SA 3.0
2016-08-04T09:37:54.313Z
2016-08-04T18:39:01.820Z
2017-04-13T12:40:15.377Z
ProfRob
<p>How big are the tidal influences of planets (say, Jupiter) on the sun? Do they lead to any measurable effects, for example, do they influence the sunspots, or are they observable in any other way?</p>

CC BY-SA 3.0
2016-08-04T11:44:26.567Z
As a rule of thumb, the tidal force is roughly equivalent to the size of the object visually (from the center of the planet, so holding a tennis-ball up to your eye doesn't count). From the sun, Jupiter is just a point in the sky, similar to the size it appears from earth, so the tidal effects are pretty tiny.
userLTK
CC BY-SA 3.0
2016-08-04T13:32:24.683Z
There are conjectures that Jupiter and the other giant planets are responsible for the solar cycle. The frequencies of the solar cycle and of Jupiter's orbit are similar, but it's important to keep in mind that these are just conjectures.
David Hammen
CC BY-SA 3.0
2016-08-05T17:14:34.717Z
@userLTK Care to explain what you mean more? I don't understand what you mean by a tidal *force* being equivalent to the *size* of the object as it appears from the center of the planet.
zephyr
CC BY-SA 3.0
2016-08-06T00:59:29.667Z
@zephyr mathematically, it's because both vary by the power of 3.   If, theoretically, the Moon was 1/2 the size across (1/2 the diameter) but the same density and twice as close, it would have 1/8th the mass but it would look the same size in the sky.   Being twice as close, that to the 3rd power is 8 times a strong, so they cancels out, both factors being cubed.    Density also matters, but just to the first power.   The visible arc in the sky is cubed, so that's a kind of a rule of thumb for tides (unless the object is super dense like a Neutron star or super close like the space station).
userLTK
CC BY-SA 3.0
2016-08-06T01:15:57.343Z
@zephyr but if you hold a tennis ball in the air, that can appear bigger than the moon - that's all I meant, the tennis ball's tidal effect on the Earth is essentially zero, even if visually it looks bigger than the moon held out at arms length.   The visual distance needs to be taken with a grain of salt, it's actually the distance between the centers of the two objects, which for distant objects like even the moon, is quite similar to the angular diameter.
userLTK
<the-sun><tidal-forces>
Tidal influences of planets on the sun
dirkt
# Tidal influences of planets on the sun

(Asked by: username_0 on 2016-08-04)

How big are the tidal influences of planets (say, Jupiter) on the sun? Do they lead to any measurable effects, for example, do they influence the sunspots, or are they observable in any other way?

**Tags:** the-sun, tidal-forces
> **Comments:**
> **username_1**: As a rule of thumb, the tidal force is roughly equivalent to the size of the object visually (from the center of the planet, so holding a tennis-ball up to your eye doesn't count). From the sun, Jupiter is just a point in the sky, similar to the size it appears from earth, so the tidal effects are pretty tiny.  
> **username_2**: There are conjectures that Jupiter and the other giant planets are responsible for the solar cycle. The frequencies of the solar cycle and of Jupiter's orbit are similar, but it's important to keep in mind that these are just conjectures.  
> **username_3**: @userLTK Care to explain what you mean more? I don't understand what you mean by a tidal *force* being equivalent to the *size* of the object as it appears from the center of the planet.  
> **username_1**: @zephyr mathematically, it's because both vary by the power of 3.   If, theoretically, the Moon was 1/2 the size across (1/2 the diameter) but the same density and twice as close, it would have 1/8th the mass but it would look the same size in the sky.   Being twice as close, that to the 3rd power is 8 times a strong, so they cancels out, both factors being cubed.    Density also matters, but just to the first power.   The visible arc in the sky is cubed, so that's a kind of a rule of thumb for tides (unless the object is super dense like a Neutron star or super close like the space station).  
> **username_1**: @zephyr but if you hold a tennis ball in the air, that can appear bigger than the moon - that's all I meant, the tennis ball's tidal effect on the Earth is essentially zero, even if visually it looks bigger than the moon held out at arms length.   The visual distance needs to be taken with a grain of salt, it's actually the distance between the centers of the two objects, which for distant objects like even the moon, is quite similar to the angular diameter.  

---

## 2 answers

### Answer by username_6 on 2016-08-04. Score: 9 | Accepted answer

The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$a\_{\rm tidal} = 2\frac{Gm R}{r^3},$$
where $R$ is the radius of the Sun.

You can work out that Jupiter has the largest tidal effect on the Sun, *very closely followed* by Venus, and both produce tidal forces that are only a factor of $\sim 2.2$ times greater than that produced by the Earth on the Sun.

The distortion of the Sun could be modelled as following the consequent equipotentials (well inside the Roche lobe) in a binary system, but this is not analytically solvable and I could not locate any straightforward approximations. One way to judge the size of the effect would be to compare the tidal acceleration with the centrifugal acceleration at the Sun's surface $R\omega\_{\rm Sun}^2$.

So $a\_{\rm tidal}$ for Jupiter is $3.7\times10^{-10}$ m/s$^2$, whereas the centrifugal acceleration (at the solar equator) is 7 orders of magnitude larger at 0.006 m/s$^2$. Given that the Sun is [almost spherical](https://physics.stackexchange.com/questions/208344/why-is-the-sun-almost-perfectly-spherical/208348#208348) to 1 part in 100,000, because the centrifugal acceleration is only 20 millionths of the Sun's surface gravity, we can safely assume that tides due to planets that are 7 orders of magnitude weaker do not distort the shape of the Sun to any degree worth worrying about.

I was surprised to find that the potential influence of planetary tides on solar activity is still an area of recent research. Suggestions that planetary tides might be responsible for sunspots or the sunspot cycle on the Sun go back a long way. [Grandpierre (1996)](http://adsabs.harvard.edu/abs/1996Ap%26SS.243..393G) suggested that planetary tides influence flows in the Sun and could be responsible for the 11-year solar cycle. A NASA report by [Hung (2007)](http://ntrs.nasa.gov/search.jsp?R=20070025111), although not peer-reviewed, does suggest a link between solar activity and planetary positions. The claim is made that major solar flares happen "underneath" the locations of the major tide-inducing planets (Jupiter, Venus, Earth, Mercury [in that order]).

I very quickly found a peer-reviewed paper in "Solar Physics" by [Seker (2013)](http://link.springer.com/article/10.1007/s11207-013-0288-6), which discusses all this and reviews the evidence on sunspots. They conclude that the appearance of new sunspots is *not* correlated at all with the positions of the planets. However, they do raise the caveat that because the magnetic fields producing the spots are produced at the tachocline between the radiative core and convective envelope of the Sun and take several days to buoyantly rise. Thus the longitude at which the spots appear and where they are generated, with respect to the positions of the planets, can change significantly.

I guess the jury is still out, but this is quite "fringe" material. The solar cycle is certainly thought to be a product of the dynamo mechanism that produces the magnetic field. However, the location of sunspots definitely cannot be predicted with such models at the moment. Interest in this topic has been revived recently by claims of correlations between magnetic activity and the orbital positions of hot Jupiter-type exoplanets and even the suggestion that tidal effects or interactions between stellar and planetary magnetic fields might be responsible for increased magnetic activity (see [Lanza 2014 for a recent review](http://arxiv.org/abs/1408.6049)). For these close-in exoplanets of course the tides are many orders of magnitude larger than in the solar system (see equation above).

> **Comments:**
> **username_7**: Any idea of the ballpark magnitude for the tidal effect of Jupiter on the *shape* of the sun? It's probably small and difficult to measure, but possibly a slight deviation from an axisymmetric oblate shape might break some degeneracies or otherwise be detectable with solar seismography maybe? Or possibly the orbit of Solar Probe Plus (see [here](http://astronomy.stackexchange.com/q/16707/7982) and [here](http://space.stackexchange.com/q/17562/12102))?  
> **username_6**: @uhoh utterly negligible - see edit.  
> **username_7**: Thanks! I remember something like 25 years ago the [Eöt-Wash Group](https://www.washington.edu/research/pathbreakers/1990c.html) was stacking a few lead blocks around a torsional pendulum to try to cancel the *gravitational quadruple moment of the hill* next to the Nuclear Physics building because (as Eric Adelberger enthusiastically explained to me) it was overwhelming the experiment. To them the words "...to any degree worth worrying about." would have some extra zeros as well. Anyway, I'll try to formulate a question based on "potentially measurable"-like language.  

---

### Answer by username_4 on 2016-08-04. Score: 1

The height of the tidal bulge raised on the Sun by the Earth is about $10^{-13}$ times the radius of the Sun, which is about a 20th of a millimetre.

(This presumes that the Sun is made of a frictionless inertialess fluid: it is a measure of the *driving force* behind the tides).

> **Comments:**
> **username_5**: This answer needs explanation, and maths.  

---

