<p><strong>Warning:</strong> I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.</p>

<p>First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts: </p>

<blockquote>
  <p>There are three requisites for a dynamo to operate: </p>
  
  <p>1) An electrically conductive fluid medium</p>
  
  <p>2) Kinetic energy provided by planetary rotation</p>
  
  <p>3) An internal energy source to drive convective motions within the fluid.</p>
  
  <p>In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.</p>
</blockquote>

<hr>

<p>The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other "space weather" events.</p>

<hr>

<p>To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by uhoh, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather "pushing" stream of charged that will "push" away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they "drape" themselves onto the magnetic field-less object and form the "magnetosphere" of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.</p>

<p>If you are looking for numbers, a hefty equation to calculate the electrical
magnetism oh an object is the Boit-Savart Law. </p>

<blockquote>
  <p>If you have a
  current <span class="math-container">$I$</span> flowing through an infinitesimal length
  <span class="math-container">$\mathrm{d}\vec{l}$</span> then the field at a point <span class="math-container">$\vec{r}$</span> is given by
  <span class="math-container">\begin{equation} \mathrm{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{I
&gt; \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation}</span> The
  field does still drop of with the inverse square of the distance, but
  the vector nature of the current makes the formula rather more
  complicated. Generally this expression has to be integrated over the
  length of a wire carrying the current. This can get quite complicated
  and so this method is generally only used for simple situations such
  as the field around a magnetic dipole.  Having found the field, the
  force on a infinitesimal length of current can be found from.
  <span class="math-container">\begin{equation} \vec{F} =
&gt; I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation}</span> This is again
  analogous to the electrostatic case but with added complication due to
  current being a vector.</p>
  
  <blockquote>
    <p><a href="https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction">https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction</a></p>
  </blockquote>
</blockquote>

<p>This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation: </p>

<p><span class="math-container">$$F=\frac{\mu q_{m1} q_{m2}}{4\pi r^4}$$</span>
where</p>

<ul>
<li><span class="math-container">$F$</span> is force attraction in newtons.</li>
<li><span class="math-container">$q_{m1}$</span> and <span class="math-container">$q_{m2}$</span> is the magnitudes of magnetic poles in ampere-meters.</li>
<li><span class="math-container">$\mu$</span> is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.</li>
<li><span class="math-container">$r$</span> is the separation in meters.</li>
</ul>

<p>Thus, the attraction between the Earth and the Sun can be calculated in the following means:
<span class="math-container">$$\implies \operatorname{B}(\text{Earth})=q_{m1}=7.84027\times 10^{22}$$</span>
<span class="math-container">$$\implies \operatorname{B}(\text{Sun})=q_{m2}=3.48526\times 10^{29}$$</span>
<span class="math-container">$$\implies \mu\approx 10^{-6}$$</span>
<span class="math-container">$$\implies r=\operatorname{AU}=149597870700$$</span>
<span class="math-container">$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$</span>
<span class="math-container">$$\implies F=\boxed{4.34165\text{ newtons}}$$</span>
If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.</p>

<p>I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.</p>

CC BY-SA 4.0
2019-02-22T01:21:35.077Z
2019-03-13T05:17:08.137Z
2019-03-13T05:17:08.137Z
Max0815
<p>Modestly I suggest that due to vectorial nature of magnetic fields, random bursts of plasma in the sun doesn't generate &quot;a well defined north-south poles structure&quot; as we have on earth. So random nature of solar magnetic field can't exert a stable or well defined (don't confuse with solar wind) magnetic force on earth. (sorry if my suggestion is inmature, and apologize for my bad english).</p>

CC BY-SA 4.0
2021-03-06T10:56:27.423Z
2021-03-06T10:56:27.423Z
Pilindongo
<p>In SI units the <a href="https://en.wikipedia.org/wiki/Force_between_magnets" rel="nofollow noreferrer">force between magnets</a> is 
<span class="math-container">$$\approx 10^{-7}\frac{m_Sm_E}{r^4}$$</span>
<span class="math-container">$$=~10^{-7}\frac{(3.5 \times10^{29}) (7.84\times 10^{22})}{(1.5 \times 10^{11})^4} = 5.4 $$</span> Newtons.</p>

<p>Magnetic dipole moments <span class="math-container">$m_E$</span> from <a href="https://arxiv.org/pdf/1007.4497.pdf" rel="nofollow noreferrer">here</a> and <span class="math-container">$m_S$</span> from <a href="https://www.conservapedia.com/Sun" rel="nofollow noreferrer">this impeccable "scientific" source</a>. From Figure 4 <a href="https://ned.ipac.caltech.edu/level5/March03/Vallee2/Vallee2_3.html" rel="nofollow noreferrer">here</a>, <span class="math-container">$m_S$</span> is about <span class="math-container">$10^7 \times m_E$</span> so the value for <span class="math-container">$m_S$</span> looks about right.</p>

CC BY-SA 4.0
2019-02-24T21:53:23.833Z
So you imply to use the top equation to get the attraction between the Earth and the Sun? How would you calculate the variables though? And how did you derive the variables to get the second equation?
Max0815
CC BY-SA 4.0
2019-02-24T22:20:12.767Z
$1.5 \times 10^{11}$ meters is the Earth-Sun distance.
Keith McClary
CC BY-SA 4.0
2019-03-13T05:14:01.963Z
`-1` Your "impeccable scientific source" for $m_S$ gets its value from *Creation Research Society Quarterly* and *Estimated*. The Sun's magnetic field is complex and dynamic. Can you find a value from a real scientific source?
uhoh
CC BY-SA 4.0
2019-03-13T18:03:11.943Z
@uhoh Sorry, I was being sarcastic there (I shouldn't do that). The first link is a good source for $m_E$ and the third allows one to read the ratio $m_E/m_E$ from the graph. I could use the values from the graph but I would have to figure out Gauss m$^2$ units. You might ask Max0815 where his values come from - he added that part of his answer after I posted mine.
Keith McClary
CC BY-SA 4.0
2019-02-24T21:47:08.563Z
2019-03-13T18:25:00.193Z
2019-03-13T18:25:00.193Z
Keith McClary
<p>How much magnetism is there between the Earth and Sun? Would the distance from the Sun differ for not the Earth iron core and how much?</p>

CC BY-SA 4.0
2019-02-16T10:39:36.430Z
It's an interesting question but the situation is more complicated because the solar wind has such an impact on the https://en.wikipedia.org/wiki/Interplanetary_magnetic_field
uhoh
CC BY-SA 4.0
2019-02-22T03:11:08.473Z
Compared to gravity, all other interactions between Sun and Earth are negligible.
Florin Andrei
CC BY-SA 4.0
2019-02-22T04:15:58.097Z
@FlorinAndrei yes. Gravity is the main interaction between the sun and its satellites.
Max0815
<the-sun><orbit><earth><magnetic-field>
How much magnetic attraction is there between the Sun and Earth?
Muze
# How much magnetic attraction is there between the Sun and Earth?

(Asked by: username_0 on 2018-09-02)

How much magnetism is there between the Earth and Sun? Would the distance from the Sun differ for not the Earth iron core and how much?

**Tags:** the-sun, orbit, earth, magnetic-field

> **Comments:**
> **username_1**: It's an interesting question but the situation is more complicated because the solar wind has such an impact on the https://en.wikipedia.org/wiki/Interplanetary_magnetic_field  
> **username_2**: Compared to gravity, all other interactions between Sun and Earth are negligible.  
> **username_3**: @username_2 yes. Gravity is the main interaction between the sun and its satellites.  

---

## 3 answers

### Answer by username_3 on 2019-02-22. Score: 3 | Accepted answer

**Warning:** I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.

First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts:

> There are three requisites for a dynamo to operate:
>
> 1) An electrically conductive fluid medium
>
> 2) Kinetic energy provided by planetary rotation
>
> 3) An internal energy source to drive convective motions within the fluid.
>
> In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.

---

The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other "space weather" events.

---

To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by username_1, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather "pushing" stream of charged that will "push" away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they "drape" themselves onto the magnetic field-less object and form the "magnetosphere" of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.

If you are looking for numbers, a hefty equation to calculate the electrical
magnetism oh an object is the Boit-Savart Law.

> If you have a
> current $I$ flowing through an infinitesimal length
> $\mathrm{d}\vec{l}$ then the field at a point $\vec{r}$ is given by
> \begin{equation} \mathrm{d}\vec{B} = \frac{\mu\_0}{4\pi} \frac{I
> > \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation} The
> field does still drop of with the inverse square of the distance, but
> the vector nature of the current makes the formula rather more
> complicated. Generally this expression has to be integrated over the
> length of a wire carrying the current. This can get quite complicated
> and so this method is generally only used for simple situations such
> as the field around a magnetic dipole. Having found the field, the
> force on a infinitesimal length of current can be found from.
> \begin{equation} \vec{F} =
> > I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation} This is again
> analogous to the electrostatic case but with added complication due to
> current being a vector.
>
> > <https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction>

This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation:

$$F=\frac{\mu q\_{m1} q\_{m2}}{4\pi r^4}$$
where

* $F$ is force attraction in newtons.
* $q\_{m1}$ and $q\_{m2}$ is the magnitudes of magnetic poles in ampere-meters.
* $\mu$ is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.
* $r$ is the separation in meters.

Thus, the attraction between the Earth and the Sun can be calculated in the following means:
$$\implies \operatorname{B}(\text{Earth})=q\_{m1}=7.84027\times 10^{22}$$
$$\implies \operatorname{B}(\text{Sun})=q\_{m2}=3.48526\times 10^{29}$$
$$\implies \mu\approx 10^{-6}$$
$$\implies r=\operatorname{AU}=149597870700$$
$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$
$$\implies F=\boxed{4.34165\text{ newtons}}$$
If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.

I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.


---

### Answer by username_4 on 2021-03-06. Score: 0

Modestly I suggest that due to vectorial nature of magnetic fields, random bursts of plasma in the sun doesn't generate "a well defined north-south poles structure" as we have on earth. So random nature of solar magnetic field can't exert a stable or well defined (don't confuse with solar wind) magnetic force on earth. (sorry if my suggestion is inmature, and apologize for my bad english).


---

### Answer by username_5 on 2019-02-24. Score: 0

In SI units the [force between magnets](https://en.wikipedia.org/wiki/Force_between_magnets) is
$$\approx 10^{-7}\frac{m\_Sm\_E}{r^4}$$
$$=~10^{-7}\frac{(3.5 \times10^{29}) (7.84\times 10^{22})}{(1.5 \times 10^{11})^4} = 5.4 $$ Newtons.

Magnetic dipole moments $m\_E$ from [here](https://arxiv.org/pdf/1007.4497.pdf) and $m\_S$ from [this impeccable "scientific" source](https://www.conservapedia.com/Sun). From Figure 4 [here](https://ned.ipac.caltech.edu/level5/March03/Vallee2/Vallee2_3.html), $m\_S$ is about $10^7 \times m\_E$ so the value for $m\_S$ looks about right.

> **Comments:**
> **username_3**: So you imply to use the top equation to get the attraction between the Earth and the Sun? How would you calculate the variables though? And how did you derive the variables to get the second equation?  
> **username_5**: $1.5 \times 10^{11}$ meters is the Earth-Sun distance.  
> **username_1**: `-1` Your "impeccable scientific source" for $m_S$ gets its value from *Creation Research Society Quarterly* and *Estimated*. The Sun's magnetic field is complex and dynamic. Can you find a value from a real scientific source?  
> **username_5**: @username_1 Sorry, I was being sarcastic there (I shouldn't do that). The first link is a good source for $m_E$ and the third allows one to read the ratio $m_E/m_E$ from the graph. I could use the values from the graph but I would have to figure out Gauss m$^2$ units. You might ask username_3 where his values come from - he added that part of his answer after I posted mine.  

---

