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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?

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    $\begingroup$ It's an interesting question but the situation is more complicated because the solar wind has such an impact on the en.wikipedia.org/wiki/Interplanetary_magnetic_field $\endgroup$ – uhoh Feb 16 at 10:39
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    $\begingroup$ Compared to gravity, all other interactions between Sun and Earth are negligible. $\endgroup$ – Florin Andrei Feb 22 at 3:11
  • $\begingroup$ @FlorinAndrei yes. Gravity is the main interaction between the sun and its satellites. $\endgroup$ – Max0815 Feb 22 at 4:15
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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 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.

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.

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In SI units the 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 and $m_S$ from this impeccable "scientific" source. From Figure 4 here, $m_S$ is about $10^7 \times m_E$ so the value for $m_S$ looks about right.

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  • $\begingroup$ 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? $\endgroup$ – Max0815 Feb 24 at 21:53
  • $\begingroup$ $1.5 \times 10^{11}$ meters is the Earth-Sun distance. $\endgroup$ – Keith McClary Feb 24 at 22:20
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    $\begingroup$ -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? $\endgroup$ – uhoh Mar 13 at 5:14
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    $\begingroup$ @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. $\endgroup$ – Keith McClary Mar 13 at 18:03

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