The titular question should say it all. Let's assume they're essentially touching each other (so the distance between them is 0 based on their commonly stated radii). At that exact moment, how does the magnetic force based on their magnetic moments compare to the gravitational force between them?

Edit for resolution:

Here are some details, the values I found online for their magnetic moments; no need to calculate the gravitational force, that's around $10^{27}~{\rm N}$ in that scenario, I'm really mostly interested in the magnetic force:

magnetic moment of Jupiter = $1.55\times 10^{20}~{\rm Tm}^3$

magnetic moment of Saturn = $4.6\times 10^{18}~{\rm Tm}^3$

distance between centers of mass ($r_{\rm Jup} + r_{\rm Sat}$) = $128143~{\rm km}$

So what I'm wondering is essentially what the magnetic force is so I can compare it to the gravitational force. I don't know how to calculate this.

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    – Buzz
    Commented May 30, 2023 at 22:47

1 Answer 1


Would the magnetic force come close to the gravitational one?

No. Nowhere near close.

Wikipedia gives a different value for the magnetic dipole moment of Jupiter -- $2.8 \times 10^{20} \text{ T} \cdot \text{m}^3$ -- so I'll use that instead of your value.

For simplicity, model the planets as point magnetic dipoles located at their centers, with the given magnetic dipole moments. (If instead you model them as uniformly magnetized balls, you get the same result. See this paper.)

The magnitude of their magnetic attraction is then

$$F_\text{mag}=\frac{3\mu_0m_1m_2}{2\pi d^4}$$

where $\mu_0$ is the magnetic constant, $m_1$ and $m_2$ are the two planets' magnetic dipole moments, and $d$ is the separation of their centers.

If you try evaluating this formula, you will discover that the units don't work out! This is because some planetary astronomers (and Wikipedia articles on the magnetospheres of Jupiter and Saturn) don't use conventional SI units for a magnetic dipole moment, which are $\text{A}\cdot\text{m}^2$.

You can read more about this messiness here. You have to divide the magnetic moments given in $\text{T} \cdot \text{m}^3$ by $\mu_0/(4\pi)$ in order to get magnetic moments in the proper SI units of $\text{A}\cdot\text{m}^2$.

Putting in the numbers, I find that the magnetic attraction is $2.9\times 10^{14}$ newtons compared with $4.4\times 10^{27}$ newtons for the gravitational attraction, so it's negligible in comparison with gravity.


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