Of the same mass. It has been pointed out many times on this board that replacing the sun with a black hole of equivalent mass would have no effect on Earth’s orbit, but it seems to me that the magnetar’s field would cause the Earth’s orbit to decay rapidly, and flip so that the magnetic poles were aligned properly. Am I right? How long would this take?
Thanks,
Rob
Assuming that the magnetic forces are significant compared to the gravitational forces (I’m not sure on this; I’ll have to run some numbers), then what you’d have would be a force that falls off as a higher power of distance, rather than 1/r[sup]2[/sup] (something like 1/r[sup]6[sup], I think). Orbits with such a force law are in fact unstable, and spiral either in or out. Given the situation of the Earth in its orbit and the magnetic field suddenly turning on, we’d be in a spiraling-in situation.
As for how long it’d take, it’d be on the order of a year, but again, I’d have to calculate it.
I suddenly regret selling you that sun-sized magnetar the other week, and I ESPECIALLY regret throwing in that star destroyer-class frigate as a freebie. When will I learn?
In spite of inverse square law, the force pulling the Earth would go up by orders of magnitude, as the Earth’s core is largely iron.
Honestly, such a massive change would probably disrupt the crust totally.
Presumably it would announce it’s existence and hunger and taunt us to run.
Ohhhhhh, wait. You said “magnetar”. My bad.
Why would the magnetic forces not be significant compared to the gravitational forces? Also, where does 1/r[sup]6[/sup] come from? I thought that only applied to the strong force.
Thanks,
Rob
The strength of a magnetic (dipole) field falls off proportionally to r[sup]-3[/sup], while the gravitational field falls off as r[sup]-2[/sup]. So for a magnetar with a given mass and magnetic dipole strength, there’s going to be some radius beyond which the gravitational fields are greater than the magnetic fields. The question is whether Earth would be beyond that radius.
It doesn’t apply to the strong force — that dies off exponentially. I’m not sure where the exponent in the r[sup]-6[/sup] comes from, off the top of my head, though. Maybe something to do with the induced dipole in the Earth’s core?
Right, that was assuming an induced dipole. On the other hand, though, the Earth’s core, despite being mostly iron, is probably above the Fermi point, so it might not induce a significant dipole. But the Earth already has a dipole of its own; that might be more significant.
OK, I finally got around to getting off my butt (well, on my butt) and digging up the numbers. Using the Earth’s existing dipole, and a magnetar with 10^15 G at 10 km, I find that the magnetic force between the two is less than a newton (about 1/3 newton, but I was really only working to order-of-magnitude). There’s a 1/r^4 in the force law that’s a killer.
And a magnetar at the distance of the Sun would have an even weaker field than the Earth’s existing one, so I think we can say that the induced dipole moment of the Earth would be even more negligible, even if we do assume that the Earth is made up of ferromagnetic material.
Am I right in thinking the smallest possible magnetar has a greater mass than the sun?
Also, what would a magnetar do to our atmosphere?
Unclear. We don’t know the equation of state of neutron star material, and so don’t know what the maximum and minimum masses possible are. Depending on the model, masses can range as low as .7 or .8 solar masses on the low end, or as high as a bit over 2 on the high end. Magnetars might have tighter bounds than that, due to their formation mechanisms, but we can say even less about that. So a magnetar with the same mass as the Sun can’t be ruled out, at least in our current understanding.