Greg Egan on Nostr: The LISA gravity wave detector will comprise three spacecraft, all in free-all in ...
The LISA gravity wave detector will comprise three spacecraft, all in free-all in heliocentric orbits, that always lie on the vertices of an equilateral triangle. But how can they do that?
The answer is: they can't! Not precisely. Of course, 3 spacecraft at 120° intervals around a perfect circular orbit could do it, but the LISA craft need to be much closer together.
So, the question is: how do they fly in a formation that *approximates* this goal?
To see how, the best way is to imagine what forces you would feel if you were inside a hollowed-out asteroid in a circular orbit, tidally locked so that one axis always pointed towards the sun.
Call the sunwards axis the x-axis, the direction pointing around the orbit the y-xis, and the direction out of the plane of the orbit the z-axis.
Along the z-axis, you would feel a “tidal squeeze” of –GM/r^3 z pulling you back towards the plane of the orbit. A free particle displaced along the z-axis would bounce up and down as if it was on a spring, with a periodic motion (not coincidentally) the same as the period of the orbit.
In the xy plane, things would be a little more complicated. You would feel a “tidal squeeze” of –GM/r^3 y along the y-axis, but that would be exactly cancelled out by a centrifugal force of GM/r^3 y due to the tidally locked rotation of the asteroid, which has an angular rate of ω = √(GM/r^3).
Along the x-axis, you would feel a “tidal stretch” of 2GM/r^3 trying to tear you away from the centre of the asteroid, *plus* a centrifugal force of ω^2 x = GM/r^3 x, for a total of 3GM/r^3 x.
The answer is: they can't! Not precisely. Of course, 3 spacecraft at 120° intervals around a perfect circular orbit could do it, but the LISA craft need to be much closer together.
So, the question is: how do they fly in a formation that *approximates* this goal?
To see how, the best way is to imagine what forces you would feel if you were inside a hollowed-out asteroid in a circular orbit, tidally locked so that one axis always pointed towards the sun.
Call the sunwards axis the x-axis, the direction pointing around the orbit the y-xis, and the direction out of the plane of the orbit the z-axis.
Along the z-axis, you would feel a “tidal squeeze” of –GM/r^3 z pulling you back towards the plane of the orbit. A free particle displaced along the z-axis would bounce up and down as if it was on a spring, with a periodic motion (not coincidentally) the same as the period of the orbit.
In the xy plane, things would be a little more complicated. You would feel a “tidal squeeze” of –GM/r^3 y along the y-axis, but that would be exactly cancelled out by a centrifugal force of GM/r^3 y due to the tidally locked rotation of the asteroid, which has an angular rate of ω = √(GM/r^3).
Along the x-axis, you would feel a “tidal stretch” of 2GM/r^3 trying to tear you away from the centre of the asteroid, *plus* a centrifugal force of ω^2 x = GM/r^3 x, for a total of 3GM/r^3 x.