8
$\begingroup$

Is there a necessary relation between orbital period and the time of precession? In the case of Earth, the orbital period is one year, while a complete precession cycle takes thousands of years. Is a planet that completes a precession cycle within one of its "years" possible? Besides one of the poles having perpetual night and the other perpetual day, what else would this entail?

$\endgroup$
1
  • 1
    $\begingroup$ I guess it would have to be light or slow, because fast heavy gyroscope usually has low procession. $\endgroup$
    – Mołot
    Commented Nov 24, 2016 at 16:56

4 Answers 4

1
$\begingroup$

The planet has some spin, with a rate that can be freely chosen. Given some outside torque, the spin axis will precess. The details of the angular momentum, the rotational inertia, and the mass distribution, along with the details of the force doing the torquing, will determine the speed of the precession.

Now what does any of that have to do with the orbital period? Well, the sun is one source of torque, and if that’s what you are counting on, you won’t be able to get the rates to match.

Say for example the precession period comes out to 15000 years. So, put the planet far enough away so the orbital period is also 15000 years.

You see why the sun won’t give much torque, wince it is so far away it will apply very little force. I suggest getting a sattelite of the planet to cause the torque since it will be a steady influence that always goes with the planet.

$\endgroup$
1
$\begingroup$

For an Earthlike world in an Earthlike orbit you'd need about 1.4*10^27Nm of torque annually to get the axis of rotation to precess 360 degrees per year which is what you're asking for, that's going to produce 1.4*10^27j of waste heat. To put that in perspective one megaton is 4.18*10^15j so you get the equivalent of 291,666,666,666 million tons of TNT worth of waste heat annually (it's almost exactly 0.1% of the sun's total constant output). Earth weighs 5.97*10^24kg so that's 234.5j per kilogram per annum for the entire planet from core to exosphere, that will heat the whole planet by roughly 0.25K a year every year, you'll melt the whole planet inside of 6800 years and evaporate it en masse after only 11800 years. That assumes you start from a completely solid lump of pure silica which the Earth is not, it will take far less time to vapourise an Earthlike world. You can of course extend the orbit and thus slow down the precession reducing the energy inputs needed.

$\endgroup$
0
$\begingroup$

yes, but it still would be slow. both the orbital period and the precession cycle. precession cycle have, and will almost always be slow. A fast precession cycle would cause too much instability.

$\endgroup$
1
  • $\begingroup$ I would really like to see your answer improved. While this may be absolutely correct, you haven't provided any facts, information or reasons why orbital periods and precession cycles need to be slow for this to work. I'd just like to know why. This is the sizzle, not the steak. $\endgroup$
    – a4android
    Commented Nov 26, 2016 at 7:37
-1
$\begingroup$

Okay...I'm going to take a stab at this question...but I offer this caveat: the last time I studied astronomy was the 8th grade.

I think you could probably achieve this by having a rotating axis of exactly 90 degress. If I understand these mechanics correctly, this would essentially give a precession which would parallel the planet's orbit, as the north and south poles would constantly align with the same point in space. The planet would still rotate, so you would still have day and night, but the time between the two would be split equally all year round. It might also affect the intensity of the seasons, as well as the wind currents, and ocean tides, so those are things to consider if you go this route.

Alternatively, you could try playing around with the size of the orbital path of the planet you are creating relative to the tilt of the axis.

Hope this helps.

$\endgroup$
4
  • $\begingroup$ I really can’t follow what you are saying. If the precession is nil, how is that equal to the orbital period? $\endgroup$
    – JDługosz
    Commented Nov 28, 2016 at 4:12
  • $\begingroup$ The precession is basically a circle "drawn" on the heavens by the northern and southern axis poles. If the axis poles were not tilted - were straight up and down, it would not produce a circle, but a dot. $\endgroup$
    – Cadence
    Commented Nov 28, 2016 at 5:28
  • $\begingroup$ Now, the more I think about it, the less I'm convinced it actually presents a 'nil' precession, but a precession which actually tracks the annual orbit of the planet in question. As I said, I haven't studied a lot of astronomy in the last, oh, thirty years or so, but I did look up some youtube videos that explained this. (I'm answering the question because I liked the challenge of it :) It might be easier to visualize my explanation if you do a video search for precession. $\endgroup$
    – Cadence
    Commented Nov 28, 2016 at 5:33
  • $\begingroup$ I edited my answer to reflect my new thoughts on this. $\endgroup$
    – Cadence
    Commented Nov 28, 2016 at 5:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .