Science Quiz

[Pirate]

Hey! the science quiz is alive again!

Well, I'd say the moon roughly follows the elipse of Earth's orbit, but because the moon is orbiting the Earth, it follows the ellipse in fits and starts and it wobbles in and out, up and down. The end result probably looks like a lopsided corkscrew, if it in fact looks like anything. That's about as specific as I know how to get.

 

[betazed]

Well, when I think about it the moon's orbit around the sun got to be a cycloid. Could be either a epicycloid or hypocycloid. I am not sure which one.

Of course it will not be either one of the exactly because earth's orbit is not exactly circular but slightly elliptical.

 

[Pirate]

Cycloids are formed (briefly) by tracing a point on a circle that is rolling inside(or outside) another circle. You get the "sharp" points when there is no relative motion between the two circles.

Does this happen with the moon? The moon just kind of gets dragged around the sun by the earth. Since the moon's rotation around the earth doesn't necessarily relate to the earth's rotation around the sun (as they would have to for a cycloid) do you get the sharp points? I envision that they would be stretched out.

 

[betazed]

Originally posted by Pirate
Cycloids are formed (briefly) by tracing a point on a circle that is rolling inside(or outside) another circle. You get the "sharp" points when there is no relative motion between the two circles.

Does this happen with the moon? The moon just kind of gets dragged around the sun by the earth. Since the moon's rotation around the earth doesn't necessarily relate to the earth's rotation around the sun (as they would have to for a cycloid) do you get the sharp points? I envision that they would be stretched out.

We are basically talking about a rotating circle going in a circular path. So the sharp points you talk about are only there if the point whose locus we are trying to plot is on the edge of the rotating circle. If the point in inside the circle then we do not get sharp points.

I know it is hard to visualize. I will try to draw it up in Mathematica and post the picture.

 

[betazed]

Actually google is better than my mathematica

http://mathworld.wolfram.com/ProlateCycloid.html

http://mathworld.wolfram.com/CurtateCycloid.html

So the orbit of the moon will be a prolate or a curtate cycloid closing in on itself.

Anybody know the name of such a curve?

 

[Pirate]

Yes, I was envisioning a curtate cycloid, I just didn't know the terminology. (curtate because I think the velocity of the moon around the earth is less than the velocity of the earth around the sun so the moon would never be in retrograde)

But theoretical cycloids are planar, so unless the plane of the moon's orbit is in the exact same plane as the earth's orbit that cycloid will have a 3rd dimension to it - making the actual path more helical.

 

[betazed]

Originally posted by Pirate

But theoretical cycloids are planar, so unless the plane of the moon's orbit is in the exact same plane as the earth's orbit that cycloid will have a 3rd dimension to it - making the actual path more helical.

Good point. So what we really have is a curtate cycloid curving in on itself and oscillating up and down in the earth sun plane. Phew!

Someone needs to draw this out. Anybody care to venture the parametric equations for this curve? Should be easy (could be the next question )

 

[betazed]

What nobody want to write down the equations? Gee, I thought we were full of nerds here.

Here it is.

In two dimensions (that is without taking Pirate's oscillation into account) the parametric equations are { a Cos(t) + Cos(b t), a Sin[t] + Sin (b t) }. I plotted this out in Mathematica.

The nature of the curve depends markedly on the ratio of a to b. For a typical moon orbit I took a = 60 and b = 13 to make the figure look like a cycloid. In reality a will be very large and the figure will be almost a circle.

In three dimensions we just add another z component to the equation and give it a sinusoidal oscillation (since it is a small osccillation anyway) and we get the following.

I magnified the z direction 15 fold to get a better perspective and assumed the moon goes round the earth 12 times in the time the earth goes round the sun (a good first approximation).


The mathematica commands to generate these are

ParametricPlot[{60Cos[t] + Cos[13 t], 60 Sin[t] + Sin[13 t]}, {t, 0, 2\[Pi]}]

and

ParametricPlot3D[{60Cos[t] + Cos[13 t], 60 Sin[t] + Sin[13 t], 15 Sin[12t]}, {t, 0, 2\[Pi]}]

 

[Guildenstern]

I magnified the z direction 15 fold to get a better perspective and assumed the moon goes round the earth 12 times in the time the earth goes round the sun (a good first approximation).

It's a decent enough first approximation. Actually there are about 13 lunar months in a year; a lunar month is slightly less than 28 days. 13 times 28 is 364, which is pretty close. In reality it's slightly less close than that, because lunar months line up again about every 19 years (i.e., about every 19th Halloween is a full moon, for example).

However, whether there are 12 or 13 lunar months does not change the name of the curve we're talking about, whatever it is.

 

[slothman]

Betazed can get it. I was just wondering if many people would think it to spiral and cross itself while in actually it doesn't. A good page is:
http://www.math.nus.edu.sg/aslaksen/teaching/convex.html
Basically it doesn't because the orbit of the Earth is so large compared to the moon that there's too much space. If the EM distance was .5 AU's then it might.

 

[betazed]

Great, I get to ask a question in this thread after a long time.

How are the masses of extra solar planets determined?

Think hard and carefully before you answer.

 

[Guildenstern]

By "extra solar planets", do you mean planets that orbit other stars?

 

[betazed]

Originally posted by Guildenstern
By "extra solar planets", do you mean planets that orbit other stars?

Correct.

 

[Baleog]

The plane of the extra solar planet's (ESP) orbit must be in our own plane so that we first detect a dimming of light coming from the star as the EXP passes btween us and the star (that's how they're found first of all).
The star exerts an attractive gravitational force on the EXP which gives it it's orbit. But in turn the EXP also exerts an attractive force on the star causing it to have it's own minute 'orbit'. As the force of the EXP on the star is much smaller than the star on the EXP only the most massive and most inward planets to the star cause an effect on the star that we can detect. From our point of view this effect manifests itself as a wobble in the star's position. Via Newton's equations the mass of the ESP may be roughly determined.

 

[betazed]

Originally posted by Baleog
The plane of the extra solar planet's (ESP) orbit must be in our own plane so that we first detect a dimming of light coming from the star as the EXP passes btween us and the star (that's how they're found first of all).

Occultation can only detetct the existence of planets and not their mass. Also, this was not how ESPs were first discovered. The wobble that you descrine below was how the first ESP was discovered by Swedish astronomers.

The star exerts an attractive gravitational force on the EXP which gives it it's orbit. But in turn the EXP also exerts an attractive force on the star causing it to have it's own minute 'orbit'. As the force of the EXP on the star is much smaller than the star on the EXP only the most massive and most inward planets to the star cause an effect on the star that we can detect. From our point of view this effect manifests itself as a wobble in the star's position. Via Newton's equations the mass of the ESP may be roughly determined.

Hmmm.... While you have got the basic concept right, the devil as they say is in the details and you have missed some crucial details. Also, This method can only determine the minimum mass. Can anyone tell why? So to determine the actual mass we have to measure something else.

 

[Baleog]

Originally posted by betazed
Hmmm.... While you have got the basic concept right, the devil as they say is in the details and you have missed some crucial details. Also, This method can only determine the minimum mass. Can anyone tell why? So to determine the actual mass we have to measure something else.

I suppose that we don't actually know how far away the planet is from the star so we can't accuratly apply Newton's theory.

We could make up a range of possible values for the distance hence giving a range of possible masses. However it doesn't occur to me at the moment how to come up with such a range. Possibly it is related to the other measureable quantity?

Maybe the percentage dimming of the star's light as the planet passes between it and our line of sight might help to make up some kind of relationship between the volume of the planet and it's distance from the star?

 

[betazed]

Nope. You are on the wrong track. maybe it is google time (for some research).

 

[Baleog]

Right. To find out how much a gravitating object is gravitating you need to know it's mass and how far away it is to the object it is 'gravitating'.

But as we know how much it is gravitating (the wobble) we have two unknowns. The mass and distance. So we need to determine the distance to find mass. I'll kick myself when I find out.

The period of the wobble could be used in conjunction with how often the planet orbits (which we know by the dimming of the star, once for each orbit) via Keppler to get the distance between star and planet and hence mass of planet?

 

[betazed]

@Baleog: As a hint first think why the wobble method would determine a minimum mass and never the actual mass. Then try to determine what other data you would need to determine the missing data in the wobble method.

 

[Baleog]

Don't worry I give up. This is why I fail astrophysics.