Are there any physicists in the house? I need your help..

[JeremiahTheGreat]

I wouldn't normally ask for help on assignments, but since this was simply a "since we've finished this term already, i'll give you a bumma of a question to keep you quiet"..

The teacher wants us to plot the path of the MOON around the SUN... and its driving me nuts!!! He said something like the curve of the plot never deviats from the same position or something, and when we said it would be a circle, he wouldn't tell us the real answer!!!

Any Suggestions?


Thanks

 

[flood]

pick a point on a bicycle wheel and follow it as it travels along the ground. Each cycle will be a catenary curve. (not to be mistaken for a sinusoidal or quadratic path.)
Put several of these together and then imagine them put together in a circle. (there are two arrangements possible depending on the direction of rotation of the moon around the earth)

 

[Armitage]

Plot it to scale and he can't argue with you ... just draw a circle around a point and label the radius as 1.5x10^8 Km

The path of the moon is actually a sinusoid imposed on the path of the earth which is an ellipse), but the magnitude of the sinusoid is trivial (384400 Km) wrt to the distance from the earth to the sun. If you drew the earth-sun orbit as a 4" radius circle (so you could fit it on your 8.5" x 11" exam paper), the magnitude of the maximum distance of the moon from the earth would be about 1/32" ... to small to draw with a number 2 pencil

Actually, the path of the earth is a sinusoid imposed on an elliptic orbit also. The path of the orbit is actually traced out by the center of mass of the earth moon system. Both the earth and the moon orbit this center of mass, but due to the size difference, the C.M. is well within the earth so the orbit shows up as more of a wiggle.

 

[rummyPPG]

I don't know what the name for the curve would be, but using simple superposition you should be able to get an equation to toss into MatLab to generate a plot. The earth is eliptical so its position in a x-y coordinate system (choose the major axis to contain the +/- x directions) would be:
x = a1 cos [2pi/period(earth)*(t)]
y = b1 sin [2pi/period(earth)*(t)]

I'm not positive about the period stuff, i'd have to think about it more carefully, but its key to use the superposition with the next equation correctly.

Now assume that the earth is stationary and just consider the moon going around it. If it has an eliptical path as well then its position in the same x-y coordinate system as before would be:
x = a2 cos [2pi/period(moon)*(t)]
y = b2 sin [2pi/period(moon)*(t)]

So that means at any given time you can figure out the earths position relative to the sun and then add in the moon's position relative to the earth to get a posistion of the moon relative to the sun
x = a1 cos [2pi/period(earth)*(t)] + a2 cos [2pi/period(moon)*(t)]
x = b1 sin [2pi/period(earth)*(t)] + b2 sin [2pi/period(moon)*(t)]

The trick is to get the periods correct so that the rotate right (i.e. it shouldn't take the moon one revolution around the earth for the same time it takes the earth around the sun, otherwise the picture will not be right; the big ellipse won't have enough "squiglyness" in it)

Good luck and post any plot pics if you get it done.

Thanks

 

[damn1]

LMAO!
You guys are all correct, ellipsoidal nature of planetary motion.. For superior precision look up just how much of an ellipse the Earth's orbit really is, and then worry about superimposing the catenary of the moon's motion onto the Earth's orbit. It gets better, because it's not the center of the Earth that traverses the elliptical path, but a center of mass of the Earth/Moon system. At this point you oughtta say, ola, crappola !

 

[Armitage]

Originally posted by: damn1
LMAO!
You guys are all correct, ellipsoidal nature of planetary motion.. For superior precision look up just how much of an ellipse the Earth's orbit really is, and then worry about superimposing the catenary of the moon's motion onto the Earth's orbit. It gets better, because it's not the center of the Earth that traverses the elliptical path, but a center of mass of the Earth/Moon system. At this point you oughtta say, ola, crappola !

Ok, so what exactly are you adding to this discussion?
And it won't be a catenary you superimpose on the ellipsoid, it will be a sinusoidal curve.

FWIW, the eccentricity of the earth's orbit is about 0.0167, so the earth is about 5e6 Km closer to the sun @ perigee then @ apogee. That translates to about 1/8" difference between apogee & perigee on the 4" scale drawing I described previously. Still far more accurate then you could draw freehand on an exam paper.

 

[Armitage]

Originally posted by: rummyPPG
I don't know what the name for the curve would be, but using simple superposition you should be able to get an equation to toss into MatLab to generate a plot. The earth is eliptical so its position in a x-y coordinate system (choose the major axis to contain the +/- x directions) would be:

x = a1 cos [2pi/period(earth)*(t)]
y = b1 sin [2pi/period(earth)*(t)]

I'm not positive about the period stuff, i'd have to think about it more carefully, but its key to use the superposition with the next equation correctly.

The problem with these parametric equations is that the speed of an object on an elliptic orbit is not constant. An orbiting body moves faster @ perigee then @ apogee. More specifically, by Kepler's 2nd law, a line joining the primary object to the secondary object always sweeps out equal areain equal time.

If you neglect the eccentricity of the orbit, which is reasonable as a first approx, your stuff looks good.

Now assume that the earth is stationary and just consider the moon going around it. If it has an eliptical path as well then its position in the same x-y coordinate system as before would be:

x = a2 cos [2pi/period(moon)*(t)]
y = b2 sin [2pi/period(moon)*(t)]

Minor note, but these equations give the position of the moon wrt the earth, not "the same coordinate system as before".

So that means at any given time you can figure out the earths position relative to the sun and then add in the moon's position relative to the earth to get a posistion of the moon relative to the sun
x = a1 cos [2pi/period(earth)*(t)] + a2 cos [2pi/period(moon)*(t)]
x = b1 sin [2pi/period(earth)*(t)] + b2 sin [2pi/period(moon)*(t)]

The trick is to get the periods correct so that the rotate right (i.e. it shouldn't take the moon one revolution around the earth for the same time it takes the earth around the sun, otherwise the picture will not be right; the big ellipse won't have enough "squiglyness" in it)

That bit will be taken care of as long as you have the right periods for both, which is easy. Getting the stuff syncronized, such that the moons path crosses the earths path at the right time & direction is a bit trickier.

Good luck and post any plot pics if you get it done.

Thanks

Another note, the moon's orbit is inclined by about 5 degrees to the earth's equator, which is inclined roughly 20 degrees to the earth's orbital plane ... just to complicate things a bit more

 

[Tbirdkid]

hahahahahahaha..... dude..... i gotta go back to college and actually care..... hahahahahah

 

[rummyPPG]

ergeorge

The problem with these parametric equations is that the speed of an object on an elliptic orbit is not constant. An orbiting body moves faster @ perigee then @ apogee. More specifically, by Kepler's 2nd law, a line joining the primary object to the secondary object always sweeps out equal areain equal time.

Umm the parametric equations won't actually give out a constant speed:

x = a1 cos [2pi/period(earth)*(t)]
y = b1 sin [2pi/period(earth)*(t)]

dx/dt = -a1*(2pi/period(earth)) sin [2pi/period(earth)*(t)]
dy/dt = b1*(2pi/period(earth)) cos [2pi/period(earth)*(t)]

dy/dx = - b1/a1 cot [2pi/period(earth)*(t)]

speed = |dy/dx|

However it is duly noted that this is not the correct speed of the bodies either, as they obvoiously don't go near an infinite speed as cotangent does near 0 and pi (and 2pi and 3pi ...)

But regardless, the teacher asked for a plot of the curve. He made no requests for the accurate speed of the earth or moon so if I was taking the class, I'd use the simplest method possible to produce the curve, which i believe is the superposition of the parametric equations with the periods (and phases as you pointed out) correctly adjusted.

If you remember what the speed is for an eliptical path (or the velocity would actually be better), we can integrate that to get a completely accurate equation for the function, which I'd be totally game for.

Another note, the moon's orbit is inclined by about 5 degrees to the earth's equator, which is inclined roughly 20 degrees to the earth's orbital plane ... just to complicate things a bit more

If I was doing this problem I'd be graphing it in 2D unless explicitly told otherwise. I don't know if I'm right on this, but it seems if you're in 2D and you take an ellipse and rotate it out of the plane on its major (or minor i suppose) axis, then it will fall back into the plane into an ellipse with a shorter minor or major axis, depending on which one you rotated.

The way I visualize this is to have a coat hanger in an elliptical shape sitting on a desk and you take a top view snapshot of it to get an accurate picture of the ellipse in that plane (of the desk). Then you leave the camera where it is, take the desk away while holding the coat hanger stationary then rotate it on one of its axises (sp?). Take another snapshot with the unmoved camera (still looking at the 2D plane of the desk) and i think you see another ellipse with one axis shrunk. I could be wrong, but I don't know what else you'd do, short of heading 3D, and that doesn't sound that appealing to me.

Side note: Is this problem still due, or has that passed and we're discussing for the sake of physics? just curious

 

[Armitage]

Originally posted by: rummyPPG
ergeorge

The problem with these parametric equations is that the speed of an object on an elliptic orbit is not constant. An orbiting body moves faster @ perigee then @ apogee. More specifically, by Kepler's 2nd law, a line joining the primary object to the secondary object always sweeps out equal areain equal time.

Umm the parametric equations won't actually give out a constant speed:

x = a1 cos [2pi/period(earth)*(t)]
y = b1 sin [2pi/period(earth)*(t)]

dx/dt = -a1*(2pi/period(earth)) sin [2pi/period(earth)*(t)]
dy/dt = b1*(2pi/period(earth)) cos [2pi/period(earth)*(t)]

dy/dx = - b1/a1 cot [2pi/period(earth)*(t)]

speed = |dy/dx|

Oops, you're correct, but not for the reason you show here.

Speed = dPOSITION/dT not dY/dX

So, taking your derivatives for x & y velocity, we can find the magnitude of the velocity as:

dPOSITION/dt = V = sqrt((dx/dt)^2 + (dy/dt)^2)
Skipping some steps...
V = (2 pi/ period) sqrt( (a^2 sin^2(2 pi t/period) + b^2 cos^2(2 pi t/period))

Note that if the curve is circular (ie. a = b), then V is constant. In the general case where a != b, V is not constant, but it does not vary in the way that an orbital velocity does because the velocity described here must be symmetric about both the major & minor axis of the ellipse, while an orbital velocity is only symmetric about the minor axis.

However it is duly noted that this is not the correct speed of the bodies either, as they obvoiously don't go near an infinite speed as cotangent does near 0 and pi (and 2pi and 3pi ...)

Yea, that singularity was my first clue that your velocity equation was wrong. Your parametric position equations correctly describe an ellipse, so the velocity should have been sane

But regardless, the teacher asked for a plot of the curve. He made no requests for the accurate speed of the earth or moon so if I was taking the class, I'd use the simplest method possible to produce the curve, which i believe is the superposition of the parametric equations with the periods (and phases as you pointed out) correctly adjusted.

Yea, for the sake of this assignment, I'd make the assumption that a = b (circular orbit) for both earth & moon and simplify everything enough to get a generic plot that shows the major features.

If you remember what the speed is for an eliptical path (or the velocity would actually be better), we can integrate that to get a completely accurate equation for the function, which I'd be totally game for.

Ahh, now you want a propagator! Unless you have some insight into the Kepler equation that the rest of mathematics hasn't had in the last 400 years, that's a bit beyond the scope of this post Of course, there is lots of software that can do it for you.

Another note, the moon's orbit is inclined by about 5 degrees to the earth's equator, which is inclined roughly 20 degrees to the earth's orbital plane ... just to complicate things a bit more

If I was doing this problem I'd be graphing it in 2D unless explicitly told otherwise. I don't know if I'm right on this, but it seems if you're in 2D and you take an ellipse and rotate it out of the plane on its major (or minor i suppose) axis, then it will fall back into the plane into an ellipse with a shorter minor or major axis, depending on which one you rotated.

The way I visualize this is to have a coat hanger in an elliptical shape sitting on a desk and you take a top view snapshot of it to get an accurate picture of the ellipse in that plane (of the desk). Then you leave the camera where it is, take the desk away while holding the coat hanger stationary then rotate it on one of its axises (sp?). Take another snapshot with the unmoved camera (still looking at the 2D plane of the desk) and i think you see another ellipse with one axis shrunk. I could be wrong, but I don't know what else you'd do, short of heading 3D, and that doesn't sound that appealing to me.

LoL. No I would definitely stick to 2D. I was just being difficult!

 

[rummyPPG]

Oops, you're correct, but not for the reason you show here.

Speed = dPOSITION/dT not dY/dX

So, taking your derivatives for x & y velocity, we can find the magnitude of the velocity as:

dPOSITION/dt = V = sqrt((dx/dt)^2 + (dy/dt)^2)
Skipping some steps...
V = (2 pi/ period) sqrt( (a^2 sin^2(2 pi t/period) + b^2 cos^2(2 pi t/period))

Note that if the curve is circular (ie. a = b), then V is constant. In the general case where a != b, V is not constant, but it does not vary in the way that an orbital velocity does because the velocity described here must be symmetric about both the major & minor axis of the ellipse, while an orbital velocity is only symmetric about the minor axis.

You're completely correct. I was trying to simplify too much and thinking in 1D, pretending that the celestial bodies moved just up-down, which is obviously incorrect. Good thing i was of a clearer mind when I did my last calc tutoring sessions a month ago.

Question: Is there any way to figure out the area of a "pie" piece of an ellipse given an equation(s) for the ellipse and two points on the ellipse?

 

[Armitage]

Originally posted by: rummyPPG

Question: Is there any way to figure out the area of a "pie" piece of an ellipse given an equation(s) for the ellipse and two points on the ellipse?

Yes there is, but I'll have to look into my astro notes to find it. I should have it there, because that's needed in using Kepler's 2nd law.