# Need Help on twin-planet problem!

1. Sep 24, 2004

### lubi

I met a question when I try to solve a twin-planet problem in "Classical mechanics" by Goldstein 3rd edition. Chap3-17. As followings:
One of the classic themes of science fiction is a twin planet (“Planet X”) to Earth that is identical in mass, energy, and momentum but is located on the orbit 180° out of phase with Earth so that it would be hidden by the Sun. However because of the elliptical nature of the orbit it would not always be completely hidden. Assume there is such a planet in the same Keplerian orbit as Earth in such a manner that it is in aphelion (furthest from the Sun) when Earth is in perihelion (closest to the Sun). Calculate to first order in eccentricity e the maximum angular separation of the twin and the Sun as viewed from Earth. Could such a twin be visible from Earth? Suppose the twin planet were in an elliptical orbit having the same size and shape as that of Earth, but rotated 180° from the orbit of Earth, so that Earth and the twin would be in perihelion at the same time. Repeat your calculation and compare the visibility in the two situations.
Would you like to give some hint to solve this problem?
I will be very appreciated for your great help.

2. Sep 26, 2004

### tony873004

180 degrees away from Earth is called the Lagrange 3 point. But the L3 point is not stable, and a planet there will wander out of the L3 point, and more than likely collide with Earth. So the maximum angular seperation from the Sun could be 180 degrees, as its perturbed orbit will likely carry it past Earth, both interior and exterior to Earth's orbit many times before the ultimate collision takes place.

But if you'd like to pretend that L3 is stable, you must compare the mean anomolies of both fake earth and real Earth with their true anomolies, find the minimum angle that (fake earth, Sun, real Earth) can form. In the same triangle, the angle (Sun, real Earth, fake earth) will be your angular seperation between the Sun and fake earth. There's probably an easy way to do it if the book gives it as a problem. Does the text offer any clues.

I'm pretty sure that the second scenerio, with both at perihelion at the same time, 180 apart, you will have 0 degrees of seperation for the entire orbit since they're basically mirroring each other's actions. But there will be seperation in the other scenerio. But even if the seperation is enough that fake earth will emerge from behind the solar disk, I doubt the seperation will be enough for the fake earth will climb out of the Sun's glare. Although I imagine they'd want you to ignore the glare.

Last edited: Sep 26, 2004
3. Sep 26, 2004

### enigma

Staff Emeritus
L3 is not in the same orbital track as the secondary planet, which I believe is presupposed for this problem.

lubi, it's an interesting question.

I'll think about it a bit. My preliminary guess is that the solution may lie by playing around with the true anomaly - eccentric anomaly relationship. I could easily be wrong, though.

4. Sep 26, 2004

### tony873004

Actually, you're right. For the second scenerio (both at perihelion at the same time) they're not in the same orbital track.

Although for the first scenerio (PlanetX at aphihelion while Earth is at perihelion) they are in the same track.

Scenerio 1: Angular seperation will occur (eccentricity and sizes exaggerated):
http://orbitsimulator.com/orbiter/PlanetX1.GIF

Scenerio 2: Angular seperation will not occur (eccentricity and sizes exaggerated):
http://orbitsimulator.com/orbiter/PlanetX2.GIF

5. Sep 27, 2004

### BobG

The twin planets will always have a Mean Anomaly 180 degrees apart (Mean Anomaly could almost be thought of as identifying the objects location in the orbit by time). The better way to identify their position is by radians - they're mean anomalies are always pi radians apart.

The maximum difference in Eccentric Anomaly (in radians) will equal the eccentricity of the orbit. (M=E-e sin E). It will occur when Eccentric Anomaly is pi/2 radians (90 degrees) or 3pi/2 (270 degrees) [absolute value of the difference].

Since the Earth's orbit around the Sun has an eccentricity of .0167, the maximum difference between Eccentric Anomaly and Mean Anomaly is .0167 radians, or .0167*pi/180 degrees.

You technically have to convert the Eccentric Anomaly to True Anomaly to find how far apart True Anomaly and Mean Anomaly are. But, with an eccentricity of only .0167, you can safely say that both objects will be offset from Mean Anomaly by a maximum (absolute value) of about 1 degree. Your eccentricity has to be pretty high to get big differences between Eccentric Anomaly and True Anomaly - the tangents of the half-angle of the two are related by [(1+e)/(1-e)]^(1/2)

Both will hit maximum at the same time - one positive and one negative. That gives you a 1-1-178 triangle (approximately). Your maximum separation is very close to .0167*pi/180, or slightly less than 1 degree. The Sun's apparent width is about 0.5 degrees, .25 degrees either side of the direct line joining the center of the Earth and Sun.

The twin planet technically lies in the line of sight of the Earth (but, as Tony said, it would be too close to the Sun to be visible to the naked eye).

Last edited: Sep 27, 2004
6. Oct 16, 2004

### Jenab

I assume that the two planets follow the same orbit, but that they are pi radians apart in mean anomaly. They are pi radians apart in true and eccentric anomalies only when each is at either apside.

Let me put Earth initially at perihelion and Planet X initially at aphelion.

I notice, first, that the angular separation, subtended at Earth, between Planet X and the Sun, is

D = pi - (Q2 - Q1)

Where Q1 is the true anomaly of Earth and Q2 is the true anomaly of Planet X. This follows from the geometry of the triangle having vertices at Earth, the sun, and the point of intersection between the Earth-Planet X line and the orbit major axis.

Since Q2>Q1, one would minimize (Q2-Q1) in order to maximize the angular separation D.

Now, Earth starts at perihelion, whereas Planet X is starting from the aphelion of the same orbit, which means that Earth is catching up to Planet X in true anomaly and will continue to do so until...

Q2 = 2 pi - Q1

Meaning that...

Dmax = 2 Q1 - pi

To make the math easier, we can recognize that the eccentric anomaly (u) has the same geometrical symmetry that the true anomaly does:

u2 = 2 pi - u1

(Understand, though, the the mean anomaly does not have that symmetry!)

The angular separation, subtended at Earth, between the sun and Planet X is maximized when Earth and Planet X are equidistant from (and on the anti-solar side of) the minor axis of their mutual orbit. Accordingly, the mean anomaly interval traveled by Earth from perihelion equals the mean anomaly interval traveled by Planet X from aphelion.

M2 = M1 + pi

Logic time's over. Time to crunch some numbers.

M1 = u1 - e sin u1

M1 = u2 - e sin u2 - pi

u2 - e sin u2 - pi = u1 - e sin u1

u2 - u1 + e (sin u1 - sin u2) = pi

2 pi - u1 - u1 + e {sin u1 - sin (2 pi - u1)} = pi

2 pi - 2 u1 + 2 e sin u1 = pi

u1 - e sin u1 = pi/2 = M1

Well, what have we here? The maximum angular separation, subtended at Earth, between the sun and Planet X, seems to occur when Earth is 1/4 of the way around its orbit from perihelion, in terms of time or mean anomaly.

(Added later: You know, if I'd been thinking more clearly, I'd have known from symmetry that the point of equidistance from the minor axis, with both planets on the anti-solar side thereof, must occur when they are one-quarter of the orbital period from their respective apsides. After all, the times in which the two planets travel are equal, and the arc of one planet, reflected across the major axis, simply completes the half of the orbit which the other planet travels. I was a doofus for doing the work, and I was lucky not to have made any mistakes while I was at it.)

You can now use Newton's method to get u1. Since the eccentricity of Earth's orbit is 0.01671022, the eccentric anomaly will work out to about 1.587504 radians.

You can convert from u1 to Q1:

x1 = a (cos u1 - e)

y1 = a sin u1 (1-e^2)^0.5

Q1 = arctan2(y1 , x1)