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Imagine another Earth-sized planet, in the exact same orbit as Earth, but 180 degrees out-of-phase. In this arrangement, at all times, you would be able to draw a single straight line through space that starts with Earth, crosses through the Sun at the midpoint, and ends at Opposite-Earth.

Would our current systems of observation (telescopes, Voyager, etc etc) be capable of perceiving the existence of Opposite-Earth?

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6 Answers 6

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We could certainly be able to infer its existence through tracking the motions of the planets we can see and then simulating the gravitational motions of those planets. Adding a similar sized planet as ours on the opposite side of the orbit would have a gravitational interaction with all the other planets and would lead to a discrepancy between observations and the simulation.

I certainly would hope that noticing that there's a missing mass situated near us probably would lead someone in the past to send off a telescope/probe to the L3 point to find it (though I imagine the L4 or L5 points would also work). Interestingly enough, the Wikipedia link above adds the following comment,

Sun–Earth L3 was a popular place to put a "Counter-Earth" in pulp science fiction and comic books, despite the fact that the existence of a planetary body in this location had been understood as an impossibility once orbital mechanics and the perturbations of planets upon each other's orbits came to be understood, long before the Space Age; the influence of an Earth-sized body on other planets would not have gone undetected, nor would the fact that the foci of Earth's orbital ellipse would not have been in their expected places, due to the mass of the counter-Earth.

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    $\begingroup$ but a low mass hollow earth space station could be at L3, covered in RAM and vanta-black. $\endgroup$
    – JEB
    Commented Aug 4, 2023 at 17:01
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    $\begingroup$ FWIW, I don't think the Wikipedia quote is exactly correct. Sun-Earth L3 is where you can place a negligible-mass object and get an almost-stable orbit (in the theoretical case of only 3 bodies in the universe and perfectly circular orbits, it will last forever). An Earth-mass object should be an equal distance from the sun, while L3 is farther (I originally said, closer, but I got that wrong). Maybe the consensus is to still call it L3, but the Newtonian 3-body solution with two identical objects orbiting a central one antipodally is a different solution from that of Lagrange. $\endgroup$ Commented Aug 5, 2023 at 18:52
  • $\begingroup$ Hey, sorry I've moved the accepted answer, but @Mike's answer was ultimately too useful to ignore. You may want to check it out as it is very cool. $\endgroup$ Commented Aug 9, 2023 at 9:08
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    $\begingroup$ @ConnieMnemonic it's your prerogative to mark/unmark answers as accepted; no need to apologize. $\endgroup$
    – Kyle Kanos
    Commented Aug 9, 2023 at 20:02
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You have some good answers already, but neither touch on a critical issue with this proposal.

The wikipedia about Lagrange points clarifies that the Lagrange point in question (L3) is not stable. (See the comments for the higher-level discussion about how the Wikipedia assumes the object to be very light compared to Earth, but the question does not)

enter image description here

So if you tried to put your opposite-Earth there, a tiny error in the positioning of the two planets would grow over time and after not-so-long (millions of years probably since $M_{\text{earth}}/M_{\text{sun}}=3\times10^{-6}$) the two planets would no longer be on opposite sides of the sun. Eventually one would be sent farther away and one would be sent closer to the sun or they would collide (less likely).

Note that JWST sits at another unstable Lagrange point (L2), and in order to remain there it occasionally does a small propellant burn to correct its orbit.

Only L4 and L5 are stable, so natural objects are only ever found sitting in these Lagrange points. There are many in Jupiter's L4 and L5 Lagrange points.

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    $\begingroup$ Lagrange points describe the stable and unstable orbit of a third light object interacting gravitationally with two more massive objects. The underlying assumption is broken in the case of a system Sun-Earth_1, Earth_2. $\endgroup$ Commented Aug 4, 2023 at 14:49
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    $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 You are correct, wikipedia is considering the new object to be much lighter than the Earth. I do think it's unlikely that increasing the mass ratio will make these points go from unstable to stable. In fact I believe the potential plot still applies... it's the potential due to the earth and the sun on the new object (unaffected by the new object). But technically we don't know until we do a lot of math or find someone else who did it. (ignore my deleted comment where I thought incorrectly that a thesis was showing that it remains unstable for all mass ratios) $\endgroup$
    – AXensen
    Commented Aug 4, 2023 at 15:10
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    $\begingroup$ FWIW, the JWST is in a halo orbit, with a mean distance of ~640,000 km from the L2 point. Here's a plot (with a 3 day time step) I just made, using JPL data. i.sstatic.net/HpijS.png Also see space.stackexchange.com/a/57832/38535 $\endgroup$
    – PM 2Ring
    Commented Aug 4, 2023 at 22:18
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    $\begingroup$ I think it drifts away faster than that--you're thinking of a three-body situation but we have Jupiter out there tugging on things. $\endgroup$ Commented Aug 5, 2023 at 3:18
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Just another piece of information, complementing the issue of stability correctly raised by @AXensen.

I do not know if the configuration you suggested is stable, secularly stable, or unstable. What I can add, by checking numerically, is that if the initial conditions for Earth_2 are not exactly the initial conditions for Eart_1, rotated by $180$ degrees around the position of the Sun, the resulting orbit of the two Earths around the Sun can be stable over thousands of orbits, but the three bodies do not remain on a straight line, and, with a difference of the aphelion distances of less than 0.1 %, deviations are large enough to allow to see one Earth from the other.

Taking into account the perturbations due to the remaining part of the Solar System, even without adding the presence of the Moon, completely symmetric dynamics does not seem a probable event.

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    $\begingroup$ Nice investigation. If the configuration is unstable, it will take more like millions of orbits for that to be obvious. And that'll be hard to show with a simulation (hard to be sure numerical issues are sufficiently removed to resolve this effect). The instability will be due to the influence of the two planets on each other. But the gravitational forces on each planet is dominated by the sun $M_{\text{earth}}/M_{\text{sun}}=3\times10^{-6}$. $\endgroup$
    – AXensen
    Commented Aug 4, 2023 at 16:55
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    $\begingroup$ @AXensen Actually, this configuration may require millions of orbits to reveal the instability. Laplace's restricted three-body system (three equal masses at the vertices of an equilateral triangle) shows its instability quickly. However, the main point is that the probability of the symmetric starting point is extraordinarily small. and almost every small perturbation would allow to see directly the other Earth. $\endgroup$ Commented Aug 4, 2023 at 21:22
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    $\begingroup$ actually I just realized... this effect your describing, where a small difference in orbit radius causes them to get out of phase is the instability in question. The fact that the second-earth doesn't orbit the lagrange point when slightly displaced from it is exactly the instability in question. Also, I said "it will take millions of orbits" and you replied "Actually it will require millions of orbits". Seems like a typo $\endgroup$
    – AXensen
    Commented Aug 4, 2023 at 21:45
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    $\begingroup$ @AXensen It was not a typo. One instability may be revealed after a very short time, and others may require longer times (it has t do with the value o Lyapounov's exponents). This one does not seem to manifest in a very short time. $\endgroup$ Commented Aug 5, 2023 at 6:11
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    $\begingroup$ Millions of years is still short in astronomical time. Unless it was placed there recently, we should have seen it by now. $\endgroup$ Commented Aug 5, 2023 at 9:17
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With telescopes it maybe difficult, and I personally don't know much about cataloguing of the night sky. However satellites can detect such planets. One way could be to observe mysterious gravitational pull on incoming asteriods and comets from outer solar system, or even 'mars' or 'venus' in that hypothetical system. Best example is, Neptune was not discovered by observation directly but by doing similar caluclations about unusual orbits and gravitational effects on other bodies which indicated the 8th planet.

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    $\begingroup$ Thank you for your response. Would it be reasonable to say that if Opposite-Earth existed, we would have noticed by now? $\endgroup$ Commented Aug 4, 2023 at 13:50
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    $\begingroup$ @ConnieMnemonic In the range of inner planets (planets up to the orbit of Mars) the only planet ever hypothesized was a planet significantly inside the orbit of Mercury, being referred to as the planet 'Vulcan'. If it would exist it would be a planet with a period of revolution in the order of weeks. A planet orbiting exactly opposite to the Earth was never hypothesized, as such a planet would introduce anomalies in the orbits of the other inner planets, and anomalies like that were not observed. $\endgroup$
    – Cleonis
    Commented Aug 4, 2023 at 14:09
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    $\begingroup$ From even a small space telescope it's not hard to see very close to the Sun. In this answer to What is this white dot and strange line in SOHO image? there's a GIF of the Pleiades passing 4° from the Sun but approaches as close as 1° would be easy to see wit this 11 cm aperture telescope + chronograph (a small blocking disk at the focal plane). $\endgroup$
    – uhoh
    Commented Aug 4, 2023 at 21:42
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Yes, you would be able to see it for most of the year with binoculars or a small telescope

While the other answers are correct, they haven't answered the question. And the answer is yes, the other Earth would be easily directly detectable.

The Earth's orbit is an ellipse, and a consequence of this is that a planet sharing every orbital parameter other than the orbital phase (180 degrees off) will not be opposite the Sun for very long. The velocities of the planets at perihelion and aphelion will be quite different

$$v_{peri, aph} = \sqrt{\frac{GM_*}{a} \frac{1 \pm e}{1 \mp e} }$$ With $v_{peri}$ = 30.291 km/s and $v_{aph}$= 29.296 km/s. So there is a 1 km/s difference between the planets' speeds, which will lead to the following situation:

enter image description here

Where while the Sun blocks the view of one planet for a while, it is not blocked forever. (Here, the eccentricity of the earth's orbit (0.01671) has been grossly exaggerated.)

We can crudely estimate how long after peri/aphelion it would take for one planet to be visible from the other by computing

$$t_{vis} \approx \frac{R_{*}}{v_{peri} - v_{aph}}\approx 8 \ \rm{ days}$$

This completely ignores the effect of non-circularity of the orbital path, but uses the differences in velocity.

We can check this result easily with an orbital simulation (code at end) and obtain the following result:

enter image description here

this shows that one planet would be visible from the other for all but a few days per year (and also confirms the above approximation very well). It would be at a maximum of about 1.5 degrees from the limb of the Sun.

Ok, but could it be seen by eye? The magnitude of the Earth is -4, and observed from 2 AU away it reduces to -2.5, see [**] so just a bit fainter than Jupiter. At this angular separation, (about a thumb's width held at arm's length) it would be challenging to see behind the glare of the Sun but easily done with binoculars or a small telescope after sunset, particularly if using a coronagraph to block the Sun out. The following image shows almost the exact angular separation Earth #2 would be at from the Sun, except this time it is Venus, which is about twice as bright (and saturating the detector)

https://spaceweather.com/images2022/22oct22/divebomb2.gif

[**]https://astronomy.stackexchange.com/questions/41226/formula-to-calculate-the-apparent-magnitude-of-earth-from-arbitrary-distances

import time
import numpy as np
import matplotlib.pyplot as plt
import sys

G = 6.67430e-11 #Newton's gravity constant
M = 1.9891e30 #Mass of Sun in kg
AU = 1.495978e11 #1 astronomical unit in meters
R_Sun = 0.00465047*AU #Radius of sun in meters
day = 24*60*60 # day in seconds

def distance_from_line_to_origin(a, b, c):
    return np.abs(c) / np.sqrt(a**2 + b**2)

def two_point_line(x1, y1, x2, y2):
    """takes a two point form line and returns
    the coefficients of the form a*x + b*y + c = 0
    using y-y1 = (y2-y1)/(x2-x1) * (x-x1)"""
    m = (y2 - y1) / (x2 - x1)
    a = m
    b = -1
    c = y1 - m * x1
    return a, b, c

def f(r):
    """derivative function to pass to rk4
    pass state vector r = [x, y, vx, vy]"""
    x, y, vx, vy = r
    rcubed = np.sqrt(x**2 + y**2)**3

    fx = vx
    fy = vy
    fvx = -G*M*x/rcubed
    fvy = -G*M*y/rcubed
    return np.array([fx, fy, fvx, fvy])

def rk4_step(r=None, h=None, f=None):
    """Takes a single step using the RK4 algorithm.
    r = state vector
    h = stepsize
    f = dx/dt function"""
    k1 = h*f(r)
    k2 = h*f(r+0.5*k1)
    k3 = h*f(r+0.5*k2)
    k4 = h*f(r+k3)
    return 1.0/6*(k1 + 2*k2 + 2*k3 + k4)

def run_rk4_fixed(initial_state=None, initial_h=None, tmax=None):
    """Runs a fixed RK4 integrator from the initial state and time 0
    until the max time.
    initial_state = initial state vector
    initial_h = stepsize (fixed)
    tmax = max time"""
    r = initial_state
    h = initial_h
    xpoints = []
    ypoints = []
    t=0
    while t<tmax:
        r1 = r + rk4_step(r=r, h=h, f=f)
        r = r1
        xpoints.append(r[0])
        ypoints.append(r[1])
        t = t+h
    return np.array([xpoints, ypoints])

if __name__ == "__main__":
    h = 1.0e3 #step size
    tmax = 366*24*60*60 #total time
    times = np.arange(0, tmax, h)

    a = 1.0000*AU
    e = 0.01671 #unitless

    #initial condition of Earth #1 at perihelion
    xperi = -a * (1 - e)
    yperi = 0
    vperi_x = 0
    vperi_y = -np.sqrt( G * M * ((1 + e)/(1 - e)) / a)
    s_earth1 = np.array([xperi, yperi, vperi_x, vperi_y])

    #initial condition of Earth #2 at aphelion
    xaph = a * (1 + e)
    yaph = 0
    vaph_x  = 0
    vaph_y = np.sqrt( G * M * ((1 - e)/(1 + e)) / a)
    s_earth2 = np.array([xaph, yaph, vaph_x, vaph_y])
    #r0 = np.array([x0, y0, vx0, vy0])

    print("Fixed RK4")
    t0 = time.time()
    xpos_e1, ypos_e1 = run_rk4_fixed(initial_state=s_earth1, initial_h=h, tmax=tmax)
    xpos_e2, ypos_e2 = run_rk4_fixed(initial_state=s_earth2, initial_h=h, tmax=tmax)
    t1 = time.time()

    print("Fixed RK4 took ",t1-t0," seconds to complete with a step size of \
    , ", h, " meaning a total steps of ",str(len(xpos_e1)))
    fig, ax = plt.subplots(figsize=(10, 10))
    plt.title('RK4, fixed step, '+str(len(xpos_e1))+' total points')
    plt.plot(xpos_e1/AU, ypos_e1/AU, alpha = 0.5)
    plt.plot(xpos_e1/AU, ypos_e1/AU, 'k.', label='Earth 1', markersize=0.01)

    plt.plot(xpos_e2/AU, ypos_e2/AU, alpha = 0.5)
    plt.plot(xpos_e2/AU, ypos_e2/AU, 'k.', label='Earth 2', markersize=0.02)

    plt.plot([0],[0], 'yo',label='Sun')
    plt.gca().set_aspect('equal', adjustable='box')
    plt.xlabel('X position [AU]')
    plt.ylabel('Y position [AU]')
    plt.legend()
    plt.show()

    dists = []
    for idx, t in enumerate(times):
        a, b, c = two_point_line(xpos_e1[idx], ypos_e1[idx],
                                 xpos_e2[idx], ypos_e2[idx])
        dists.append(distance_from_line_to_origin(a, b, c))

    angles = np.array(dists)/AU*180/np.pi
    fig, ax1 = plt.subplots()
    ax1.set_xlabel('Time since perihelion (approx Jan. 4) [days]')
    ax1.set_ylabel('Distance from center of Sun [Solar Radii]')
    ax1.plot(times/day, np.array(dists)/R_Sun, color='black')
    plt.fill_between(times/day, np.repeat(1, len(times)), alpha=0.3,
                     color='gray', label='Hidden behind Sun')
    plt.legend()
    ax2 = ax1.twinx()  # instantiate a second axes that shares the same x-axis
    ax2.set_ylabel('Angle from center of Sun [deg]', color='black')
    ax2.plot(times/day, angles)#, color=color)
    ax2.tick_params(axis='y')#, labelcolor=color)
    plt.savefig('another_earth.png', dpi=300)
    plt.show()
   
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  • $\begingroup$ I just had to mark this as the accepted answer. Sorry to the previous accepted, but this is just so clearly presented, detailed and useful to my understanding. Really outstanding, thanks man. $\endgroup$ Commented Aug 9, 2023 at 9:07
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    $\begingroup$ You make some valid points, but I think your orbit sim isn't quite right (and the code looks like it got cut off). This is a 3 body system, so you probably should include the effects of the Earth & counter-Earth on each other. BTW, you can get more accuracy by using the gravitational parameters, rather than mutiplying G by mass. That article has a table, but JPL has a better one: ssd.jpl.nasa.gov/astro_par.html $\endgroup$
    – PM 2Ring
    Commented Aug 10, 2023 at 10:54
  • $\begingroup$ You're right that I neglected the three-body dynamics of the system (which other commenters have rightly pointed out will "quickly"--eg, 10 million years--lead to unstable outcomes). However, this will be irrelevant to the overall "single period" result as these dynamics build up over time; the force of earth-2 on earth-1 is of order 1 million times weaker than that of the sun. You could also argue that the choice of setting the initial conditions at aph/perihelion is arbitrary. PS--thanks for the code tip, I fixed it. $\endgroup$
    – Mike
    Commented Aug 10, 2023 at 18:34
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@GiorgioP-DoomsdayClockIsAt-90 wrote:

Taking into account the perturbations due to the remaining part of the Solar System, even without adding the presence of the Moon, completely symmetric dynamics does not seem a probable event.

The following provides further support to the above:

Our entire solar system also has a barycenter. The sun, Earth, and all of the planets in the solar system orbit around this barycenter. It is the center of mass of every object in the solar system combined.

Our solar system’s barycenter constantly changes position. Its position depends on where the planets are in their orbits. The solar system's barycenter can range from being near the center of the sun to being outside the surface of the sun. As the sun orbits this moving barycenter, it wobbles around.

Therefore, your Earth opposite ours will be seen from our Earth from time to time.

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