If another planet was opposite Earth, would we be able to observe it?

Question

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?

Answer 1

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.

Answer 2

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)

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.

Answer 3

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.

Answer 4

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.

Answer 5

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:

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:

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://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()
   

Answer 6

@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.