1
$\begingroup$

Apparent magnitude is meant with respect to the crew of the vehicle. For the trajectory, please pick any that can be considered normal.

Also, it would be nice to have comparisons with other objects of similar apparent magnitudes with respect to observers on Earth.

I was close to post this on Astronomy.SE, but I think it is more related to space, especially when thinking about the implications for space travel. Please feel free to edit tags.

$\endgroup$
  • $\begingroup$ I am not sure what your question has to do with "cubic meters per astronaut" discussed in your link at t=59m45s perhaps there is a different cue time for discussion of views of Mars and Earth from the spacecraft? $\endgroup$ – uhoh Jun 13 at 14:56
  • $\begingroup$ Different but related: How might the experience of a trip to Mars differ in comfort and heath impact from a stint on the ISS? $\endgroup$ – uhoh Jun 13 at 14:57
  • $\begingroup$ I'm not sure a general answer to this question is possible. Planets are illuminated, not luminous, so the amount of light depends on the angle from the Sun to the planet to the spacecraft. (Much like the light from the moon depends on its phase.) Not only will this angle will vary throughout the trajectory, but different starting conditions will yield different results. So I'm not sure you can generalize this problem. $\endgroup$ – DrSheldon Jun 13 at 15:30
  • 1
    $\begingroup$ @uhoh At 1:00:04, Richard Davis talks about the Earth being a tiny blue dot and Mars simultaneously being a tiny red dot. The linked time is the beginning of the context of this statement. I'm wondering how "tiny" exactly they can be, and think that the apparent magnitude is an appropriate measure. $\endgroup$ – Everyday Astronaut Jun 13 at 20:16
  • 1
    $\begingroup$ Looking at this video: youtu.be/r1_B7k6JYNs , because the Earth has a lower orbit than the transiting spacecraft, the earth’s day side is turned away rather quickly into the flight. This happens before the 1/4 mark of the transfer. Then, the earth would be lost in sun glare for a while, then visible again from the side. $\endgroup$ – CourageousPotato Jun 14 at 5:26
4
$\begingroup$

As pointed out in the comments, there are many things to consider, but let's try to pull out reasonable assumptions for an estimation. Apparent magnitude being meant with respect to the crew of the vehicle, we can ask your very interesting question in a more physical way: at what point on a trajectory from Earth to Mars are flux densities of both planets equal using a visible spectral filter ?

Visible flux densities from Earth and Mars are therefore the key parameters to determine and hard to get: they depend on the time of the year, the state of the sun, the state of the atmosphere of the planet, and numerous second-order phenomena.

Let me then simplify this out using the convenient planetary equilibrium temperature equivalent, that is we assume both Mars and Earth are perfect black bodies at thermal equilibrium. Actually it is far from being a rough approximation for most applications. We will also state that the balance of radiative energy in visible wavelenghts regarding the whole spectrum is the same for both planets.

Then, we can use Stefan-Boltzmann law to equal flux densities from the two planets: $$ \frac{4\pi R^{2}_{Mars} \sigma T^{4}_{Mars}}{4\pi D^{2}_{M-O}}=\frac{4\pi R^{2}_{Earth} \sigma T^{4}_{Earth}}{4\pi D^{2}_{E-O}} $$ With $D_{M-O}$ and $D_{E-O}$ the distances between the observer and Mars or the Earth respectively. This is equivalent to :

$$ \frac{D_{E-O}}{D_{M-O}}=\frac{R_{Earth}T^{2}_{Earth}}{R_{Mars}T^{2}_{Mars}} $$

With $R_{Mars}=3389*10^{3}m$, $R_{Earth}=6371*10^{3}m$ and equivalent equilibrium temperatures $T_{Mars}=255K$ and $T_{Earth}=206K$, this gives you :

$$ \frac{D_{E-O}}{D_{M-O}} \approx 1.227 $$

This means the flux densities will be equal when the observer will be 1,23 times closer to Mars than to earth. Since Earth-Mars distance $D_{E-M}=D_{M-O}+D_{E-O}$, this can also be written $\frac{D_{M-O}}{D_{E-M}}=0.45$. When will this occur ? Let's consider a classic potential trajectory with a Hohmann transfer orbit from Earth to Mars.

Hohmann transfer orbit trajectory

I will not develop the maths here since it is of little interest, but you can develop from Kepler's equations of motion the same parameter $\frac{D_{M-O}}{D_{E-M}}$. Here is the result:

$$ \frac{D_{M-O}}{D_{E-M}}=\frac{a_{M}^{2}-2*a_{M}*a*(1-e*cos(2\pi x))*cos(\phi+2\pi x(\frac{T}{T_{M}}-1))+a^{2}(1-e*cos(2\pi x))^{2}}{a_{M}^{2}+a_{E}^{2}-2*a_{M}*a_{E}*cos(2\pi x(\frac{T}{T_M}-\frac{T}{T_E})+\phi)} $$

With $a_M, a_E$ and $a$ the semi-major axis of Mars, Earth and the transfer orbit; $T_M, T_E$ and $T$ the period of the orbit of Mars, Earth and the transfer; $e$ the excentricity of the transfer orbit, $\phi$ the phase angle between Mars and Earth for the Hohmann transfer window and $x$ the fraction of time over the period of the Hohmann transfer orbit : $x=\frac{t}{T}$ and the observer arrives on Mars when $x=0.5$.

To solve the problem you just have to find $x$ so that $\frac{D_{M-O}}{D_{E-M}}=0.45$. Here is a plot :

Graphical resolution

The solution is for $x=0.165$ which means approximately 85 days after departure, when you have performed one third of the travel. At that time you will be more or less at a distance from Mars twice the distance from Earth to Mars at the nearest. So the apparent magnitude of both Mars and Earth will have nothing particular, quite close to the average apparent magnitude of Mars throughout time.

$\endgroup$
  • 1
    $\begingroup$ Welcome to Space! While I applaud the effort you made for this answer, you treat the planets as luminous objects that give off light from all angles, rather than illuminated objects that have a "dark side". Consider what the spacecraft sees halfway through your proposed trip. They will be on the dark side of the Earth and see very little of it, but will see the fully-illuminated side of Mars. $\endgroup$ – DrSheldon Jun 13 at 23:15
  • $\begingroup$ If the black body radiation was replaced with average albedo (reflected light) and shadowed/unshadowed area, this would be a great answer. $\endgroup$ – CourageousPotato Jun 14 at 5:20
  • $\begingroup$ @DrSheldon you're right Doctor ! Let me work this out again and I'll try to include this geometrical aspect and edit my answer soon. Thank you ! $\endgroup$ – Arnaud BUBBLE Jun 14 at 5:41
2
$\begingroup$

I've done a quick numerical integration for a circular Earth orbit and elliptical orbits for Mars and the Ship on a Hohmann transfer ellipse. I normalized to 1 AU so 1 year corresponds to a time of $2 \pi$.

The vis-viva equation gives

$$v^2(r) = \left(\frac{2}{r}-\frac{1}{a}\right),$$

and period from this answer

$$T = 2 \pi \sqrt{a^3}.$$

I used absolute magnitudes $H$ for Earth and Mars of -4.0 and -1.6 respectively. Those would be the visual magnitudes if you stood next to the Sun and viewed each planet at a distance of 1 AU.

I used equations from Wikipedia for the apparent magnitude $V$ and the phase integral of the phase angle $q(\alpha)$:

$$V = H + 5 \log_{10}(\frac{r_{Sun} r_{Obs}}{1 AU^2}) - 2.5 log_{10}(q(\alpha))$$

$$q(\alpha) = \frac{2}{3}\left( \left(1 - \frac{\alpha}{\pi} \right) \cos(\alpha) + \frac{1}{\pi} \sin(\alpha) \right) $$

A more thorough and accurate calculation might be done based on Computing Apparent Planetary Magnitudes for The Astronomical Almanac.

Results (approximate)

  • Total time is 0.649 years or 237 days (your milage may vary)

  • Mars starts at about +1 magnitude and steadily brightens throughout the trip.

  • Earth starts out extremely bright and becomes invisible as it passes conjunction with the Sun at 0.205 years or about day 75. It then rapidly brightens to about -2 magnitude and stays near that for the rest of the journey.

  • They reach equal brightness the first time at about 0.178 years or day 65 at -1 magnitude.

  • After Earth passes conjunction with the Sun it rapidly brightens again, and for a long stretch between 0.27 0.32 years (100 to 120 days) both Mars and Earth are roughly -2.7 magnitude. Earth remains bright, fading only to about -2 magnitude at the end of the journey.


quickie orbit Earth to Mars

quickie orbit Earth to Mars apparent magnitudes

quickie orbit Earth to Mars apparent magnitudes

def deriv(X, t):
    x, v = X.reshape(2, 3, -1)
    acc = -x * (((x**2).sum(axis=1))**-1.5)[:, None]
    return np.hstack((v.flatten(), acc.flatten()))

def dotem(a, b, axis):
    return (a*b).sum(axis=axis)

def phase_angle(x_sun, x_observer, x_body):
    a, b = x_sun-x_body, x_observer-x_body
    cos_angle = dotem(a, b, axis=0) / (np.sqrt((a**2).sum(axis=0)) * np.sqrt((b**2).sum(axis=0)))
    return np.arccos(cos_angle)


def q(angle):
    return (2./3.) * ((1 - angle/pi)*np.cos(angle) + np.sin(angle)/pi)

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

halpfi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

H_earth, H_mars = -4.0, -1.6   # roughly

a_earth   = 1.0 # AU

peri_mars = 1.38 # AU
apo_mars  = 1.67 # AU
a_mars    = 0.5 * (peri_mars + apo_mars)

a_ship    = 0.5 * (a_earth + peri_mars)

T_earth = twopi * np.sqrt(a_earth**3) # years
T_ship  = twopi * np.sqrt(a_ship**3) 
T_mars  = twopi * np.sqrt(a_mars**3)

v0_earth = np.sqrt(2./a_earth   - 1./a_earth) 
v0_ship  = np.sqrt(2./a_earth   - 1./a_ship) 
v0_mars  = np.sqrt(2./peri_mars - 1./a_mars)

X0 = np.array([a_earth, 0, a_earth, 0, -peri_mars, 0] +
              [0, v0_earth, 0, v0_ship, 0, -v0_mars])

times = np.linspace(0, 0.5*T_ship, 1001)

answer, info = ODEint(deriv, X0, times, full_output=True)

x_earth, x_ship, x_mars = answer.T[:6].reshape(3, 2, -1)
x_mars = x_mars[:, ::-1] * (np.array([1, -1])[:, None])
x_sun  = np.zeros_like(x_earth)

r_ship_earth = np.sqrt(((x_ship - x_earth)**2).sum(axis=0))
r_ship_mars  = np.sqrt(((x_ship - x_mars )**2).sum(axis=0))

r_sun_earth  = np.sqrt(((x_sun  - x_earth)**2).sum(axis=0))
r_sun_ship   = np.sqrt(((x_sun  - x_ship )**2).sum(axis=0))
r_sun_mars   = np.sqrt(((x_sun  - x_mars )**2).sum(axis=0))

phase_angle_earth = phase_angle(x_sun, x_ship, x_earth)
phase_angle_mars  = phase_angle(x_sun, x_ship, x_mars)
q_earth = q(phase_angle_earth)
q_mars  = q(phase_angle_mars)

V_earth = H_earth + 5*np.log10(r_ship_earth * r_sun_earth) - 2.5*np.log10(q_earth)
V_mars  = H_mars  + 5*np.log10(r_ship_mars  * r_sun_mars)  - 2.5*np.log10(q_mars)

if True:
    plt.figure()
    for x in (x_earth, x_ship, x_mars):
        plt.plot(x[0],     x[1]          )
        plt.plot(x[0,:1],  x[1,:1], 'ok' )
        plt.plot(x[0,-1:], x[1,-1:], 'ok')
    plt.plot([0], [0], 'oy', markersize=12)
    plt.title('Earth, Ship and Mars (AU)', fontsize=16)

    plt.show()

if True:
    plt.figure()

    plt.subplot(4, 1, 1)

    colors = '-b', '-g', '-r'
    for r in (r_sun_earth, r_sun_ship, r_sun_mars):
        plt.plot(times[1:-1]/twopi, r[1:-1])
    plt.title('distance from Sun (AU) Earth, Ship, Mars', fontsize=16)

    plt.subplot(4, 1, 2)

    colors = '-b', '-r'
    for phase_angle, color in zip((phase_angle_earth, phase_angle_mars), colors):
        plt.plot(times[1:-1]/twopi, degs*phase_angle[1:-1], color)
    plt.title('phase angle (degs), Earth, Mars', fontsize=16)

    plt.subplot(4, 1, 3)

    colors = '-b', '-r'
    for q_object, color in zip((q_earth, q_mars), colors):
        plt.plot(times[1:-1]/twopi, q_object[1:-1], color)
    plt.title('q(phase_angle), Earth, Mars', fontsize=16)

    plt.subplot(4, 1, 4)
    for V, color in zip((V_earth, V_mars), colors):
        plt.plot(times[1:-1]/twopi, V[1:-1], color)
    plt.title('Earth, Mars apparent magnitue from Ship', fontsize=16)
    plt.xlabel('time (years)', fontsize=14)
    plt.ylim(-20, 10)

    plt.show()
$\endgroup$
  • 1
    $\begingroup$ Wow, thanks uhoh! Especially for the code. Just to be clear: does this account for the dark side of the planets? $\endgroup$ – Everyday Astronaut Jun 15 at 19:04
  • $\begingroup$ @EverydayAstronaut yes that's taken into account by q(phase_angle) explained further in the linked Wikipedia article. It goes to zero at phase_angle = 180 degrees (conjunction with the Sun). I found problem with my dot product (forgot to normalize) and a missing minus sign, this looks much better now. Thanks for the ping! $\endgroup$ – uhoh Jun 16 at 0:50
  • 1
    $\begingroup$ Still don't have time to go through the code lines, but this looks a lot less surprising now. Thumbs up! $\endgroup$ – Everyday Astronaut Jun 17 at 5:41
  • $\begingroup$ @EverydayAstronaut I'll give it another go-over in the next day or so. I wish there were an independent way to check this, or a verifiable source to quote. $\endgroup$ – uhoh Jun 17 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.