Say there is a satellite in orbit around Mars, and it sends a signal at such a time that it is more or less behind the Sun. In that case, Mars would be "Occulted" by the Sun.

That signal is then received on Earth at a ground station. (Place won't matter)

What would be the doppler effect introduced to the signal if the signal is of 2.4 GHz?

Here, there are a lot of things to consider, as not only the relative motion between the planets and satellite matters, but also the relativistic doppler due to Sun's mass would as well, if the signal is anywhere between 5 to 10 Solar radii from the Sun at its point of closest approach.

How do I calculate the "Theoretical Doppler" the signal will have in the above scenario?

This is a schematic of the geometry -

enter image description here

Date: 24 July 2012 Satellite: Mars Express Depicts: Artist's impression of radio sounding of the solar corona with Mars Express Copyright: ESA/AOES Medialab

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    $\begingroup$ It's an interesting question (but maybe more on-topic in space exploration). Have you tried simply adding up all the contributions, only considering the projections of the motion onto the line-of-sight from the satellite to the receiver on Earth? Please share with us your thought processes on solving this problem. $\endgroup$ Commented Dec 25, 2023 at 15:32
  • $\begingroup$ Hi, I feel that the major issues would be relative motion of the satellite and the planet Mars, relative motion of this system with respect to the Earth, relativistic doppler due to the Sun, and the doppler due to atmosphere of the two planets. However, I am not sure on what/ how to proceed, and am also unsure whether Mars would have an impact at all, considering its thin atmosphere. $\endgroup$ Commented Dec 25, 2023 at 16:36
  • $\begingroup$ 2.4 or 2.42, there is an inconsistency in the question $\endgroup$
    – James K
    Commented Dec 26, 2023 at 16:35
  • $\begingroup$ My bad. Take it as 2.42 GHz $\endgroup$ Commented Dec 26, 2023 at 17:31
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    $\begingroup$ There is a formula given by Krisher in his 1993 paper - doi.org/10.1103/PhysRevLett.70.2213 , which seems like the solution to the problem, but not sure, and it looks kinda complex. $\endgroup$ Commented Dec 27, 2023 at 2:24

3 Answers 3


There will several contributions:

As Earth and Mars are both orbiting the sun, they will have some radial velocity, but in this geometry, most of their motion with respect to each other will be transverse, and so this might be quite small. Depending on the exact position in the orbits of the Earth and Mars it could be a relative velocity up to about 1km/s.

The motion of the satellite around Mars, and the motion of the ground station due to the rotation of the Earth will have contributions. The satellite may be moving at 3.4 km/s relative to Mars. This could change the frequency of the signal by up to 0.0011% (so between 2.419973 and 2.420027 GHz). The rotation of the Earth can move the ground station at up to 0.45 km/s at the equator so this is a relatively small contribution.

You also ask about the relativistic shift due to gravity. When a signal passes from a high potential to a low potential it will be blue-shifted. The fact that the signal passes close to the sun is irrelevant, it is only the potential at the start and end that are relevant to this calculation. This induces only about a 0.0000003% change in the signal, and the change doesn't depend much on the relative positions of Mars and Earth (Mars does have an elliptical orbit, so when it is furthest from the sun the effect will be slightly larger). This is much smaller than the Doppler effect due to the motion of the satellite.

The passage of the signal next to the sun will cause a deviation in the position of the signal, by up to 1.75 arc-seconds. However the motion of the satellite around Mars will cause greater changes motion. The disc of Mars in that geometry will be about 4 arc-seconds across.

In practical terms, when Mars is less than 10 solar radii from the sun, it is too close to the sun to communicate. The solar corona is a source of radio waves, so in this geometry, the satellite would not be in contact with the ground station at all.

The refraction by the atmosphere of Earth doesn't change the frequency. It does change the wavelength (making it shorter) but it isn't a blue-shift, as the frequency remains the same.

(formulae and calculations: Doppler shift depends on relative radial velocity $v$ the resulting frequency is $\frac{c\pm v}{c}\times 2.42\, GHz$. Gravitational redshift depends on potential, which varies in proportion to $1/r$, so the change in frequency will be a factor of $(1/r_m-1/r_e)\times GM/c^2$). Note this depends only on the orbital radii $r_{m,e}$. The bend ing the signal is frequency independent, and the value of 1.75 arcsec is that obtained in the famous 1916 experiment of the shift in light during an eclipse. The size of Mars's disc is from Stellarium.)

  • $\begingroup$ The radial velocity of Mars relative Earth ranges from -14 km/s to 17 km/s. i.sstatic.net/5BBiy.png See astronomy.stackexchange.com/a/50004/16685 for the plotting script. $\endgroup$
    – PM 2Ring
    Commented Dec 26, 2023 at 10:44
  • $\begingroup$ Indeed, it depends on the reletive positions of the planets. At solar conjunction, the distance from Earth to Mars is at (or close to) a maximum, so the line of sight relative velocity is small. At other times the relative velocity of the planets is the largest contribution. $\endgroup$
    – James K
    Commented Dec 26, 2023 at 14:04
  • $\begingroup$ There is a formula given by Krisher in his 1993 paper - doi.org/10.1103/PhysRevLett.70.2213 , which seems like the solution to the problem, but not sure, and it looks kinda complex. $\endgroup$ Commented Dec 26, 2023 at 17:35

In such a theoretical test system testing for the Doppler effect, we have to account for various types of redshifts, majorly 2 of them:

  1. The Doppler effect (to account for the redshift taken by the radial orbital motion of the Earth around the Sun i.e Revolution)

  2. Gravitational redshift - When the signal nears the Sun and gradually begins its way out of the gravitational well or the hill sphere even though it has not escaped, the light begans losing energy and with the equation $ E=h\nu $ where h is the planks constant and $\nu$ is the frequency we can observe that the frequency will also decrease with the energy

There are other redshifts as well like cosmological redshift but their effect are almost nil, additionally the close encounter to the sun during it's perihelion can potentially gravitationally lens it slightly, leading to the Shaipro Delay, this can cause additional delay in the reception of the signal, this may lead to redshift as well but due to it's distance being away, it is less influential to the redshift .

Now considering the gravitational redshift we can use the formula $z=GM/Rc^2​​$ where G is the gravitational constant, M is the mass (assume 1 solar mass), R is the radius and c is the speed of light (299,792,458 m/s), we can assume the radius to be 5-10 solar radii away from the center and get the gravitational redshift, to get the observed frequency by the receiver at mars we need to apply the following formula - $fGrav​=f×(1+z)$

So in order to find out the redshifted frequency we need to solve the Doppler effect after the gravitational redshift, so we need to use the formula $f'=fGrav * ​(c±vMars​/c±vEarth​​)$ where there are 2 different relative velocities between the source and the receiver, if you want the simpler approach you can rather combine it into a single relative velocity by either adding the relative velocities if it is getting closer (blueshift) or you may substract if they are moving away from each other. The resulting frequency is the result of the redshift by the Doppler effect. If we were calculating without redshift instead of $fGrav$ we would use $fEarth$. In this case, we have used gravitational redshift as well so the final result is $f'$. But It is also essential to incorporate the movement of the satellite as well so you can add or minus the velocity on to the vMars if it is getting closer or away to get more accurate measurements regarding this 3 body system.

Bonus Result and Code:

P.S I tried making a python program to calculate the supposed gravitational redshift, here is the code:

g = 6.674*10**-11
m = 1.9891 * 10**30
c = 299792458
r = 3.478*10**9
z = g * m / (r * c**2)
f = 2.4 #GHz
fgrav = f * (1+z)

This yields 2.400001019255533 GHz which is the value for gravitational redshift, you can modify the values according to your needs

Also to get the overall redshift including the doppler effect by adding $f1 =fgrav∗​(c±vMars​/c±vEarth​​)$ in the code if you know the velocity of mars and earth respectively by probably using the skyfield.almanac module, and knowing whether they are approaching or moving away

Also it is important to note that the dust grains maybe coming from the dust storms on mars may redshift the signal but it is rather hard to predict

Thank you, Hope it helps you!

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    $\begingroup$ Your explanation and calculation of the gravitational redshift is incorrect. The effect is very small and it is a net blue shift. $\endgroup$
    – ProfRob
    Commented Dec 26, 2023 at 9:05

So, the final solution is using the Krisher's Formula to account for all of the effects, except the atmospheres, which you have to compensate for separately.

def dot_product(a, b):
dot_products = [sum(i*j for i, j in zip(rowA, rowB)) for rowA, rowB in zip(a, b)]
return dot_products

def krisher(n,v_IDSN,v_MOM, r_IDSN_S,r_MOM_S,r_IDSN_E,r_MOM_E,r_IDSN_M,r_MOM_M):

    U_IDSN = (GM_Sun/r_IDSN_S) + (GM_Earth/r_IDSN_E) +(GM_Mars/r_IDSN_M)
    U_MOM = (GM_Sun/r_MOM_S) + (GM_Mars/r_MOM_M) +(GM_Earth/r_MOM_E)

    t1 = dot_product(n,(v_MOM-v_IDSN))/c
    t2 = 0.5* (dot_product(v_MOM,v_MOM) - dot_product(v_IDSN,v_IDSN))/(c*c)
    t3 = (dot_product(n,v_MOM)*dot_product(n,v_IDSN))/(c*c)
    t4 = ((dot_product(n,v_MOM))**2)/(c*c)
    t5 = (U_MOM - U_IDSN) /(c*c)

    freq = f_sc* (1- t1-t2-t3+t4-t5)
    # print(f_sc*t1,"\n",f_sc*t2,"\n",f_sc*t3,"\n",f_sc*t4,"\n",f_sc*t5)
    return freq

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