Cees Bassa's tweet shows a plot of received frequency versus time, with the Doppler shift of the Moon's center of mass subtracted. For a lunar orbit the resulting plot would show sinusoidal-like oscillations around zero, and that's what happens up until just before 20:10.

The tweet notes that the path not only passes through zero, but climbs to a large positive shift in a continuous ramp, almost to the same shift seen when it was passing over the Moon's north pole coming straight at us.

A naive interpretation would be that the spacecraft made a powerful u-turn and was flying straight at us at the moment of loss of contact, but that's quite counterintuitive.

What does this plot mean, and how can it be compatible with a trajectory that got at least close to landing on the Moon?

enter image description here

This is the full @isro #Chandrayaan2 Doppler curve as observed by @radiotelescoop and compared against JPL Horizons predictions. Why does the Doppler curve of the lander cross that of the moon during descent? When landed they should be equal. Can @dsn_status experts explain this?

You can also see Scott Manley's video showing some of the same doppler data below, after about 04:10

  • 1
    $\begingroup$ What's the "moon prediction", if it's an ephemeris from Horizons, then it must correspond to the Moon center of mass ? More clarity is required on what the basis of that curve is. $\endgroup$
    – ASRI_306
    Commented Sep 9, 2019 at 7:45
  • $\begingroup$ @ASRI_306 I believe that's exactly what it means, based on common sense and comments in the the thread below the original tweet, but I can't say yet for sure, thus my question. $\endgroup$
    – uhoh
    Commented Sep 9, 2019 at 8:02
  • $\begingroup$ I'm not sure how that can be co-plotted you know, I mean we're talking here about measuring the Doppler curve of a body without a RF source. It's a reflection compared to an active RF source emitting something. I can't make sense of it. $\endgroup$
    – ASRI_306
    Commented Sep 9, 2019 at 8:05
  • $\begingroup$ @ASRI_306 my guess is that an absolute frequency is assumed, so the received frequency divided by the absolute frequency gives a radial velocity $v/c = \Delta f/f_0$. Then a correction is subtracted, which is the rate of change of the round-trip path length from DSN (or IDSN) to spacecraft to receiver based on JPL's Horizons' prediction (the reference to Horizons in the discussion) due to the radial velocity of the Moon wrt Earth and the motion of the two ground stations. I think the parameter might be called two-way light time or something similar, but I'm out of my depth here. $\endgroup$
    – uhoh
    Commented Sep 9, 2019 at 8:12
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    $\begingroup$ Hey @uhoh, Cees bassa posted a corrected doppler plot, which is the correct one, and he superimposed that with the predicted descent profile here. twitter.com/cgbassa/status/1171333649461108736 $\endgroup$
    – ASRI_306
    Commented Sep 10, 2019 at 9:07

1 Answer 1


The plot shows the raw, received frequency of the signal from the spacecraft. If the output signal from the spacecraft would be of a fixed frequency, we could directly extract the relative velocity towards/away from us. Unfortunately it's not as simple as this.

E.g. this Twitter thread and the text in the Python code used to make the plot explains what actually happens:

The transmitter on most spacecraft don't produce their own frequency signal but simply copy the frequency they receive from a ground station (the so-called carrier signal). To keep the transceiver on the spacecraft operating at its sweet spot, the ground station tries to compensate for the Doppler effect on the carrier signal received by the spacecraft by shifting its frequency based on the expected relative movement. During the regular orbit this compensation was pretty exact so that the received signal only contains the Doppler shift from the the link back from the spacecraft and can be converted directly into a velocity (1 kHz shift roughly corresponds to 130 m/s line-of-sight speed at this frequency).

As soon as descent started the compensation didn't work any more: The ground station sent out a frequency that would be correct for the regular orbit, but not for the now changing trajectory before landing. As a result the frequency now shifts twice as much as it should be: First because of the Doppler effect due to the difference between actual and expected orbit and then again (as before) due to the actual movement of the spacecraft. This means, the difference between received frequency and expected frequency (shown as blue line) is roughly twice as large as it should be.

Thus: The actual Doppler shift of the probe doesn't change from -10 kHz to +10 kHz, but roughly from - 10 kHz to -2 kHz. So, the probe is still approaching the Moon and is not accelerating towards us. This is also shown in the plot linked by @ASRI_306 which as been corrected for this effect: https://twitter.com/cgbassa/status/1171333649461108736


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