note: This is a question about relativistic effects. I've included some detail about the spacecraft and its orbit for background, but the question is about relativistic effects and their observability.

When the Parker Solar Probe passes within 8.5 solar radii of the Sun, it will be moving really fast.

A perihelion and aphelion of about 6.6 and 109 million kilometers suggests a semi-major axis of 57.8 million kilometers. The standard gravitational parameter for the Sun is:

$$GM_{Sun} \ = 1.327\times 10^{20} m^3/s^2$$

So using the vis-viva equation:

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

the orbital velocity at perihelion will be about 195,000 m/s, or about 25 times faster than a satellite in LEO. This is close to one part per thousand of the speed of light! I think this will be a new record for a spacecraft velocity relative to the solar system.

Depth of a gravitational potential well for an object of mass $m$ is approximately given by:

$$u_G \ = \ -GM_{Sun}\frac{1}{r}$$

At 6.6 million kilometers the spacecraft will be 23 times deeper in the Sun's gravitational well than it would be at 1 AU.

Will there be any particularly unique relativistic effects detectable for the Parker Solar Probe during it's close flyby of the Sun?

Just for example, during one close flyby, how much time will its on-board clock gain or lose compared to some convenient standard solar system timescale? I know there are several different choices, "GPS time" might be one but I don't want to arbitrarily choose an inconvenient one.

enter image description here

above: Parker Solar Probe cropped, from here.

enter image description here

above: screen shot from NASA's

Solar Probe Plus Fact Sheet

note: the link seems to be dead now, but there is plenty of related information available at

enter image description here

above: Illustration of the Parker Solar Probe's orbit, from here.

migrated from Jul 27 '17 at 13:06

This question came from our site for spacecraft operators, scientists, engineers, and enthusiasts.

  • 1
    Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Oct 18 '17 at 12:58
  • 2
    The "standard solar system timescale" we use for reference is an ephemeris time, specifically TDB.… I checked and our current spacecraft clock kernel doesn't include any relativistic correction. In practice the time folks will be updating based on actual on-board time every contact. I'll let you know in a year :) – jtniehof Mar 22 at 16:12
  • @jtniehof that's good to know! I'd always read it as TBD thinking they weren't sure how to do it yet and were still working on it ;-) Okay folks, we'll hear more about this in the future. Watch This Space!! – uhoh Mar 22 at 22:39
  • 1
    @jtniehof in the mean time I've added a supplementary answer to get an idea of how big the shift will be. – uhoh Aug 30 at 6:23

I've talked to one of the engineers who is involved with timekeeping on the Parker Solar Probe mission. I have not yet talked with the principle timekeeping lead, so this answer may have to be revised.

There seems to be two relativistic effects that will affect PSP due to the Sun's gravity well. The first is clock error due to relativistic effects near the Sun, and the second, as mentioned by DuffBeerBaron above, is spacecraft ranging for navigation.

Of the two, the effect on ranging is the more significant. The relativistic effects on the clock seem to be very small compared to oscillator drift, which is very carefully tracked. Corrections made for oscillator drift will also correct for any relativistic effects.

The effect on ranging calculations is due to what is called the Shapiro delay. From the paper linked below:

Shapiro delay is an increase in the travel time (the OWLT) of a signal passing through the gravitational field of a massive body, in this case the Sun. It was first confirmed by Irwin Shapiro in the 1960s using measurements of radar distances to Venus and Mercury. The Shapiro delay is important not only to the timekeeping as discussed here, but as a consideration in the orbit determination process itself for all deep-space missions. Both the radiometric range measurements and the Doppler velocity measurements are markedly affected by the Shapiro delay, so this delay is routinely included in deep-space navigation calculations. This effect is important for all portions of these missions; it is not limited to solar conjunctions, but it is more pronounced near those events.

Note that this effect is not due to the spacecraft itself being inside the Sun's gravity well; it is due to the radio transmission through the gravity field. Because of this all deep-space missions are affected by it; it is not a unique consideration for PSP. The paper cited used data from two missions: MESSENGER, at Mercury, very close to the Sun, and New Horizons, out at Pluto, very, very far from the Sun. But both had to transmit through solar conjunctions, sending the radio signal right through the Sun's intense gravitational field.

The paper linked below describes the Shapiro delay measured by MESSENGER and New Horizons timekeeping experiments, the second is the Wikipedia article about it.

Sorry for not including any mathematical equations in my answer! That is beyond my area of expertise! I will update this answer with any further information I obtain.

MESSENGER Onboard Timekeeping Accuracy

Shapiro delay From Wikipedia

  • This is really interesting, thank you! I'm not sure we should call it an "error" since it's actually correct. Error would come from trying to use non-relativistic orbital mechanics. – uhoh Jan 9 at 3:29
  • 2
    I had that thought also. In this case "error" is a term of art, which includes any "deviation" that has to be accounted for in the final data product. The relativistic effects are included in the timekeeping "error budget" while assessing whether or not the spacecraft capabilities and operations meet the timekeeping requirements. But you are right, it is not really an error. Its a natural fact. – That60sKid Jan 9 at 13:50

Even missions to Phobos have to consider relativity, but that's because of the necessary sensitivity of any instrument trying to measure Phobos's gravity. So my cop-out answer is it depends on what you consider significant. I'd assume that most instruments would be measuring pretty large properties and would not have to consider relativity, but anything that needs to be super precise would have to consider it. I can't imagine the mission doesn't plan to measure relativity with some instrument.

Others are more qualified to give you numbers as to how much time will be "lost", but I know the equation for gamma, the factor by which time slows, is

$$\gamma=\frac{1}{\sqrt {1-v^2/c^2}}.$$

Therefore, traveling at 195 km/s, the probe would experience time about 0.2 parts per million slower than "normal".

Then using the equation for gravitational time dilation:


Gravitational time dilation also slows time by about 0.2 parts per million. I'm not positive how these two interact, however. I doubt it's as simple as addition.

  • Info on how to format equations appreciated - I'm pretty new here, and don't have time to take a look at the moment. – Deimophobia May 31 '17 at 15:39
  • 1
    Take a look and see if this is how you'd like it. For equations stackexchange has MathJax support. There is an informal "tutorial" here":… – uhoh May 31 '17 at 15:54
  • 1
    Does this calculation take into account the gravity well, and all of the acceleration? I'm not sure under which conditions non-inertial conditions it can be used correctly. – uhoh May 31 '17 at 15:56
  • @uhoh Gravitational time dilation, by itself, has about the same effect. Post updated with equation. Not sure how they interact with each other. – Deimophobia May 31 '17 at 16:03
  • 1
    @uhoh will do, later today. – Deimophobia May 31 '17 at 16:37

From here (or here if you are ambitious) the lowest order terms to the relativistic frequency shift of a clock in orbit around a gravitational body are:

$$ \frac{\Delta f}{f} \approx -\frac{\Phi}{c^2} - \frac{v^2}{2c^2} = -\frac{GM}{r c^2} - \frac{v^2}{2c^2},$$

where the first term is the gravitational shift and the second is time dilation. Plugging in the vis-viva equation where $a$ is the semimajor axis:

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


$$ \frac{\Delta f}{f} \approx -\frac{GM}{c^2}\left(\frac{2}{r}-\frac{1}{2a} \right).$$

With a perihelion of 6.6 million km and a semimajor axis of 57.8 million km, that gives frequency shifts of 4.3E-07 (half a ppm) and 1.3E-08 at perihelion and aphelion, respectively.

To calculate the approximate total shift in elapsed time you would have to integrate over the orbit. With a period of about 3 months (87.7 days) I get a total shift of about 0.3 seconds per orbit:

Python script:

Parker Solar Probe around the Sun (approx)

  • From chatting with the flight operations folks, I should note that the magnitude of the frequency shifts during orbital maneuvers has been on the order of ~50 Hz due to the change in speed and direction of the spacecraft (i.e., the shift from before to after). The maneuver speed changes so far have been relatively small compared to the spacecraft speed (relative to Earth), so I imagine the $\Delta f$ near perihelion will be much larger. In any case, the two main maneuvers to get it on track with the first Venus flyby went very well so it's on its way... – honeste_vivere Aug 30 at 14:17
  • @honeste_vivere first sentence: "...frequency shift of a clock..." Absolutely nothing here about radio. – uhoh Aug 30 at 15:25
  • No, I meant the frequency shift of the transponder signal, not the clock. – honeste_vivere Sep 1 at 17:19
  • @honeste_vivere so not really directly related to the question, got it. – uhoh Sep 1 at 18:22
  • Well if you do not consider relativistic Doppler-shifts relevant to "...what relativistic effects will it experience and how large will they be?" then I suppose yes, it is not related to the question. They use the Doppler-shift to determine the attitude/orbit information after each maneuver to determine the speed difference between the source (spacecraft) and the receiver (Earth). So it is a relativistic effect and I provided a magnitude. – honeste_vivere Sep 2 at 12:46

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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