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What is the current speed of the parker probe which is just launched today(After Final boost(detached by the rocket)).

I was just guessing as it will attain max speed of 194 Km/sec, that's why does the velocity of the probe mainly comes due to this boost or it will be mainly because of the gravity assists?

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  • $\begingroup$ there is some helpful information here and here and a bit of math here $\endgroup$
    – uhoh
    Aug 12, 2018 at 11:26
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    $\begingroup$ Relative to what? When? $\endgroup$
    – Mark Adler
    Aug 12, 2018 at 15:22
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    $\begingroup$ I didn't find enough facts to make up a detailed answer... Essentially all maneuvers of the upper stage and fly-bys are to slow down the probe, not to make it faster (with respect to the Sun). $\endgroup$
    – asdfex
    Aug 12, 2018 at 17:04

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I added this to a special page to Where is Roadster, you can find the Parker Solar Probe info here.

The current location is 7,338,815 miles (11,810,682 km, 0.079 AU) from Earth, moving away from Earth at a speed of 27,043 mi/h (43,522 km/h, 12.09 km/s).

By the way, that is CRAZY fast. For comparison, to get that far Elon's Tesla Roadster took about 2 months if I recall correctly.

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  • $\begingroup$ Isn't new horizons at about 17km/s? I was wondering who the parker probe will dethrone for highest orbital velocity relative to the sun. $\endgroup$ Aug 23, 2018 at 12:29
  • $\begingroup$ Errr... voyager. Not horizon. $\endgroup$ Aug 23, 2018 at 12:44
  • $\begingroup$ I need to add the orbital velocity with respect to the sun for PSP. Bottom line, Voyager and any probe in the outer solar system is actually slower then almost any other probes. The fastest previous probe was a sun probe, I forget which one exactly... Helios maybe? $\endgroup$
    – PearsonArtPhoto
    Aug 23, 2018 at 13:26
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    $\begingroup$ Odd-- where does the 17km/s come from-- a bad article? Okay, wait, no I get it. those speeds were during the gravity assists - from the rare quality Quora answer. $\endgroup$ Aug 23, 2018 at 13:30
  • $\begingroup$ How fast is Juno at perijove? $\endgroup$ Aug 23, 2018 at 13:58
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At 9 hours after launch, the Parker Space Probe's current speed is about 12.4 km/s with respect to the center of the Earth. This is smaller than it was a bit over 8 hours ago when the third stage burn had completed. I calculate the vehicle's velocity was about 14.7 km/s just after third stage cutoff.

The vehicle's current velocity (9 hours after launch) with respect to the center of the Sun is about 17.2 km/s. Note that this is a good deal smaller than is the ~30 km/s velocity of the Earth about the Sun. The primary purpose of this launch and of each of the Venus flybys is to reduce the vehicle's velocity with respect to the Sun. Reducing velocity with respect to the Sun at aphelion increases the vehicle's velocity with respect to the Sun at perihelion (closest approach to the Sun).

The hoped-for 194 km/s velocity will result from drastically reducing velocity with respect to the Sun at Venus. The consequences of that reduction in aphelion velocity are (a) a drastic reduction in the perihelion distance, and (b) a drastic increase in perihelion velocity.

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Technically this answer doesn't specify what slows it down, but it does explain how it does.

I've never done anything with orbital mechanics before today, but I got bored and read a few articles, linked below and made a scale model of the Parker solar probe's final orbit in my program I wrote from scratch!

enter image description here

Click Here for Animation

What you see here is the arguments of the orbit of the parker solar probe:

https://en.wikipedia.org/wiki/Parker_Solar_Probe

0.043 AU is the distance from the sun it will end up at, this is the first oblong orbit that you see. 0.738 AU is the final distance it will be travelling after the 6 fly-bys of venus it makes.

  • With each flyby of venus it lowers the lowest point of the orbit.
  • When it starts at earth, it's going to be at approximately 1 AU away from the sun!
  • The animation shows the difference between the starting orbit (Earth) and it's final one.

Note: AU is arbitrary units in the graph, don't pay attention to the scaling for AU.


If you look at the points on the graph, every single individual point is a specific unit of time. The only important thing to know is that the time it takes to go between any two dots is the same.

  • Notice how the dots are super far apart closer to the sun!
  • This means that it takes a the same amount of time to traverse more distance.
  • This is also known simply as "going faster".

At it's final orbit, closest to the sun, it will be going 700,000 km/h (wikipedia).

  • The Earth revolves around the sun at about ~107200 km/h (wikipedia).
  • The parker solar probe will be lowering one end of its orbit to a measly 7.3% of earth's lowest.
  • This is done by using Venus' gravity to repeatedly slow (yes, slow!) the highest orbital point.
  • By slowing the highest point, you stay there for a longer time and a smaller amount at the lowest.
  • However, you travel the same amount of distance on both sides of the orbit!

This is called eccentricity (cue the graph to the bottom left, and the outer ellipse):

  • "Eccentric" is the word to describe "non-circular" orbits, or ones with a low point.
  • Earth is about .01673 eccentric, that's basically nothing, we're a perfect circle.
  • Parker solar probe will be approximately .846 eccentric by my calculations (probably off).

Now, lastly, lets look at that velocity graph.

  • Notice how the axis of the graph shrinks dramatically when we go from "Parker" to "Earth".
  • The maximum and minimum vector velocities shrink with a less "eccentric" orbit.
  • With a perfectly circular orbit, you're not changing speed at all (simplified greatly).
  • Look at how evenly the dots are spaced, all line segments are equal!
  • This means they're equal in time (length) and speed (distance between two dots over time)!

PS: Anybody please correct anything glaringly wrong with this answer, I want to learn too.


Resources:

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  • $\begingroup$ Nice work! :) I tried to click on the picture but I get a static jpg...something went wrong? $\endgroup$
    – BlueCoder
    Aug 23, 2018 at 7:52
  • $\begingroup$ I've got a few questions about labels: - what is the "mean orbit"? Just a circular orbit? (If that's the case, the first graph could be resized to a square to emphasize this). - Some words on the Anomaly plot (I think it's referred in your eccentricity part but it's not very clear)? What are "mean" anomaly, "true" anomaly and "eccentic" anomaly? $\endgroup$
    – BlueCoder
    Aug 23, 2018 at 7:57
  • $\begingroup$ Links to the articles used to make the model would make a nice addition too :) $\endgroup$
    – BlueCoder
    Aug 23, 2018 at 7:58
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    $\begingroup$ @bluecoder heres the animation: i.imgur.com/6F55Apg.gifv also that page, at the bottom, has the articles I used and the link to my code on github. I can put it in the answer when Im on a computer instead of my phone. physics.csbsju.edu/orbit/orbit.2d.html will explain well how you turn the mean orbit into the eccentric one. $\endgroup$ Aug 23, 2018 at 12:23
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    $\begingroup$ @BlueCoder here's the direct link to the code: github.com/carusocomputing/OrbitPlotter (Just 3 classes, 1 unused for the most part) $\endgroup$ Aug 23, 2018 at 13:32

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