Can anybody explain why is the speed of Webb slowing down as it is moving away from earth?

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    $\begingroup$ Can you explain why your speed drops when you jump up? Same reason, gravity. $\endgroup$ Commented Dec 28, 2021 at 8:22
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    $\begingroup$ This is pedantic, but - the speed isn't slowing down. The speed is dropping, and the JWST is slowing down. $\endgroup$ Commented Dec 28, 2021 at 15:59
  • $\begingroup$ My gosh, the knowledge here! How fast can my head spin in Earth's gravity from reading all this point-counterpoint? - Tom $\endgroup$ Commented Jan 10, 2022 at 3:59

5 Answers 5


JWST is like rolling a ball up a hill to have it coast to a stop at the top. Low Earth orbit is the "bottom" and the Sun-Earth L2 Lagrange point is the "peak" at the "top".

enter image description here

A note about speed and distance.

Speed is how much distance you're covering per unit of time. JWST's speed, as of this writing is, 0.7772 km/s. Every second it moves 0.7772 km along its path. When you're driving at 100 km/h you're moving 100 km in one hour down the road.

Distance how far JWST has traveled along its curved path to L2, just how in a car your distance traveled is along the road. JWST's distance from the Earth is its "altitude".

enter image description here

The frame of reference for JWST is Earth's orbit around the Sun. In this frame, JWST's speed was 0 just before launch. As it travels, its speed is measured as how fast it is proceeding along its path to L2 (not how fast it is moving away from the Earth), and it will be 0 again once it reaches L2 (more or less, read on about halo orbits).

(Since JWST is coasting it is technically going in a straight line in space; it's space which is curved.)

See NASA's Where Is Webb? for more, and click on each individual metric for details.

Here's a little more about the L2 Lagrange Point.

After JWST fires its upper stage to leave Earth orbit it will be coasting for the next 30 days into a higher orbit around the Sun. As it coasts, the gravity of the Earth and the Sun slow it down. As it moves away from the Earth, Earth's gravity gets weaker and weaker. After 1.5 million km Earth's gravity will be just a tiny fraction of what is at the surface.

This orbit is very special.

If you draw a line from the Sun through the Earth and 1.5 million km more you reach L2. At L2, JWST will be in orbit around the Sun. Normally objects in a higher orbit move slower and take longer to orbit the Sun, so normally JWST would take longer to orbit the Sun than the Earth. But at L2 the pull of Earth's gravity speeds JWST up just the right amount so its orbit is exactly 1 year. This keeps JWST always in the same position relative to the Earth and the Sun; JWST is orbiting the Sun, but its distance to the Earth will remain the same, and it will always have the Earth between it and the Sun.

Here is what the Earth, Sun, and Earth-Sun Lagrange Points look like relative to the Sun.

enter image description here


Since L2 is a gravitational "peak" it's not stable; at L2 any slight push in any direction starts you going downhill. Instead, JWST will orbit L2 in a halo orbit around L2. This is not entirely stable, but it is more stable. JWST is orbiting L2 and L2 orbits the Sun.

Watch James Webb Space Telescope Launch and Deployment and Scott Manley's video What Makes Lagrange Points Special Locations In Space for more about Lagrange Points.

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    $\begingroup$ No, the JWST velocity relative to Earth will not be zero -- it will be orbiting the Earth once a year, too. It will also be orbiting the L2 point itself, which gives it another component of velocity relative to the Earth. $\endgroup$
    – Dave Tweed
    Commented Dec 29, 2021 at 12:20
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    $\begingroup$ @DaveTweed Yes, I chose to leave the details for the Scott Manley video. I started going into halo orbits and realized it's complicated and only tangentially related to the question. How about "JWST is orbiting the Sun, but its distance to the Earth will remain the same, and it will always have the Earth between it and the Sun." $\endgroup$
    – Schwern
    Commented Dec 29, 2021 at 21:44
  • $\begingroup$ I still haven't understood with respect to which reference point the NASA distances and speeds of JWST refer to. Your rotating frame makes my head spin even more! $\endgroup$
    – Ng Ph
    Commented Dec 30, 2021 at 20:56
  • $\begingroup$ @NgPh If you click on "Crusing Speed" here there is an explanation. It's using speed and distance traveled along its flight path, like how you'd calculate travel distance along a road rather than in a straight line. $\endgroup$
    – Schwern
    Commented Dec 30, 2021 at 22:37
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    $\begingroup$ @Schwern - "but the speed is along the path, not relative to the Earth" - the speed can be along a path, but speed is always relative to something. There's no absolute speed. $\endgroup$ Commented Jan 4, 2022 at 21:02

Well I think that once the engines are off the only force that affects James Webb is the gravitational force (from mainly the Earth and the Sun) so although this force decreases with distance is the only force. So its speed will decrease slower with time


Copernicus discovered this phenomena this long before Newton came up with his laws of motion. Copernicus motion

As a space craft moves to a higher orbit, it trades velocity for altitude. For the red dot to get to the same path as the blue dot, it has to accelerate twice. Meaning the blue dot has even more energy than the red dot, even though it's going slower. Kind of unintuitive.

JWST still has as much energy as when it left LEO (minus what it spent on course corrections), but it has traded a lot of speed for altitude.

BONUS: In the next video, coins are put in a spiral machine. It simulates orbital mechanics. Notice how the coins at the top are going slower in bigger arcs than the coins at the bottom, which are going faster in smaller circles.

spiral coin machine

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    $\begingroup$ This is not wrong but it's missing the three-body-ness of the problem. JWST is not bound to Earth at all; it has more than escape velocity and is in a heliocentric orbit. Yes as it moves closer to its final halo orbit it is moving up in the Earth's gravitational well but it is also moving up in the Sun's gravitational well as well. It's much better to express this with JWST orbiting the Sun, not Earth. $\endgroup$
    – uhoh
    Commented Dec 30, 2021 at 21:28
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    $\begingroup$ @uhoh Judging by the numbers on jwst.nasa.gov/content/webbLaunch/whereIsWebb.html?units=metric, it's actually currently moving slower than escape velocity for its Distance from Earth, by about 300 m/s, which is as intended, because it's headed to L2 and they don't want to flip it over to slow down when it gets there. $\endgroup$
    – notovny
    Commented Dec 30, 2021 at 22:49
  • $\begingroup$ @notovny ya that's a good point and I should have said something more complicated. The concept of "escape velocity" only makes sense in a two-body context and I shouldn't have used it here. $\endgroup$
    – uhoh
    Commented Dec 30, 2021 at 22:56
  • $\begingroup$ @notovny (and uhoh), interesting analysis. I have noted down the altitude and speed provided by Arianespace just before separation (1390 km and 9.6 km/s, resp.). Using your proposed comparison with the notion of "escape velocity (at radial r)", which must be legit because close to Earth JWST was in still the realm of "2-body", my calculation gives ~ 510 m/s (at separation event). So, at some time this difference of "speeds" has diminished. Apparently it is slowly increasing again, as can be checked on the NASA site. $\endgroup$
    – Ng Ph
    Commented Dec 31, 2021 at 10:56

imagine a car going litle uphill ( or on strait road) at 100kmph and then you just move your foot from the accelarator pedal the car will start to slow down .When the JWT got detached from the launch rocket or vehicle earths and suns gravity slows its speed just like the car example


I did some calculations and found that the Sun's gravitational force on JWST seems to be around 39.6N. The centrifugal force of JWST in orbit around Sun seems to be 39.6N as well so they seem to cancel out. The Sun has no affect in slowing down JWST.

The Earth's gravitational pull force on JWST at 1,200,000 km away seems to be about 0.046N at that position and the speed of JWST is 0.3683 km/s at that point according to NASA.

At that speed, Earth's gravity is acting to slow down JWST. By knowing that F=Ma or a=F/M and V=at the time is about 600 days. We know that JWST will slow down in the next 15 days so there must be thrusters slowing down JWST.

  • $\begingroup$ -1. I would like to see your calculations. Since Webb's orbit between Earth and L2 is close to circular, I think you need some very specific calculations to show that it is not closing down due to the 40 N force from the Sun. Also, I do not understand your last sentence. "We know that JWST will slow down in the next 15 days". Do you mean it will slow down in 15 days when it fires its engine to "enter orbit" around the L2 point? Or are you saying that the engines have been firing continuously since leaving Earth to cause the speed to decrease? $\endgroup$
    – JohnHoltz
    Commented Jan 10, 2022 at 15:42
  • $\begingroup$ Centrifugal force: F = m v² / r. Speed of earth around sun: 67,000mph or 29,777m/s. Radius of orbit R=91,432,000Miles or 146.29*10^9 meters. Radius of Webb R=147.51*10^9m. Mass of Webb= 6,500Kg. F= 6,500*29,777^2/147.51*10^9=39.06N. Gravity force is mostly from Sun as Earth is negligible, Fgs=G m1m2/r^2= 6.674*10^-11 1.989*10^30*6,500/(147.51*10^9)^2=39.65N. Earth gravity on Webb , Fge=6.64*10^-11*6,500*5.972*10^24*/1,224,000,000^2=1.72N Earth gravity on Webb. It i actually bigger than 0.046N which I calculated previously, made a mistake there apparently. In that case it will take 16 days. $\endgroup$
    – fil
    Commented Jan 14, 2022 at 15:52
  • $\begingroup$ So actually the earth gravity is slowing down Webb no thrusters. $\endgroup$
    – fil
    Commented Jan 14, 2022 at 15:53
  • $\begingroup$ So actually the earth gravity is slowing down Webb no thrusters. Calculations why it will take 16 days: Speed of Webb is 297.2m/s. Acceleration a=f/m=1.72/6500=2.64*10^-4m/s^2. v=at or time t=v/a=297.2/0.000264=1,122,383s=13 days. So it is actually 13 days not 16 days. $\endgroup$
    – fil
    Commented Jan 14, 2022 at 16:04
  • $\begingroup$ If you add the Sun's effect of about 0.6N since Sun gravity is bigger than centrifugal force by that much than it will take 9 days. $\endgroup$
    – fil
    Commented Jan 14, 2022 at 16:08

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