2
$\begingroup$

This tweet

The second (crucial) engine firing of the Block DM-03 stage in today's #SpektrRG launch to come in an hour, when the spacecraft almost completes its first orbit around the Earth. Here is my estimate of the ground track: https://t.co/GC8WmAMGct

contains the first image below, and the link contains the others as well.

  1. Why does it look like SpektrRG makes a u-turn in the ground track image?
  2. How does it manage to orbit the Earth at a steady 6 degrees above the equator?

Images from Russian Space Web's Proton sends Spektr-RG into deep space


Russian Space Web' Proton sends Spektr-RG into deep space

below: click for full size

Russian Space Web' Proton sends Spektr-RG into deep space Russian Space Web' Proton sends Spektr-RG into deep space

$\endgroup$
9
$\begingroup$

Why does it look like SpektrRG makes a u-turn in the ground track image?

An L2 injection trajectory looks a lot like going straight away from Earth -- i.e. the ground-track position in inertial space is changing only slowly. Meanwhile, the Earth is rotating Eastward underneath that point, so the ground track appears to move Westward across the surface of the Earth.

How does it manage to orbit the Earth at a steady 6 degrees above the equator?

It's not orbiting around the Earth; it's going almost straight away from a point 6 degrees above the Equator.

Compare the "DM-03 Burn 2" and "Spacecraft Separation" checkpoints on the ground track against the same checkpoints in your lower right image.

It's going from a 170km x 1927km orbit, which is not very eccentric at all¹, to a 500km x million-km orbit, which is practically a straight line by comparison.

Equidistant-time marks on the ground track would help with the visualization (or synchronized animation of both the ground track and perspective views) -- at spacecraft separation the ground track would be going ~10 km/sec across Earth's surface, but the Westward track after the U-turn is Earth rotation speed, ~400 m/sec.

¹ Remember to add Earth radius to the given orbit altitudes.

| improve this answer | |
$\endgroup$
  • $\begingroup$ so the third image showing what looks like SpectrRG at a very high inclination and latitude is probably inaccurate? $\endgroup$ – uhoh Aug 7 '19 at 1:38
  • 1
    $\begingroup$ No, the inclination is in fact 51º. At the "spacecraft separation" point it's over high latitude at the start of the ground track "U-turn" and going really fast, and on the southbound leg of its orbit. Its velocity, however, is largely upward rather than horizontal. As it gets farther from Earth, two things happen: it slows down due to Earth's gravity (small effect), and it sweeps less Earth-surface-angle per time unit because it's getting further away (large effect). ... $\endgroup$ – Russell Borogove Aug 7 '19 at 1:46
  • 2
    $\begingroup$ ... Note that the outbound leg of the L2 trajectory takes three months; if L2 was right over the equator, that means the ground-track position would spiral from 6ºN to equator over 90 Earth revs, but < 1 rev is being drawn here. (Sun-Earth L2's Earth-latitude changes over time +/- 23º, I don't know where the final latitude will be, but it's going to be slow going in any case.) $\endgroup$ – Russell Borogove Aug 7 '19 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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