3
$\begingroup$

This answer to Why aren't satellites launched into a Polar LEO to get into Polar Heliocentric Orbit? leaves me wondering if it could be done today without the trip to Jupiter.

Could a replica of the original Ulysses spacecraft be put into its 79.11° inclination heliocentric orbit directly from Earth by any launch vehicle currently available?

Here "directly from Earth" means not going to another planet first.

If not, could this be done with something likely to be available in the near future?

$\endgroup$
2
3
$\begingroup$

Barely, but it involving shenanigans even less practical than a Jupiter flyby.

The lowest delta-v cost trajectory without any planetary flybys from Earth to a 1.35AU-5.4AU 79.11° orbit is the following:

  1. Do a burn in LEO, reaching a solar system escape trajectory. Cost: 8750m/s
  2. At the edge of the solar system, do an inclination change and set the periapsis to 1.35AU. Cost: ~0m/s
  3. At 1.35AU, lower the apoapsis to 5.4AU. Cost: 3830m/s

The first burn is quite comparable to what was done for the New Horizons spacecraft. It didn't accelerate quite as much (as it too used a Jupiter flyby), but it was considerably lighter, allowing for an extra mass ratio of 1.3 for the upper stage.

At the moment, we don't have anything much beefier than the Atlas V launching New Horizons, so the extra 3830m/s can't be crammed in anywhere else if we use chemical rocket engines.

But the final apoapsis lowering can conceivably be performed by an ion engine, as 1.35AU is well within where solar panels provide enough power for that. From Dawn acceleration data, this would take two or three passes.

In the end, this ends up taking much more time than doing a flyby of Jupiter.


The final orbit of Ulysses is however chosen specifically because it can be achieved with a Jupiter flyby. Aiming for any 79° orbit, we could use the recipe above, just removing step 3)

Doing a heliocentric inclination change of 79° directly from LEO is more in the order of 16km/s, outside our current capabilities.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for the insightful and conclusive answer! (btw too many excellent answers per day (high EAPD) can lead to hitting the max rep cap (+200/day from one UTC 00:00 to the next?)) :-) $\endgroup$
    – uhoh
    Oct 2 '20 at 0:05

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.