2
$\begingroup$

For a book under construction, I need a starship to make a 180 degree turn rather quickly. Give him sub-light speed in a star system (no relativistic effects) and on a path to the system primary. I know he can slingshot around the star. But that will not give him a 180 degree change back to his original course.

If I give him a max thrust of 60 g and a mass of 200,000 lbs, can he utilize the Oberth effect to complete his 180 degree turn in a reasonable distance (e.g. within 1 or 2 AU?) for an intercept to a pursuing vessel on his original course? Pick any kind of star you like, I can tailor the story to fit the star.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Oberth maneuvers are used to accelerate. Is that what you're doing? Also, you're dealing with spheres, not infinitely small points. What is the trajectory? What is the starting condition for the planet vs. the sun? A "180" isn't really what you're looking for, I suspect. I'm also concerned that the acceleration will be so great that by the time you reach the primary you'll have too much speed to enter orbit. $\endgroup$
    – JBH
    Commented Aug 12, 2018 at 18:20
  • 1
    $\begingroup$ Phil - if ship A is pursued by ship B, and ship A (somehow) pulls a quick 180 degrees turn, the difference in velocities means it'll have a very hard time to intercept ship B - it'll just whoosh past it in the opposite direction (unless A can pull another 180 turn right after passing over B - but that'll probably won't work if you the maneuver is based on any kind of gravitic assist from the star) - is that what you are trying to achieve? $\endgroup$
    – G0BLiN
    Commented Aug 12, 2018 at 19:23
  • $\begingroup$ Rather quickly at 1 AU? The Earth takes 6 months to do that... $\endgroup$
    – L.Dutch ♦
    Commented Aug 12, 2018 at 19:26
  • $\begingroup$ @G0BLiN If the pursuing ship has significantly better acceleration, it would be able to overtake the other ship and still be able to slow down. This of course begs the question "what happens when they finally catch up?" It isn't as if the other ship will just stop and let them on board. They would have to slow down and match velocity with the other ship, which is essentially what happens every time a new crew goes to the Space Station. And that alone takes hours to carefully match velocities. $\endgroup$
    – Ian Johnson
    Commented Aug 12, 2018 at 23:42

3 Answers 3

Reset to default
6
$\begingroup$

Through the star.

Your ship is a Bussard ramjet. https://en.wikipedia.org/wiki/Bussard_ramjet

Bussard Ramjet https://www.deviantart.com/grahamtg/art/Bussard-Ramjet-633670900

Normally the electromagnetic collection field grabs sparse molecules which the ship uses for fusion fuel / inertial mass.

For the maneuver in question, the star is a bloated red giant. The ship goes through the star - not dead center because there will be solid stuff in the core, but through the atmosphere. The collection field of the ramjet is not used to such dense operating conditions so the engineer is going to have to be careful and reduce its size. The field will act as a brake, transferring kinetic energy to the starstuff in its path.

The ship slows and comes to a halt still within the star, then wheels about 180 degrees. Now the ramjet scoop is drinking from a firehose - much more material than it usually gets in interstellar space. The fusion core goes to 110% and star stuff is blasted as a stream out the back, accelerating the ship as fast as its inhabitants can withstand.

This is the sort of thing where the chief engineer complains "The scoop is na for gobbling hot star stuff! It canna take much more of it!"

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nice reframing of the question! :) $\endgroup$
    – G0BLiN
    Commented Aug 12, 2018 at 19:24
  • 4
    $\begingroup$ ‘How do I do a 180 using a star?’ - ‘Bounce off it.’ $\endgroup$
    – Joe Bloggs
    Commented Aug 12, 2018 at 22:29
  • 1
    $\begingroup$ Not Bussard ramjets!! We have known since the late 1970s the technology won't work. When will people learn? $\endgroup$
    – a4android
    Commented Aug 13, 2018 at 2:40
  • 3
    $\begingroup$ Nice - aerobraking in a star ! So sexy I'm not even going to check the physics. :-) $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 13, 2018 at 4:50
2
$\begingroup$

To a pretty good first approximation, the most delta V you can get from a gravitational slingshot around any object is the escape velocity at perihelion. So non-degenerate stars are good for very rougly ca. 100 miles per sec.

Given your specs, any unpowered orbit will be hyperbolic, so considerable power will be needed to do a 180.

A back of the envelope estimate is that the highest velocity that can be turned 180 is half the deltaV that the engins can provide plus escape velocity at perihelion.

Improve this answer
$\endgroup$
1
$\begingroup$

Your best option to perform a true 180 would be to use manuvering thrusters to point your main engine in the direction you are currently traveling and then fire it. This will first slow you down and will eventually begin to accelerate you in the direction you came from. This will take a long to perform as however long your ship has been accelerating since you can only change your velocity by 588m/s every second. It will then take an equal amount of time to reach that speed again, in the new direction. If your ship has been accelerating at 60g for 1 year, it would take you 1 year to stop moving and another year to reach that same speed in the new direction. There really isn’t anything “quick” about space travel...

That’s the easiest way. If you really want to slingshot through a gravity well, you would have to slow down enough to orbit the planet then fire your main engine when you have orbited around most of the planet. This is basically a U-Turn for spaceships. You won’t be on the exact same path, but you will certainly be headed back in that general direction and a couple small course corrections would fix that.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I believe this is the best answer for what I'm trying to depict, so thank you, Ian! $\endgroup$
    – Phil
    Commented Aug 14, 2018 at 1:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .