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Travel time to orbit

kurtis

SOC-10
Got a real quicky 'check my math' scenario.

I'm running my long-time player group through TTA. The group is almost entirely new to Traveller conventions, and they finally got off of Aramis last weekend. Anyway, when it was explained to them that the March Harrier doesn't exactly voom off the launch pad at 1g acccel, I was challenged by a player who ought to know better.

His thinking is that modern rockets take "hours" to reach orbit using much more than 1g acceleration. He seemed to think it would take the poor Harrier most of the day to achieve a decent orbit.

Assuming a 1G M-drive gravity powered spacecraft (unaffected by local gravity due to contragrav, etc.) lifting from Earth, my simple physics drop-out calculations show an arrival time to geostationary orbit in almost exactly 45 minutes. Can anyone else back that estimation up?

This also seems to me like a very reasonable performance number for a 1G drive from an in-game perspective.

...

As an aside, this lightly touches on a classic technology debate. For the purposes of this CT-style campaign, I will probably run this game with the assumption above, that starship gravity drives negate the local gravity field up to their drive rating, independantly of the acceleration power. Otherwise, a 1G Manuever ship can only hover in Earth gravity, without special additional acceleration (i.e., SSOM-style 'overdrive' performance, etc.).

Otherwise, my heretical tendencies lean towards a ATU style where many gravity drive ships still carry small reaction drives specifically to help boost out of gravity wells and act as power emergency backups. I like this idea for a number of reasons, not the least of which is the common artistic depictions of classic Traveller small ships with obvious exhaust nozzles.

I should save that for a MTU forum topic, though. :D
 
Earth is world size 7, right? Travel Time To Orbit according to the MegaTraveller Imperial Encyclopedia Starship Operating Procedures is 35 minutes. It doesn't show a difference of maneuver drives. Me personally, I would assume 35 minutes for 6G. Double that for 3G. If that's the case, 1G would be about 2 to 2 and a half hours to orbit of a size 7 world. That can't be right. Well, just maybe - 1G is a slow push.
 
Originally posted by Silent Cartographer:
His thinking is that modern rockets take "hours" to reach orbit using much more than 1g acceleration. He seemed to think it would take the poor Harrier most of the day to achieve a decent orbit.
While geostationary orbit is a long way out, today's space shuttle reaches a crude orbit in 10 mins, then uses its little OMS to push into a stable 'working' orbit in 35 mins later. Check here for details.

Also, you have to take into account streamlining v atmospheric density. IIRC MT said a strealined non-airframe ship could achieve a top speed of 1000kph in a standard atmosphere.

Regards PLST
 
Huh?
I’m confused. The travel times table in Starter Traveller lists the travel time to 1000km (about 600 miles)as 633 seconds (10 ½ min) at 1G. I thought the AG stuff made the ship “weightless” and the M drive pushed it “up” at a steady 1G?
 
Starship Operator's manual says:

For a size 7 world;
A Standard orbit is 140 km.
A high orbit is 700 km.

If the world has any kind of atmosphere code of 2+,
Multiply either of the above by the world UPP, to avoid friction build-up on the hull, basically taking a slower curving path through the atmosphere.

From the Standard orbit, then just boost to the Geosync orbit if you wish.


A Geostationary orbit is:
Orbit Height in KM = 5,078 X (Mass of world in Earth Masses X Rotation period in hours ^ 2)^.33 - (World Diameter in km / 2)

Megatraveller Imperial Encyclopedia 9th printing, 1987 P. 92 Step 4, Travel Times to orbit, in fact does show different times for different maneuver Drives Acceleration:

Travel Times to/from Orbit (I guess this is standard orbit, above)

Size 7 world:
1g: 35 min
2g: 25 min
3g: 20 min
4g: 18 min
5G: 16 min
6g: 14 min
 
Basic fizzix says d = .5at^2

So, t = sqrt(d/.5a)

Earth Geo Synch orbit is 24000 Miles, which is ~40000 kilometers, or 40M meters.

a = 1G, 9.8ms^2. .5 * 9.8 = 4.9

t = sqrt(40000000/4.9) = ~2857 seconds.

2857 seconds is about 47-48 minutes.

v = at, so once you reach orbit, you'll be running at about 28000m/sec velocity.

Now, that's simple basic mechanics and utterly ignores all sorts of factors having to do with being in a gravity well (like, say, gravity, friction from air resistance, slowing down if you're planning on docking with something, actually getting INTO orbit versus simply blasting off into space, etc.)

I'm sure the fellas in the trench of Mission Control have better formulae than this, but thems the basics.
 
Thanks for the replies!


That link will be useful, Hemdian, thanks tons.

Whartung, that's the formula I was using, glad I used the correct one, heh. Wikipedia listed geostationary orbit as ~35,786 km, and I was using that as a test case for highport orbit. Ignoring all the stuff like gravity, atmoshpere and such, put me right at 45 minutes.
 
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