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Acceleration/turnaround/deceleration? (Read 6733 times)
EDG
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Acceleration/turnaround/deceleration?
05/03/09 at 15:07:24
 
Say you wanted to go from Earth to Mars under constant 1G acceleration for half the trip, then in the middle you instantly flip around and decelerate for 1G. Is there way to simulate this in Gravsim?  
 
It sounds all very well on paper, but when I try to figure out travel times I assume that the initial velocity is 0 m/s and the final velocity is 0 m/s. But I can't help but feel that this isn't actually the case, because the two planets are moving at different velocities in their orbits. So I'm trying to determine how much of a difference that would make.
 
Say you're trying to get from Earth to Mars at Opposition. So the distance is 1.524-1.0 AU = 0.524 AU. I guess you launch from Earth a bit before opposition and then time it so that you arrive at Mars orbit when it's at Opposition to Earth. To calculate it you use s=ut+0.5atē , but I assume u=0 so that gets rid of the ut term. Juggling that around I get that t=SQRT(2s/a), but for s I have to use half of 0.524 since we're accelerating up to and decelerating from the midpoint. If I've done that right it takes 24.6 hours to get to the midpoint (0.262 AU from Earth), and then 24.6 hours to get to Mars from there, so the total is 49.2 hours of travel time (pretty quick!). BUt I guess you'd have to do something to actually get into Mars orbit wouldn't you?
 
I'm trying to get my head around the physics of it all...
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frankuitaalst
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Acceleration/turnaround/deceleration?
Reply #1 - 05/04/09 at 12:38:16
 
Good question Mal .  
I don't know if this is possible to solve with Gsim , in fact I don't know enough about this feature in Gsim . I guess Tony may come with a solution ?  
But , it's solvable analytically ... see annex  
The path  of the probe in this case are two parabola's as you guessed , see sketch in annex .  
After deceleration at time 2*t1 the probe should arrive at Mars's orbit , so it intersects with zero acceleration .  
I wrote down the equations which can be solved fi with an excel sheet .  
Problem is that we can't get rid of the initial y-velocity of about 30km/s . The probe should also get at the end a deceleration to 24km/s to match Mars's velocity . But this is a detail for orbit insertion .  
I come up with about 49.5 hours .  
In this time Mars has traveled about 1.08 degrees .  
The probe has travelled itself 3.93 degrees .  
This means the probe should be launched when Mars is already 3.93-1.08 = 2.85 degrees after opposition .  
I'll post the second page of the calcs.  
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frankuitaalst
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Re: Acceleration/turnaround/deceleration?
Reply #2 - 05/04/09 at 12:39:17
 
And here's the second page ...
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Re: Acceleration/turnaround/deceleration?
Reply #3 - 05/04/09 at 16:57:29
 
This is a great question.  I'll take a look at Frank's formulas.
 
The only way I've ever tried is this:
 
If Earth and Mars only stood still (without dropping into the sun), this would be easy.  Just aim straight at Mars and accelerate.  Halfway there, turn around and keep accelerating.  Your velocity will drop to 0 right as you reach Mars.
 
But, Earth and Mars do more.
 
However, the orbital velocities of Earth and Mars are pretty slow compared to the high velocities you will achieve with a constant 1g acceleration.  Earth moves at about 30 km/s around the sun.  You'll achieve that in under an hour, and the majority of your trip, you'll be travelling appreciably faster than that.
 
So... Just aim straight at Mars.
In Gravity Simulator, use the Auto Pilot:
First compute the time of your journey using kinematics and the Earth / Mars distance.
Divide this by 2 to get the time of your turn-around.
In Auto Pilot, choose "Continuous Orientation - Towards", and choose Mars
In Auto Pilot, choose "Thrust - Initiate" and 9.8 m/s/s
At your turn-around time choose "Continuous Orientation - Away", and choose Mars
At your arrival time, choose "Thrust - Cutoff".
Use the "Thrust Box" to manually tweak your velocity relative to Mars to 0.
 
 
This will get you in the general area of Mars.  It will take about 2 days to reach there, and Mars moves considerably in 2 days, so you'll miss by a few million km.  So, you do it again.  Compute the length of travel from the new position to Mars, and divide by 2 for your turn around time.  "Continuous Orientation" towards Mars, and Thrust 9.8 m/s/s.  At the turn-around point "Continuous Orientation" away from Mars, and at arrival time, Thrust Cutoff.
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Re: Acceleration/turnaround/deceleration?
Reply #4 - 05/07/09 at 08:50:04
 
Thanks all - um, looks rather more complicated than I thought it'd be! Smiley
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frankuitaalst
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Re: Acceleration/turnaround/deceleration?
Reply #5 - 05/09/09 at 06:08:56
 
Quote from Mal on 05/07/09 at 08:50:04:
Thanks all - um, looks rather more complicated than I thought it'd be! Smiley

To make life a little bit easier you can find hereby an excel sheet with the formulas... Smiley
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