# Small solution for space travel times

sonhouse
Science 01 Nov '17 13:51
1. sonhouse
Fast and Curious
01 Nov '17 13:5115 edits
To get from one point to another, say in our solar system, one solution is this:

T = time, S= distance, A= acceleration.

So starting from S=(AT^2)/2 you get T=Sqr root of (2S/A), T=(2S/A)^ 0.5

The problem with that is say you rocket from here to Mars, and accel at one g. You get there really fast, a bit over one day but with that formula, you arrive at Mars at max velocity which might be ok for a fast flyby and cameras clicking but if you want to go into orbit there, you need to decel a lot and since you would be going about 700 Km/second, you would have to do a huge delta V to get into orbit, no good.

So I looked at the idea of positive acel to the halfway point and decel the rest of the way, nothing original there.

But I played with the numbers and found if you do that, you can use a modified version of the original equation that leaves you at the destination flying by at max velocity.

New solution: T= Sqr root of 2 times (Sqr root of 2S/A)

So just use the original equation times the sqr root of 2, or 1.414ish.

T=2^0.5 * (2S/A)^0.5

So now you acel half way at one g, decel half way at one g and you end up with zero relative velocity at the destination and of course some modifications to velocity you might need to inject into a stable orbit but the main equation gets you going pretty close. And of course a bit of course correction since you can't go exactly on a straight line to the destination since you would be in the influence of solar gravity and would bend a bit but that correction is a bit above my pay gradeðŸ™‚. I am just assuming a straight line path which would work really will if you are way away from the sun.

It takes 1.414 times longer but you get there at zero delta V needed to inject into orbit.

So going 50 million Km at one g, the original formula says you get there in 101,015 seconds, a bit over one day, 28 hours and change.

But the modified formula where you go 25 million Kmat one g, decel the rest of the way at one g, it takes 142,857 seconds (1.414 times 101,015) or about 40 hours.

And of course 1 g for hours on end is not going to happen anytime soon but they are talking about the Vasimir rocket that might clock in at 1/20th g and you get there in about one month, or even a week if Mars and Earth are close, a lot faster than a chemical rocket which takes 6 or more months.

It is vital for astronauts to spend as little time in interplanetary space as humanly possible because they may run into a solar flare that could kill them, if not immediately, but later sometime from cancer.
2. apathist
looking for loot
01 Nov '17 17:27
Originally posted by @sonhouse
...
So I looked at the idea of positive acel to the halfway point and decel the rest of the way, nothing original there.

But I played with the numbers and found if you do that, you can use a modified version of the original equation that leaves you at the destination flying by at max velocity....
You lost me here. If you accel halfway there and decel the rest, the only adjustments should be to account for any velocity difference between your starting point and your destination.
3. sonhouse
Fast and Curious
01 Nov '17 18:28
Originally posted by @apathist
You lost me here. If you accel halfway there and decel the rest, the only adjustments should be to account for any velocity difference between your starting point and your destination.
Not quite true. If you are moving in a gravity well, that well will curve your direction of travel, even if just a little bit inwards towards the main mass, in this case the sun. So you may WANT to go in a perfectly straight line the gravity well of the sun stops that.

Kind of like shooting ducks with a shotgun, you have to aim where the ducks are GOING to be when the pellets meet the ducks. Otherwise you will find youself shooting behind the ducks, missing them every time. I don't know how to deal with that curved path mathematically. At the peak velocity of say 500 Km/second you spend a small amount of time in the trip trying to go in a straight line but you would still be off course if you just tried to plot a straight path from Earth to Mars. My guess is on a trip to Mars from Earth at one g, you would have to aim a bit right of Mars assuming you are going in the same general direction Earth and Mars is going. No idea of the math to do that accurately. Anyone here know how to deal with a wannabe straight line shot to Mars under 1 g of accel/decel, how and where you aim actually, to intercept Mars and not miss it by a million miles?
4. 02 Nov '17 19:40
You don't accelerate continuously in space travel. You accelerate to escape the orbit, then slingshot your way to where you want to go.
5. sonhouse
Fast and Curious
03 Nov '17 16:321 edit
Originally posted by @kazetnagorra
You don't accelerate continuously in space travel. You accelerate to escape the orbit, then slingshot your way to where you want to go.
Obviously you have to achieve orbit first but if you can acel at one g you go in almost a straight line to your target. If you could actually do that from the ground, you wouldn't get anywhere anyway at exactly one gðŸ™‚

But suppose you could do 2 g, then you would effectively acel at one g fighting Earth's gravity and when you got to around 160 km high you would be going about 1.6 Km/second