What is the most efficient flight path into orbit?

[[BL]Phalanx]

"Efficient" can mean a few different things. In this case, I don't mean "efficient" as in saving the most amount of time, or in terms of distance travelled to get to orbit or fuel consumption. I'm talking about acceleration.

What flight path into orbit would require the lowest acceleration? Then, given the altitude eventually reached and a timeframe, how would you solve for the minimum acceleration required?

[Zoink]

I don't think you're asking the right question, you can have as minimum acceleration you want (ie. a hovering spacecraft).

What you want, is for a given thrust (and for a particular craft), find the best possible trajectory to minimize fuel consumption. I'd have to do a crap load of calculation, because I don't have any simply formulas here. But you need to minimize the time to achieve orbital velocity and reach the desired height. In this case you actually looking to maximize your acceleration given a particular thrust. A regular rocket follows an arced trajectory, but this might be restricted on the flight characteristics and thrust/weight ratio of the rocket. A futuristic superlight, high thrust vehicle (maybe a scramjet/rocket combo) might have a more efficient trajectory.

[Zoink]

Hmm. I do have some formulas for optimal aircraft acceleration and climb, but I think they fall apart in space

[Wicked Pilot]

"Efficient" can mean a few different things.

I believe the meaning you are looking for here is the ratio of work done to energy expended

What flight path into orbit would require the lowest acceleration?

That's like asking what's the lowest acceleration you can use to get from New York to Chicago. You can accelerate as slow as you want, but it will take you a while to get there.

Then, given the altitude eventually reached and a timeframe, how would you solve for the minimum acceleration required?

Go back and rethink your question.

[[BL]Phalanx]

Thanks for the replies, guys, but....

I don't think you're asking the right question, you can have as minimum acceleration you want (ie. a hovering spacecraft).

Go and re-read what I originally asked. The time frame is *known*... you cannot just take your sweet time getting up to orbit. :p

Initial velocity is zero (except for the planet's rotation... let's say the craft launches in the direction of the planet's rotation at the equator so it starts with some speed), initial altitude is zero (starting on the ground), final altitude is known and final velocity is known.

What you want, is for a given thrust (and for a particular craft), find the best possible trajectory to minimize fuel consumption. I'd have to do a crap load of calculation, because I don't have any simply formulas here. But you need to minimize the time to achieve orbital velocity and reach the desired height. In this case you actually looking to maximize your acceleration given a particular thrust. A regular rocket follows an arced trajectory, but this might be restricted on the flight characteristics and thrust/weight ratio of the rocket. A futuristic superlight, high thrust vehicle (maybe a scramjet/rocket combo) might have a more efficient trajectory.

I'm not looking to minimize fuel consumption. I'm looking to minimize acceleration.

I guess the word "efficient" doesn't really apply here, since acceleration isn't really a resource like time or fuel, but I'm looking to get a low-end for acceleration.

What we do know:

We know the planet's mass and radius
We know the planet's period of rotation
From that we also know the initial velocity of a ship launching from the equator.
We know the initial altitude. It's zero, we're starting at the surface. The distance from the center of the planet is roughly the planet's radius.
We know the final altitude
We know the final velocity (which is just the speed of a craft in orbit at that final altitude over that planet)
We know the time frame.

What is the minimum acceleration required to get to orbit, given all of that?

[Darth Wong]

Go and re-read what I originally asked. The time frame is *known*... you cannot just take your sweet time getting up to orbit. :p

You originally said:

"Efficient" can mean a few different things. In this case, I don't mean "efficient" as in saving the most amount of time, or in terms of distance travelled to get to orbit or fuel consumption. I'm talking about acceleration.

What flight path into orbit would require the lowest acceleration? Then, given the altitude eventually reached and a timeframe, how would you solve for the minimum acceleration required?

Nowhere do I see a quantification of timeframe or altitude. You have given no numbers whatsoever, in fact. So how do you expect anyone to produce any numbers? You do realize that there is more than one possible altitude for "orbit", don't you?

[[BL]Phalanx]

You originally said:

"Efficient" can mean a few different things. In this case, I don't mean "efficient" as in saving the most amount of time, or in terms of distance travelled to get to orbit or fuel consumption. I'm talking about acceleration.

What flight path into orbit would require the lowest acceleration? Then, given the altitude eventually reached and a timeframe, how would you solve for the minimum acceleration required?

Nowhere do I see a quantification of timeframe or altitude. You have given no numbers whatsoever, in fact. So how do you expect anyone to produce any numbers? You do realize that there is more than one possible altitude for "orbit", don't you?

Note: "Then, given the altitude eventually reached and a timeframe..."

I do not expect others to do the work for me. I'm only asking for the way to do it. I can crunch the numbers myself later. I'm just saying that time frame and altitude is known.

[Sriad]

If you didn't care about fuel, A could be as low as you wanted as long as its enough to beat Earth's gravity. You could be accelerating at 10 m/s^2 (.4 m/s^2 net) and you'd get into orbit eventually. It would take a ricockulous ammount of fuel, of course. But you don't care about that. That's the best that can be done unless you provide specifics.

[Kuroneko]

Is it reasonable to assume that this is a rocket? I'm not sure how to model some sort of airplane-to-space craft.

[Shaka[Zulu]]

not to mention the fact that whatever thrust agent you're using has to at a minimum be able to overcome atmospheric drag until you are effectively at orbital altitudes.

If you are really serious about a minimum acceleration to orbit, you need to look outside of any kind of flight vehicle, and instead look to beanstalks, as their elevator cars dont have to worry about stuff like that, and can take as long as they want to get up high. of course, the only orbit you can really utilize with this tech would be GEO...

[[BL]Phalanx]

Heh, okay my initial wording could've been better, it seems...

I don't care about fuel consumption... but there is a time limit.

Okay, how about this. The planet is Mars. The ship starts at the equator and launches in the spin-ward direction. The initial altitude can be reasonably assumed to be zero (ship started on a runway on the ground). Thus initial distance from the planet's center is the planet's radius. The final altitude is 1921 kilometers, and final velocity is the speed of an orbit over Mars at that altitude (which should be approximately ~2837 meters/sec if I did my math right).

The time frame from ground to orbit is 18 seconds. What is the minimum acceleration requirement?

Initially I used a simple flight path, going by "the shortest distance between two points is a straight line", trying to get a low-end for the acceleration. So I just assumed a flight path straight from ground to orbit in a line aligned with the planet's radius. I'm trying to refine this some more, so I came here to ask if there were more "efficient" flight paths to get to orbit, ones that would lower the acceleration requirement. I intended to use that to generate a more conservative figure.

[Shaka[Zulu]]

ohhh... now THATS different. you are actually looking for a lower bound on acceleration to get to your desired orbit in a desired timeframe. this is quite different from an absolute minimum.

[[BL]Phalanx]

Okay so I could use math terms more precisely, but I didn't say "absolute minimum acceleration".... I said "minimum acceleration" in the context that we already know final altitude, velocity, and timeframe...

Anyways... so what is the flight path that'll give the lowest acceleration figure? I imagine that the flight paths we use in reality won't apply, since the time frame is so short... but I could be wrong.

[Shaka[Zulu]]

OK... I would suggest as an off-the-top-of-my-head number, about 20g of acceleration. That would get you up to orbital velocity in well under 20 seconds, yet still allow for some losses due to gravity and atmospheric drag -- although this is mars we are talking about, so such effects would be far less than for earth.

[Shaka[Zulu]]

also, as to the trajectory... yes you would still want something of a vertical ascent, as real LV's do, because you want to spend as little time in the grip of atmosphere as possible. at 20g, you could get out into space right quick, and worry about actually making orbit once you get there.

[[BL]Phalanx]

20 g wont' work. It'll get you up to the desired velocity, but not the desired altitude. At 20 g, assuming no drag and no gravity to slow you down, you'll only travel 31.7 kilometers in 18 seconds. The orbit is at a height of 1921 km from the surface of Mars.

[[BL]Phalanx]

also, as to the trajectory... yes you would still want something of a vertical ascent, as real LV's do, because you want to spend as little time in the grip of atmosphere as possible. at 20g, you could get out into space right quick, and worry about actually making orbit once you get there.

Well I initially assumed a perfectly vertical ascent. I also had to account for the fact that the ship would overshoot in that time frame unless it also slowed itself down half-way there.

I was thinking that with vertical ascent I'd get the most conservative acceleration figure, but I wanted to be thorough...

[Shaka[Zulu]]

20 g wont' work. It'll get you up to the desired velocity, but not the desired altitude. At 20 g, assuming no drag and no gravity to slow you down, you'll only travel 31.7 kilometers in 18 seconds. The orbit is at a height of 1921 km from the surface of Mars.

oops... my bad

well i did say it was an 'off-the-top-of-my-head' number... wasnt thinkin about altitude -- just velocity

[[BL]Phalanx]

It's okay, heh, when I first tried to do this I made the same mistake...

[Wicked Pilot]

Try this equation: d=(1/2)at^2

Let's say you want to get to 20 km in 1 minutes at constant acceleration. Put the numbers into the equation, and plug n' chug:

20,000m=(1/2)a(60s)^2

solve for a

a=11.1m/s^2, or about 1.2g.


Does this help any?

[[BL]Phalanx]

Try this equation: d=(1/2)at^2

Let's say you want to get to 20 km in 1 minutes at constant acceleration. Put the numbers into the equation, and plug n' chug:

20,000m=(1/2)a(60s)^2

solve for a

a=11.1m/s^2, or about 1.2g.


Does this help any?

Thanks, but that's what I used originally. I assumed a straight line from ground to orbit on the reasoning that the shortest path would produce the lowest acceleration needed. However, I'm open to the possibility that that wasn't the case, so I'm asking if perhaps there's another path to orbit that would take even less acceleration.

For example, going with the numbers I have:

d = (a * t^2) / 2
d = 1921000 meters
t = 18 seconds

a = (2 * d) / t^2

a = 11858 meters/sec/sec

Unfortunately, this doesn't make sense since the ship will fly waaay past an altitude of 1921 km. It'll be doing 213 km/sec away from Mars by the time it reaches that altitude. So it has to turn around half-way there and decellerate. We know the ship eventually settles into orbit, it doesn't continue to shoot off into space.

Now assuming it turned around half-way means it travelled half the distance in half the time. Acceleration is proportional to distance, but the time is squared. This doubles the acceleration required to 23716 meters/sec/sec. It ignores the initial velocity the craft may have started with due to the planet's rotation, and ignores the velocity the craft must maintain to be in an orbit at that altitude (I think it's ~2837 meters/sec at an altitude of 1921 km).

But the initial velocity is not that significant here. Even starting at the equator and launching in the spin-ward direction, on Mars you get a boost of maybe ~247 meters/sec.

Anyways, I am going for the most conservative figure I can get, so I need to know if there's a better flight path than a straight line directly up from ground to orbit.

[Wicked Pilot]

It might be easier if you use your altitude and orbital velocity to calculate your ship's kinetic and potential energies upon reaching orbit. Using that and the time it takes to reach orbit, you can work backwards, getting your acceleration figures that way.

[[BL]Phalanx]

It might be easier if you use your altitude and orbital velocity to calculate your ship's kinetic and potential energies upon reaching orbit. Using that and the time it takes to reach orbit, you can work backwards, getting your acceleration figures that way.

Hey that's a good idea, I hadn't thought of it that way, hadn't considered potential energy at all. Lemme see if I can still remember enough of what I learned about that to do this myself. If not I'll ask for help. Thanks.

[[BL]Phalanx]

I've run into a problem. Potential energy is based on the work that would be done on an object as it fell from a height, right? The force acting on it would be gravity, and the height would be the distance it would fall? That's why

PE = m * g * h

Well, the problem is, gravity isn't the same at all heights. For small changes in height this can usually be neglected, but it can't be neglected here. The gravity weakens as you gain altitude:

g = (m_planet * G) / r^2

'm_planet' and 'G' are constants, 'r' is the distance from the center of the planet. 'r' ain't constant, but changes as your altitude changes. So 'g' ain't constant.

I have a feeling that simply substituing that in for 'g' won't work, since that would only give the particular value of 'g' at that altitude, but the acceleration due to gravity is not constant throughout the fall.

I'm thinking I need to integrate something somewhere, but I'm not sure what and where, or what the limits of integration would be.

Do I just think to myself "okay, gravity changes as distance from center changes, so..."

dg/dr = (-2 * (m_planet * G)) / r^3

Bleh, I dunno... help please.

[Darth Wong]

In order to figure out minimum acceleration to achieve an orbit of given altitude in a given timeframe, you must figure out the orbital velocity (that's the velocity you need for centripetal acceleration to match gravitational acceleration). Acceleration is final velocity divided by time. That's the angular component of acceleration.

Then, use the x=0.5at^2 formula to determine minimum radial acceleration (upwards, to gain altitude). Put them together, and you've got minimum acceleration.

The use of gravitational PE affects the energy requirement (energy required to reach orbit is GPE+KE), but not the acceleration requirement, which is a simple matter of velocity and time.