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Earth & Mars Gravity Turns


Axilourous

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As the name implies, I am wondering what are the most efficient gravity turns for Earth and Mars?

I don't know if RSS and Real Life gravity turns are the same, but I am wanting real life and not RSS, unless, of course, they are the same.

Lastly, how do I calculate Delta-V?

Edited by Axilourous
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1 minute ago, munlander1 said:

1. This should be in science and space flight. No problem though. A mod like @Vanamonde can move it.

2. It would also depend on the craft itself. Like the engine isp's and the drag of the craft I guess. What I trying to say here is that there is no "best" one.

What would be the formula then? All the websites I find have not included drag in their formulas. Mars and Earth, last time I checked, both have atmospheres, meaning drag. Science and Space Flight, never even thought of puting it there lol

One other question is how to calculate dv

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Just now, Axilourous said:

What would be the formula then? All the websites I find have not included drag in their formulas. Mars and Earth, last time I checked, both have atmospheres, meaning drag. Science and Space Flight, never even thought of puting it there lol

That exceeds my knowledge, but a lot of people on here will be able to answer your question. Might have to wait till tomorrow afternoon.

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It's difficult to calculate. It depends on atmospheric pressure, of course, but also on thrust, azimuth, gravity, propellant usage, aerodynamic forces, and mass. I think that in real-life mission planning, they simulate hundreds of iterations of flight profiles until they find the best one.

Edited by Nibb31
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As a first-order approximation, treat Mars as an airless world. Its datum pressure is something like 1% of Earth's. And even on Earth, aerodynamic losses are only around 5% of your total dV budget to orbit. As such, your Mars ascent will likely be at its most efficient if you turn over fast and early, staying as flat as surface features and your engine's thrust allows you to. This will also be made easier by Mars' reduced radius and gravity when compared to Earth, and the fact that there are no weather systems worth mentioning (or worth dodging, for that matter).

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51 minutes ago, Axilourous said:

Okay, so it is difficult to calculate gravity turns, but what about calculating Delta-V

In the absence of friction, to go from one orbital altitude to another, calculate dV = SQRT( 2GM * ((1 / r1) - (1 / r2))), where r1 is your distance from the center of the gravity well at the start of your transfer and r2 is your distance at the end of it.

For increasing your altitude by 200 km above the Mars equator, you get: dV = SQRT(2 * 42,828,370,000,000 * ((1 / 3,396,200) - (1 / 3,596,200))) = ~ 1184 m/s.

But when you take off from Mars, you are not in an orbit at the level of the planet's radius. You're going much slower. So you need to add the missing speed as well.

Orbital speed at Mars equator: v(o) = SQRT(GM / r) = SQRT(42,828,370,000,000 / 3,396,200) = ~ 3551 m/s

Then you subtract Mars' equatorial rotation speed, since that is speed you already do have, and add the cost to raise your orbit: 3551 - 241 + 1184 = 4494

So about 4500 m/s.

 

Now let's type type "dV to Mars orbit" into Google... this dV map says 4100 m/s between low Mars orbit and Mars surface, although "low Mars orbit" is never defined. Looks like my approximation isn't far off, especially if they're launching into a lower orbit.

 

Edited by Streetwind
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3 hours ago, Streetwind said:

In the absence of friction, to go from one orbital altitude to another, calculate dV = SQRT( 2GM * ((1 / r1) - (1 / r2))), where r1 is your distance from the center of the gravity well at the start of your transfer and r2 is your distance at the end of it.

For increasing your altitude by 200 km above the Mars equator, you get: dV = SQRT(2 * 42,828,370,000,000 * ((1 / 3,396,200) - (1 / 3,596,200))) = ~ 1184 m/s.

But when you take off from Mars, you are not in an orbit at the level of the planet's radius. You're going much slower. So you need to add the missing speed as well.

Orbital speed at Mars equator: v(o) = SQRT(GM / r) = SQRT(42,828,370,000,000 / 3,396,200) = ~ 3551 m/s

Then you subtract Mars' equatorial rotation speed, since that is speed you already do have, and add the cost to raise your orbit: 3551 - 241 + 1184 = 4494

So about 4500 m/s.

 

Now let's type type "dV to Mars orbit" into Google... this dV map says 4100 m/s between low Mars orbit and Mars surface, although "low Mars orbit" is never defined. Looks like my approximation isn't far off, especially if they're launching into a lower orbit.

 

No, not dV to orbit, but how to calculate dV of spacecraft

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1 hour ago, Axilourous said:

No, not dV to orbit, but how to calculate dV of spacecraft

 

That's literally called "the Rocket Equation". dV = engine Isp * 9.807 * ln(stage mass ratio), calculated individually for each stage.

 

I've written a wall of text about understanding mass ratios and using the rocket equation over here. :wink:

 

 

(Will I ever run out of opportunities to link this? I think not!)

 

Edited by Streetwind
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