Escape velocity to Mars

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jhoblik

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Could we save fuel to achieve escape velocity, by periodically turn on engine to change orbit to more and more ecliptically and then hyperbolical to Mars. Use Earth gravity like slingshot.
 
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spacester

Guest
You cannot use the mass / gravity of the planet / sun you are orbiting to change the orbit. Your orbital energy is 'bound' to it already. You need to use a *third* body to pull off these tricks.<br /><br />We can use the moons of Jupiter to change our orbit around Jupiter, likewise we can use Earth's moon to change our orbit around Earth.<br /><br />For example, we can get a boost from the Moon when leaving Earth orbit. Just a boost, though, you still need to power yourself out of Earth's gravity well.<br /><br />My understanding is that the most boost one can hope for from Luna on a Mars trajectory is ~0.5 km/s, but this invloves something like 3.0 km/s of rocket-provided dV <br /><br />These maneuvers are called "slingshots" but I do not like the term. More descriptive is the term 'fly-by'. Even better is 're-direct', because in all cases what is happening is that you are changing the direction of your velocity vector. The magnitude of your velocity remains unchanged, but you can redirect it and thus achieve an orbit of a different energy. <div class="Discussion_UserSignature"> </div>
 
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darkenfast

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This is getting into an area where I am "knowledge-challenged", but I was under the impression that on a slingshot maneuver, more than a change of direction was involved, and that there was an increase in energy. There is a coresponding decrease in the energy of the object used (planet or moon's velocity in orbit). The "falling" object (the spacecraft) gets the energy by virtue of it's being pulled by gravity from the large object. If mass is expelled by thrusting at the "bottom" of the trajectory, there's a boost there (greater mass falling in than climbing out).<br />At least, that's how I understand it. Someone want to explain it better?<br />
 
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jhoblik

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I found something similar article about Hohmann transfer orbit. Could we achieve that orbit by periodically push of engine. For example during 60 days(more and more eliptical). And last push will move us to the elliptical transfer orbit.
 
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najab

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><i>The magnitude of your velocity remains unchanged, but you can redirect it and thus achieve an orbit of a different energy.</i><p>I believe this is incorrect. The momentum of the system (planet-probe) remains the same, but the planet is robbed of some momentum and the probe gains some, the difference in mass meaning that the change in velocity of the planet is imperceptible.</p>
 
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spacester

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I do not like that wikipedia explanation and I stand by my post. The planet's orbital velocity does not have anything to do with it, it is a simple matter of the mass of the planet bending the path of the probe. It's not a slingshot, it's a re-direct. Yes I know that everyone has been told it's like a slingshot, but that's wrong information. Yes the probe 'steals' angular momentum from the planet but the mechanism shown is wrong wrong wrong. Note that there are no equations given which actually allow you to calculate a trajectory change. When I get home tonight I'll dig up the equations if y'all want.<br /><br />It's all a matter of frame of reference. The probe has an orbital velocity in the sun's f.o.r., but as it crosses into the planet's sphere of influence, the probe begins falling toward the planet. If it doesn't hit atmoshere or the surface, it leaves the planet's sphere of influence at the same velocity *magnitude* as it entered - relative to the planet. Vector arithmetic then lets you find the velocity in the solar f.o.r. and you find that you've boosted your heliocentric orbital energy.<br /><br />Maybe if you google 'impact parameter' you'll find better math and diagrams. Beware entering an argument with me on this, I wouldn't have said what I said if I wasn't 100% sure. <div class="Discussion_UserSignature"> </div>
 
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najab

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><i>The magnitude of your velocity remains unchanged, but you can redirect it and thus achieve an orbit of a different energy.</i><p>If the energy of the orbit changes, then the angular momentum of the probe has changed.</p>
 
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spacester

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Absolutely! Energy and momentum must be conserved.<br /><br />It's all about the frames of reference in which you define your velocity. I think I'm writing my sentences too quickly here at work . . . <br /><br />The velocity magnitude in the planet's f.o.r. is unchanged but the direction has been altered. The velocity magnitude in the Sun's f.o.r. has most definitely been changed along with the direction, thanks to stealing angular momentum from the planet <div class="Discussion_UserSignature"> </div>
 
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najab

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Well that's the 'problem' - if we're talking about inter-planetary missions, the most convenient f.o.r is a sun-centred one.
 
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spacester

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OK the boss just left so I can slow down a sec, lol<br /><br />These things are calculated by the 'patched conic method'; all trajectories (elliptical, parabolic, hyperbolic) are conic sections. A swing-by maneuver involves leaving the heliocentric coordinates, entering the planetary coordinates and then returning to heliocentric. This is done because the gravity of the planet becomes more important than the gravity of the sun when you get close enough to the planet. While in reality this is a gradual process, math lets us make an abrupt transition from one to the other at a certain distance from the planet known as the 'sphere of influence' which is a function of the masses of the sun and planet.<br /><br />A probe is cruising along in its orbit and you can calculate its velocity vector using the vis-viva equation for magnitude and the tangent to its elliptical orbit for direction. <br /><br />And then suddenly - wham! - the probe is close to a planet and it crosses into its sphere of influence. At that instant, we mathematically transpose the heliocentric velocity vector into a planet-centered velocity vector. It's the exact same velocity, just in different coordinates. We are patching the heliocentric trajectory to the planet-centered trajectory.<br /><br />We now are looking at a hyperbolic trajectory *relative to the planet* - we're going too fast to actually orbit the planet, BUT we are close enough that the planet influences our flight path. We're in a 'hyperbolic orbit'.<br /><br />Conservation of angular momentum *of the planet-probe system* dictates that after we make our closest pass to the planet, our flight path *relative to the planet* is a mirror image of our approach. Thus, when we get back to the sphere of influence distance on our way out, we will have the same velocity magnitude *relative to the planet*. BUT we are going a different direction, because our orbit lasted a finite length of time; the planet has redirected our velocity vector.<br /><br />So here w <div class="Discussion_UserSignature"> </div>
 
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krrr

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Good. What about trying to clarify the Wikipedia article <img src="/images/icons/tongue.gif" />.
 
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henryhallam

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<font color="yellow"> You can not increase your velocity by orbiting the planet you launched from !</font><br /><br />Well, several spacecraft have made earth flybys (sometimes even more than once) to increase their energy BUT this only works after you're already in a heliocentric (solar) orbit. Then you can just treat the earth like any other planet. And depending on where you're going, this may save propellant but most of the time it will end up taking longer than a direct transfer.
 
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henryhallam

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Are you sure about that Dave? A flyby is just a hyperbolic orbit and I always understood that the total energy (kinetic + grav potential) remained the same at any point in the orbit. So when the spacecraft leaves the sphere of influence ("infinite distance") grav potential energy will be zero, i.e. same as when it entered. Therefore kinetic energy must be the same therefore same velocity <b>relative to the flyby planet</b>.<br /><br />Of course this vector is in a different direction to the initial velocity vector.<br /><br />If we let <b>Vi/p</b> be the inital velocity of the spacecraft and <b>Vf/p</b> be the final velocity relative to the planet while <b>Vi/s</b> and <b>Vf/s</b> are the velocities relative to the sun, and <b>Vp/s</b> is the velocity of the planet relative to the sun, then<br /><br /><b>Vi/s</b> = <b>Vi/p</b> + <b>Vp/s</b><br /><br />|<b>Vi/p</b>| = |<b>Vf/p</b>|<br /><br /><b>Vf/s</b> = <b>Vf/p</b> + <b>Vp/s</b><br /><br />Because the direction of <b>Vf/p</b> may be different (and must be, for a useful flyby) to <b>Vi/p</b>, you can consider the extreme case where they are in opposite directions i.e. <b>Vf/p</b> = -<b>Vi/p</b>.<br /><br />Then <b>Vf/s</b> = <b>Vi/s</b> + 2<b>Vi/p</b><br /><br />So you can get a boost of up to twice the "encounter velocity". But the magnitude of your velocity <b>with respect to the planet</b> doesn't change.
 
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spacester

Guest
"the probe begins falling toward the planet. If it doesn't hit atmoshere or the surface, it leaves the planet's sphere of influence at the same velocity *magnitude* as it entered - relative to the planet." <br /><br /><font color="yellow">No, that is not true. The probe will have a velocity vector magnitude change. </font><br /><br />Are you saying that coconuts migrate? <img src="/images/icons/wink.gif" /> <img src="/images/icons/laugh.gif" /> sorry . . . <br /><br />Um, before we argue, are you sure you're reading my statement carefully? "Relative to the planet." If you don't fire the engines, and the sphere of influence distance is the same, the flight path is symmetrical. There is nothing to cause your entering velocity - at the sphere of influence distance - to differ from your departure velocity - at the sphere of influence distance. Conservation of momentum, momentum equals r-vector dot v-vector, r magnitude unchanged, r-v angle identical . . . <br /><br />In actual practice, I believe engines are fired during the maneuver, which of course changes things.<br /><br />I can find some links if needed, but I'll wait for your response. <div class="Discussion_UserSignature"> </div>
 
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spacester

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http://www.go.ednet.ns.ca/~larry/orbits/gravasst/gravasst.html<br /><br />I do have to retract a statement I made that "the planet's orbital velocity has nothing to do with it" - you need to use that velocity to transfom the vector from one set of coordinates to another. <br /><br />I see what you're saying Dave and now I finally understand what is meant by "slingshot". The thing is, mathematically speaking, in the patched conic method, we've already transformed into a planetary coordinate system, so I cannot accept - in that mathematical context - that the planet is moving. The wikipedia article makes more sense now as well.<br /><br />Looking at the planet and probe from a heliocentric point of view, yes the magnitude changes. I've just been focused on the only way I know how to get a quantitative result, rather than just a qualitative description.<br /><br />Thank you for the input, I learned something! <div class="Discussion_UserSignature"> </div>
 
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drwayne

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I remember getting a lot of grief in my orals for working a problem in a non-inertial reference frame.<br /><br />Wayne <div class="Discussion_UserSignature"> <p>"1) Give no quarter; 2) Take no prisoners; 3) Sink everything."  Admiral Jackie Fisher</p> </div>
 
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mcs_seattle

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Adding to the list of things that amaze me about space stuff, is what it takes to plan these orbital flybys years and years in advance of the actual flyby and that just about all of them make it to where they are supposed to go to (the one to Mars that was lost due to metric/english error is the only one I know of that didn't make it).
 
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nacnud

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It's amazing what you can do with a calculator :/ <img src="/images/icons/smile.gif" />
 
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scottb50

Guest
I couldn't tell you where I will be tomorrow at 6:00pm local time, but I could figure out exactly where the Earth or any other body is, or will be at that time. The lure of astrology, pay no attention to the man behind the curtain. <br /><br />It's not magic it's numbers and the math is consistant throughout the Universe. <div class="Discussion_UserSignature"> </div>
 
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scottb50

Guest
No, I mean astrology. The seers could point to the repeating paterns of the stars and convince people that it had a meaning to their lives. They knew when Mars and Venus would both show up in the night sky and they could use this to convince people they knew what they were talking about it proved their point.<br /><br />The fact a star, that is millions of light years away has any influence on people is because they could show you this same star, or group of stars, in the place and time they said it would be there. Just like magic.<br /><br />So then we have the seers and oracles, and such, reaping the benefits of their presceince, which lead directly to the Jim Jones and Pat Robertsons we put up with today.<br /><br />I know the difference in Astronomy and astrology. One is real, the other is mythology. <div class="Discussion_UserSignature"> </div>
 
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l3p3r

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excellent explanation on previous page thanks spacester! <div class="Discussion_UserSignature"> </div>
 
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