understanding 'gravity losses'

[gunsandrockets]

There is something I don't understand about 'gravity losses'.<br /><br />Now when it comes to suborbital flight, gravity losses make perfect sense to me. When a launch vehicle is at a suborbital speed the vehicle will lose altitude unless it devotes some fraction of it's thrust to compensate for the acceleration of gravity. As the vehicle gets closer and closer to orbital speed the fraction of it's thrust needed to fight gravity shrinks, finally shrinking to zero as full orbital speed is achieved. That's easy to understand.<br /><br />What I don't understand are references to 'gravity losses' regarding vehicles already in orbit and which continue to accelerate beyond minimum orbital speed (in particular low-acceleration electric-propulsion vehicles). It makes no sense to me. How can there still be gravity losses to the delta vee of the accelerating spacecraft? It seems to me that after orbit is achieved, all of the spacecraft's thrust would increase the orbital velocity, without any gravity losses. <br /><br />For example -- take two different spacecraft in orbit around Earth which are at the same starting altitude and speed, one spacecraft with high-thrust chemical-propulsion and the other with low-thrust electric-propulsion, but both spacecraft have the same delta vee of propellant (let's say 4 km/s). The chemical-rocket expends all it's propellant in minutes while accelerating over a length of a couple thousand kilometers in orbit. After burnout as the chemical-rocket ascends away from the Earth it would gradually slow until reaching enough distance from the Earth where any change in speed from the Earth's gravity would become negligable. I don't see how the final speed of that chemical-rocket would be any different from the gentler electric-rocket. Just because the electric-rocket accelerates over a greater period of time as it slowly spirals out from Earth, why should that make any difference to the rocket's final speed?<br /><br /><br />Maybe someone here can

 

[webtaz99]

Gravity does not drop off to zero at orbital speed. For instance, the ISS is in approximately 96% of surface gravity. What happens is that the centripital acceleration needed to cause the orbit to curve balances the gravity.<br /><br />Orbital mechanics are only slighty different from what we are used to. Take any object you can pick up and toss it upwards. It slows down, stops momemtarily, then drops back down. Anything that doesn't exceed the Earth's "escape velocity" will also eventually drop back down. <div class="Discussion_UserSignature"> </div>

 

[edkyle98]

Once orbit is reached, there are effectively no gravity losses when the direction of thrust is tangential to the gravity vector (i.e. "sideways"). Traditional chemical rockets take advantage of this by applying thrust in relatively short bursts in the direction of the velocity vector at apogee or perigee (or any time for a circular orbit), when the velocity vector is tangential to gravity. <br /><br />An electric-rocket cannot perform such short "burns", and so will have to thrust during periods when the vehicle is "climbing" against gravity when the velocity vector is not tangent to gravity, suffering some gravity losses in the process.<br /><br /> - Ed Kyle

 

[radarredux]

> <i><font color="yellow"> What happens is that the centripital acceleration needed to cause the orbit to curve balances the gravity.</font>/i><br /><br />I remember an object in orbit was once described to me this way: It is still falling towards Earth, but it just keeps missing the Earth.<br /><br />Think of shooting a cannon ball horizontally say 10 feet off the ground. Gravity eventually pulls the ball to the ground (at the same speed/time as it would if the cannon ball was dropped from 10 feet). This is a basic problem set in a physics course: Given that the ball starts at 10 feet above the ground and is fired at 500 mph, (a) how much time will it take to hit the ground, and (b) how far will it travel? Then repeat the question with a horizontal speed of say 250 mph.<br /><br />The question has simplifying assumptions however: the Earth is flat and the gravity vector is always straight down.<br /><br />But the surface of the Earth is curved, so as the fired cannon ball is flying horizontally, the surface of the Earth is effectivey dropping away from it. Fire that cannon ball fast enough, and the curve of the Earth will drop away at the same rate as the ball is falling; orbital velocity is achieved.<br /><br />So the term "free fall" is actually fairly accurate. Something orbiting Earth is always falling; it just happens to always miss the Earth too.</i>

 

[gunsandrockets]

"Gravity does not drop off to zero at orbital speed."<br /><br />I didn't say that.<br /><br />"For instance, the ISS is in approximately 96% of surface gravity. What happens is that the centripital acceleration needed to cause the orbit to curve balances the gravity." <br /><br />You're not telling me anything that I don't already know.<br />

 

[gunsandrockets]

"So the term "free fall" is actually fairly accurate. Something orbiting Earth is always falling; it just happens to always miss the Earth too. "<br /><br />Free fall terminology and the cannonball analogy has always been my favorite way of understanding orbit.

 

[gunsandrockets]

"Once orbit is reached, there are effectively no gravity losses when the direction of thrust is tangential to the gravity vector (i.e. "sideways")...An electric-rocket cannot perform ... short "burns", and so will have to thrust during periods when the vehicle is "climbing" against gravity when the velocity vector is not tangent to gravity, suffering some gravity losses in the process."<br /><br />I disagree.<br /><br />I think that your example, at best, would only apply in a highly eliptical orbit. In a circular orbit I don't see the effect at all.<br /><br />Let's use the electric-rocket again, this time already in a circular orbit around Earth before the electric-engine even starts. As the electric-rocket thrusts at a tangent to it's orbit it gains velocity, so the effect is a slow spiral outward from Earth. If at any time the thrust cuts out, the rocket would still assume a circular orbit, only an orbit higher than the rocket's original circular orbit. Where is the 'gravity loss' in this example?<br /><br /><br /><br />

 

[drwayne]

Centripetal accleration can actually be more productively viewed as a "perscription" for circular motion. It is description of a particular kind of motion, it is not the result of a force.<br /><br />People tend to mix this concept with the concept of centrifugal force. Centrifugal force is known as a pseudo-force, it is a by-product of doing a problem, or riding along in an accelerating, or 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>

 

[scottb50]

Your right, if you used an engine in a circular orbit the craft will go to a higher orbit because of the higher velocity. The problem with electrics is they provide a little bit of thrust over a long time.<br /><br />Let's use the electric-rocket again, this time already in a circular orbit around Earth before the electric-engine even starts. As the electric-rocket thrusts at a tangent to it's orbit it gains velocity, so the effect is a spiral outward from Earth, to a higher stable orbit. If at any time the thrust cuts out, the rocket would still assume a circular orbit, only an orbit higher than the rocket's original circular orbit. Where is the 'gravity loss' in this example? />><br /><br />There is none. I also agree if the thrust cuts out, the rocket would assume orbit. Circular I don't know, it would depend on when it quit.<br /><br />I think the only orbits electric motors can effectively work in are highly elliptic ones. I'm still working on the effectiveness enroute to Mars. They get you there faster but you have more speed to lose once you get there. That means the propellant and engines for the braking stage have to offset the mass of the electric engines to be efficient. <div class="Discussion_UserSignature"> </div>

 

[gunsandrockets]

"I'm still working on the effectiveness enroute to Mars. They [electric-propulsion] get you there faster but you have more speed to lose once you get there. That means the propellant and engines for the braking stage have to offset the mass of the electric engines to be efficient. "<br /><br />Use aerocapture to brake into orbit around Mars.

 

[brellis]

1. Can a Solar Electric Primary Propulsion engine similar to the one used on ESA's SMART-1 mission to the moon gradually decelerate a craft approaching Mars?<br /><br />2. Can LaGrange Points be used to slow it down? I believe Smart-1 danced around some L-Points to tweak its approach to Lunar orbit. <div class="Discussion_UserSignature"> <p><font size="2" color="#ff0000"><em><strong>I'm a recovering optimist - things could be better.</strong></em></font> </p> </div>

 

[pathfinder_01]

“I think the only orbits electric motors can effectively work in are highly elliptic ones. I'm still working on the effectiveness enroute to Mars. They get you there faster but you have more speed to lose once you get there. That means the propellant and engines for the braking stage have to offset the mass of the electric engines to be efficient.”<br /><br />Well for one you don’t need a braking stage. Just reuse the engine that you used to get there. Maybe dump an empty fuel tank to reduce mass (if that makes sense). Second electric propulsion is a heck of a lot more fuel efficient than chemical which generates quite a bit of mass savings. <br /><br />The real trouble with electric is the time it takes to change velocity. Hours, days, or months. On a long trip to mars, an electrically power craft could actually take about the same amount of time traveling as a chemical one due to the distance. Of course this depends on the type of electric craft(plasma, ion) and the amount of fuel and power available and the trajectory. <br /><br />The flip side is that on shorter trips, say to the moon, an electrically powered craft would take longer to get there. However the electrically powered craft would need less mass in fuel which in turn means you could use a smaller rocket to get it into orbit. Which in turn could save money. This would be a bad way to transport men, but a cheaper way to send supplies. <br />

 

[edkyle98]

""Once orbit is reached, there are effectively no gravity losses when the direction of thrust is tangential to the gravity vector (i.e. "sideways")...An electric-rocket cannot perform ... short "burns", and so will have to thrust during periods when the vehicle is "climbing" against gravity when the velocity vector is not tangent to gravity, suffering some gravity losses in the process."<br /><br />I disagree.<br /><br />I think that your example, at best, would only apply in a highly eliptical orbit. In a circular orbit I don't see the effect at all.<br /><br />Let's use the electric-rocket again, this time already in a circular orbit around Earth before the electric-engine even starts. As the electric-rocket thrusts at a tangent to it's orbit it gains velocity, so the effect is a slow spiral outward from Earth. If at any time the thrust cuts out, the rocket would still assume a circular orbit, only an orbit higher than the rocket's original circular orbit. Where is the 'gravity loss' in this example?"<br /><br />Gravity losses are small for a spiral transfer, but not perfectly zero. The orbit is never perfectly circular. Thrusting is in the dirction of the velocity vector, which will always have a very small bias from tangential. Since this thrusting takes place over a very long period of time (months perhaps), the tiny losses accumulate.<br /><br />On the other hand, an ideal Hohmann transfer has zero gravity losses only if all of the thrusting takes place nearly instantaneously (implusively) at apogee or perigee. That cannot happen with real hardware, so some gravity losses occur with this method too.<br /><br /> - Ed Kyle

 

[gunsandrockets]

"Anytime thrust is applied tangent to a satellite's circular orbit it is no longer in a circular orbit thus the thrust (Being non instantious) is experiencing gravity loss as it fights gravity."<br /><br />How much? What % delta vee lost in this specific case? <br />

 

[scottb50]

I guess I did make that sound pretty lteral. What I meant was their effectiveness is less in a circular orbit than in an elliptical one. <div class="Discussion_UserSignature"> </div>

 

[spacester]

Are you still looking for an answer?<br /><br />I can provide a much better one than the hit-and-miss stuff I see on here. There once was a time when the second google hit on 'gravity losses' was a post of mine here. Since I do not see anything in the current google results that promotes true comprehension, and I believe I can do better, I'm tempted to make the effort.<br /><br />But after years of posting answers to these questions, I have become convinced that no one here gives a damn about true understanding, and even when they do, they're not interested is making any substantive mental effort.<br /><br />I can answer the question, but I am no longer willing to spoon feed those who are not worthy of my efforts. We're talking high school algebra here, people, yet none of you space fans can be bothered to look at an equation. Is there even a single person here with the intellectual capacity and attention span to process a better answer than what you've seen so far?<br /><br />FYI all, I feel completely justified - if not actually comfortable - in taking this new approach. Subsuming my own ego in favor of promoting better understanding of spaceflight math for five years got me exactly nowhere. I am apparently the only person here interested in making a difference somehow. Since everyone else here allows their own ego gratification to control their every word, and no one here seems to be interested in anything greater than themselves, and I do in fact miss posting here, I'm gonna come at y'all with no restraints on my ego. I'm just trying to "fit in", lol.<br /><br />Nice Guys Finish Last, and I've had enough of it. <div class="Discussion_UserSignature"> </div>

 

[scottb50]

It's about time! <br /><br />I think the level most things are discussed here are more basic concepts and understanding the nuts and bolts of how and what happens is more important than getting lost in the numbers. Obviously I want to be able to eventually put all that together if and when it becomes usable, but until there are numbers to put into an equation a basic understanding of what the equation does is enough to sketch out an idea. Once it goes beyond that stage at least you are on the right track.<br /><br />I would like to read a short description of what that equation is for and be able to enter numbers in a program. I thought that's what computers were for anyway. Silly me! <br /><br />I envy you that you have the desire to get everyone to have a facination with the math. What I would like to do is set a destination or orbit, enter the mass, available propellant and push the Engage button.<br /><br />Lets say I'm at Platform 3 and want to get to LMO and enter orbit. I enter the mass of the Vehicle and the computer tells me how many Tugs and how much propellant I will need for the round trip. Every trip is different, just like Apollo used diferent numbers for every flight. I understand how the math works, I just don't want to do the math. If I had to do that manually we wouldn't get much further than NASA has gotten. Would we?<br /><br />This thread is typical. Why are gravity loses different with electric propulsion? They're not. It takes the same number, no matter how you generate it to do the same thing. Sort of like putting Platforms to assemble and launch missions beyond LEO in elliptical orbits rather than circular ones. <div class="Discussion_UserSignature"> </div>

 

[nyarlathotep]

<font color="yellow"><br />The flip side is that on shorter trips, say to the moon, an electrically powered craft would take longer to get there. However the electrically powered craft would need less mass in fuel which in turn means you could use a smaller rocket to get it into orbit. Which in turn could save money. This would be a bad way to transport men, but a cheaper way to send supplies.</font><br /><br />Ion engines also have a design life of months and in some designs, years of firing. You can bring the tug back to LEO and strap on another couple of 'aquarius style vehicle' launched cheap as chips krypton tanks for dozens of reuses.

 

[nyarlathotep]

<font color="yellow">Lets say I'm at Platform 3 and want to get to LMO and enter orbit. I enter the mass of the Vehicle and the computer tells me how many Tugs and how much propellant I will need for the round trip.</font><br /><br />I'm like that too, but I want multiple suppliers with list prices for the tugs and tanks, both in LEO and LLO. I also want a number of suppliers for payload integration at equatorial LEO, keeps launch windows open.

 

[barrykirk]

My assumption had always been that if the vector of the thrust is kept tangent to the vector of the gravity<br />gradient during the entire burn, no matter how long it is,<br />then there are no gravity losses. Is that a correct statement?

 

[spacester]

I concur.<br /><br />But my understanding as an amatuer spacecraft guy is that it is impossible to do so on a practical basis.<br /><br />My understanding of the theory is more complete. <div class="Discussion_UserSignature"> </div>

 

[krrr]

Yes. However, there will be efficiency losses if you're not thrusting in the direction of the spacecraft's velocity vector.<br /><br />That's why continous-burning low-thrust strategies can't be optimal. Because the flight path is a spiral, the velocity vector is ever so slightly non-perpendicular to the gravity vector. Accelerating in flight direction means gravity loss, acceleration tangential to gravity means loss of efficiency.

 

[webtaz99]

"Show me the numbers". It's fine to talk theory, but try running the numbers. If gravity losses are 20%, you have a problem. If they are .0000002%, you have a mthematical curiosity. <div class="Discussion_UserSignature"> </div>

 

[krrr]

<font color="yellow">If gravity losses are 20%, you have a problem. If they are .0000002%, you have a mthematical curiosity.</font><br /><br />It's worse (or better) than that. Let's say we are on a 1000 km spiraling orbit and gaining 1 km per orbit. Then according to Pythagoras gravity loss is .0000000233%. I don't think there's an accumulation over time, it's really just a function of the angle between thrust vector and the direction perpendicular to gravity.<br /><br />Maybe I misunderstand but my conclusion is that gravity losses, while present, are negligible for continuous low-thrust acceleration, starting from a circular orbit.<br />

 

[spacester]

I was correct on that. You just seemed unable to see the problem from a new viewpoint.<br /><br />Why is it important to you that I be wrong? Is it to preserve your sense of elitism? Would it kill you to learn soemthing from someone not in the industry?<br /><br />But rather than discuss it, I'll do the same thing you guys do:<br /><br />Prove it.<br /><br />Quote my incorect words and offer us the explanation as to why I was wrong.<br /><br />Are you still unclear on how velocity can be measured in different frames of reference, or what?<br /><br />Relative to the SUN a close pass encounter most definitely changes the magnitude of the velocity.<br /><br />Relative to the PLANET a close pass encounter most definitely does NOT change the MAGNITUDE of the velocity, just the direction.<br /><br />Make my day and disagree with me on that. Please.<br /><br />I'm working up some paragraphs on the title subject but I'm working 10-12 hour days right now so it's still in the works.<br /><br />Do you have anything more to offer on this subject before I post that? Or are you just going to take cheap shots without substantiation? <div class="Discussion_UserSignature"> </div>