Basic question about spacecraft orbits

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aphh

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<p>To climb up to a higher orbit spacecraft must add delta-V. This pushes the craft away from the body being orbited and increases it's speed.</p><p>So how come a lower orbit is the faster of the two orbits?</p><p>ATV passed ISS on a 30 km lower orbit, then increased delta-V to climb to the same altitude as the ISS.</p><p>Shouldn't the relative speeds be the same for both orbits? Less delta-V, less altitude and vice versa?</p><p>Also, an object continues on a straight line unless it is affected by a force. Changing trajectory means acceleration. Hence a craft on orbit must be accelerating constantly.<br /><br />Does this mean every orbit decays eventually and the orbiting object eventually slams into the body being orbited unless delta-V is added? </p><p>I'm sure this is very basic but I'm only getting started with orbital mechanics and trying to get an idea of it before taking on the pretty serious math.&nbsp;</p>
 
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origin

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>To climb up to a higher orbit spacecraft must add delta-V. This pushes the craft away from the body being orbited and increases it's speed.So how come a lower orbit is the faster of the two orbits?</DIV></p><p>A craft doesn't have as far to travel&nbsp;in the lower orbit.&nbsp; The 'circumfrance' of the lower orbit is smaller&nbsp;than the higher orbit.&nbsp; I do not have my physics book with me, but there is a very easy formula for the relationship between the orbital velocity the distance from the body and the gravity of the body.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Changing trajectory means acceleration. Hence a craft on orbit must be accelerating constantly.Does this mean every orbit decays eventually and the orbiting object eventually slams into the body being orbited unless delta-V is added? I'm sure this is very basic but I'm only getting started with orbital mechanics and trying to get an idea of it before taking on the pretty serious math.&nbsp; <br />Posted by aphh</DIV></p><p>You are right that the orbital craft is constantly acclerating, but that does not mean that you will have the orbit decay.&nbsp; In a low orbit there is enough drag from the very scant atmosphere that the orbit will decay.&nbsp; In a perfect case of a point source of gravity in a perfect vacuum a body orbiting the gavity source will never have the orbit decay.&nbsp; Think about it, the body wants to go in a straight line but the gavity source causes it to continually fall towards the source but if there is no other force acting on the body it's orbit will not change.</p> <div class="Discussion_UserSignature"> </div>
 
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mako71

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<p>If you're in a circular orbit, around some central mass, the lower is the altitude, the larger velocity is needed to stay at that _circular_ orbit.</p> <p>When you change your orbit, you go to an elliptic one; when rising an orbit, you accelerate the speed at the periapsis (lowest point of the orbit, which is exactly the same that the radius of your current circular orbit), and thus the ship will rise higher at the apoapsis - when you reach your target altitude, your ship's velocity is smaller than the velocity needed for a circular orbit at that altitude, so you make a new acceleration ==> your ship will then be in a circular orbit at that altitude.</p> <p>When lowering the orbit, you do the vice versa; first, you use delta-v to _decelerate_ your ship, so that it moves from circular orbit to an elliptic one, and the lowest point (periapsis) of that elliptic orbit is at least the same as your target _circular_ orbit. When the ship has travelled to the lowest point, the speed of ship is much higher than the orbital velocity at that altitude - so, you need to use delta-v to decelerate your ship, so that it doesn't go back to the original height, but instead stays in a circular object at the altitude you decided.</p> <p>HTH</p><p>P.S. delta-v is the _difference_ in velocities, not necessarily an addition to a velocity (in principle, in vacuum, you'll need the same energy to decelerate that you used to accelerate). </p><p>P.P.S. Every time you transfer your ship from one circular orbit to another, you'll need two burns; the first takes you to an elliptic trajectory, and the second transfers you from that ellicptic trajectory to a circular one. If you make just one burn, your ship will stay in that elliptic trajectory, meaning that it will reach the same altitudes (both lowest and highest) periodically, to infinite. </p> <div class="Discussion_UserSignature"> <p> </p><p>________________ </p><p>reaaliaika.net </p> </div>
 
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mako71

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Less delta-V, less altitude and vice versa?</DIV></p><p>As said, less relative velocity to the central mass is needed to stay at circular orbits, when altitude (radius of the orbit) raises; delta-v is the difference of speeds.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Also, an object continues on a straight line unless it is affected by a force.</DIV></p><p>Every object is subjected to gravitational forces, because of large masses in the space ==> every path will be "curved" (except if the ship is massless).&nbsp;</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Changing trajectory means acceleration.</DIV></p><p>Yes, in the direction or another - you either add speed relative to the speed at which you are circulating the central mass, or decelerate it; both are changes to delta-v, and both are "accelerations".</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Hence a craft on orbit must be accelerating constantly.</DIV></p><p>No, you just need a short burn, to change your relative speed, to move your ship from circular to elliptic orbit and vice versa. Of course, the time of the burn is depended on the specific impulse of your motors; you will (in average) need longer burns with electro-magnetic drives than chemical drives.&nbsp;</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Does this mean every orbit decays eventually and the orbiting object eventually slams into the body being orbited unless delta-V is added?</DIV></p><p>No. Whenever you have moved your ship to one trajectory, either elliptic or circular, it will stay there.&nbsp;</p> <div class="Discussion_UserSignature"> <p> </p><p>________________ </p><p>reaaliaika.net </p> </div>
 
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MeteorWayne

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<p>Just as an interesting side note, Endeavour is orbiting in a cnear circukar orbit at ~ 17,500 mph.</p><p>The deorbit burn reduces that speed by only 209 mph, but that is enough to lower the perigee of it's orbit enough so that it enters the atmosphere.</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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mako71

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<p>MeteorWayne, :D</p><p>Another possibly interesting thing... The altitude of ISS is around 400 km (350 - 460 km). If Earth would be a size of an orange (around 10 cm), the ISS would be circulating around 3 mm above it :) </p> <div class="Discussion_UserSignature"> <p> </p><p>________________ </p><p>reaaliaika.net </p> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>MeteorWayne, :DAnother possibly interesting thing... The altitude of ISS is around 400 km (350 - 460 km). If Earth would be a size of an orange (around 10 cm), the ISS would be circulating around 3 mm above it :) <br />Posted by mako71</DIV></p><p>I use something similar in my meteor talks. I have a softball sized globe. At that scale, the height where meteors start to be seen (100 km/60 mi....how high you have to be to have flown in "space") is thinner than a credit card.<br /></p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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aphh

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<p>This is amazingly interesting subject. Next I need to brush up my math.<br /><br />I also want to know more about the reaction control systems and how they change the attitude of a orbiting craft, basically you need a reaction and then opposite reaction to turn the craft in a controlled manner. Craft's orientation in relation to sun might need to change over time to maintain surface temperatures. </p><p>Then I want to know about star trackers and how they work to provide orientation when you're lost in space. </p><p>Electricity production and communications with earth... basically one could learn what is needed to fly and operate a spacecraft <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /> </p><br />
 
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drwayne

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<p>&nbsp;A couple of minor points:</p><p>An&nbsp;object in orbit is constantly accelerating.&nbsp; In the case of a circular orbit, that acceleration is called the centripetal (or center seeking)&nbsp;acceleration "a"&nbsp;and takes the form</p><p>a = v^2/R</p><p>where v is the spacecraft velocity, and R is the oribital radius.&nbsp; This can be thought of as the "prescription" for a circular motion - this acceleration keeps the velocity vector turning in such a way as to allow circular motion.&nbsp; Remember, acceleration and velocity are vector quantities.</p><p>Also don't lose track of atmospheric drag.&nbsp; Objects in orbit experience it, and their orbits will eventually decay unless they are high enough - or, as in the case of certain satellites, they are in a weird resonance orbit.</p><p>Wayne</p> <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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aphh

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<p>* An&nbsp;object in orbit is constantly accelerating.&nbsp; In the case of a circular orbit, that<br />* acceleration is called the centripetal (or center seeking)&nbsp;acceleration "a"&nbsp;and<br />* takes the form</p><p>That's what I was thinking, because the trajectory of a orbiting object is changing constantly in relation to it's motion vector (the object "would want to continue on a straight line" but gravity does not allow it forcing change in trajectory thus creating perpetual acceleration). <br /><br />Changing trajectory (vs. motion vector) means acceleration, hence constant acceleration that you so well explained as the centripetal acceleration.</p><p>But this also made me think about energy loss, because any acceleration in my mind requires energy.</p><p>Where does the energy come for centripetal acceleration? </p>
 
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drwayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>* An&nbsp;object in orbit is constantly accelerating.&nbsp; In the case of a circular orbit, that* acceleration is called the centripetal (or center seeking)&nbsp;acceleration "a"&nbsp;and* takes the formThat's what I was thinking, because the trajectory of a orbiting object is changing constantly in relation to it's motion vector (the object "would want to continue on a straight line" but gravity does not allow it forcing change in trajectory thus creating perpetual acceleration). Changing trajectory (vs. motion vector) means acceleration, hence constant acceleration that you so well explained as the centripetal acceleration.But this also made me think about energy loss, because any acceleration in my mind requires energy.Where does the energy come for centripetal acceleration? <br />Posted by aphh</DIV></p><p>Actually, you might find it clearer if you think of things in the force/acceleration regime, rather than the energy regime.</p><p>The reason I say that, is that for a circular orbit, with drag neglected, energy is a constant.&nbsp; The magnitude of the velocity vector is constant.&nbsp; Now, you might say, "Wayne, if there is an acceleration in the system, the magnitude of the velocity has to change, doesn't it?"</p><p>What happens is this.&nbsp; For the satellite, a central force is acting, that is, the gravitational attraction between the object and the Earth.&nbsp; The force is (for a circular orbit), perpendicular to the velocity vector of the satellite, so it does not change the magnitude of the velocity vector, but it does change its direction.&nbsp; In this case, it changes it in such a way that the trajectory is curved into a circular trajectory.</p><p>The key here to remember is this - an acceleration leads to time rate of change in the velocity vector.&nbsp; That change can be in direction, or magnitude, or both.</p><p>&nbsp;Wayne<br /></p> <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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billslugg

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<p>Way back when, I was pursuing an EE degree. WE had to get 131 credits. Everybody else only needed 121. WE got one 3 credit free elective. Everybody else got like 30 credits of free electives. All the rest of our electives had to come off a list. It was a finely tuned sport to pick the easiest ones. The course in "Celestial Mechanics" was considered easy. Brutal, Brutal. </p><p>The only things I remember are the "two burns" thing and the thing about always sweeping out equal areas.</p> <div class="Discussion_UserSignature"> <p> </p><p> </p> </div>
 
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drwayne

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<p>Way back in my undergraduate days, I had the fortune to take several courses at Vassar from Dr. Henry Albers.&nbsp; The hardest was a course in which I had my first experience with numeric solution of differential equations (What the heck is Runge-Kutta I found myself asking), in this case applied to the motion of planets.&nbsp; We had to, during the course of the course, write a program to calculate the orbital position of planets, and then figure their position from Earth, and show them graphically.</p><p>I spent many hours at the Tektronix on that one, and learned a lot, about programming and math.&nbsp; :)</p><p>&nbsp;</p> <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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billslugg

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<p>Your observation parallels my experience. I made a city with road intersections. I did some kind of matrix algebra to find the shortest route. It is sort of foggy. It was probably something that occurred on a weekend. I distinctly remember that I got a "B" after presenting my code to the professor in his office.</p><p>Bill Slugg&nbsp;</p> <div class="Discussion_UserSignature"> <p> </p><p> </p> </div>
 
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spacester

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>To climb up to a higher orbit spacecraft must add delta-V. This pushes the craft away from the body being orbited and increases it's speed.So how come a lower orbit is the faster of the two orbits?ATV passed ISS on a 30 km lower orbit, then increased delta-V to climb to the same altitude as the ISS.Shouldn't the relative speeds be the same for both orbits? Less delta-V, less altitude and vice versa?Also, an object continues on a straight line unless it is affected by a force. Changing trajectory means acceleration. Hence a craft on orbit must be accelerating constantly.Does this mean every orbit decays eventually and the orbiting object eventually slams into the body being orbited unless delta-V is added? I'm sure this is very basic but I'm only getting started with orbital mechanics and trying to get an idea of it before taking on the pretty serious math.&nbsp; <br /> Posted by aphh</DIV></p><p>Hi aphh,</p><p>You've been advised to look to the force/acceleration regime for comprehension, and it is good advice. But, contrarian that I am, I am convinced that if you really want to get an intuitive grasp of what's going on, you must look to the energy regime.</p><p>I don't know your physics background, but I'll assume you know about Potential Energy and Kinetic Energy. PE comes from climbing up a gravity well, and KE is the energy of motion.</p><p>Clearly, an object needs to gain a lot of KE to achieve orbit. Way more than the piddly PE needed to get to that altitude. It needs to get moving very very fast so that its free-fall misses the planet. </p><p>Once an object achieves orbit, it has established a stable (ignoring air friction etc) means of storing energy. Not all storage of energy is done by climbing a gravity well. By virtue of being in orbit, the KE is also stored energy. This is referred to as 'Bound Energy'.</p><p>Few people seem willing to embrace the concept of 'Bound Energy' but I'm telling y'all that if you really want this orbital mechanics thing to make sense you should wrap your head around it.</p><p>There is a very famous equation for this stuff. Pretty much all the other equations derive from it. I say famous, but really that's historically speaking, because the modern space enthusiast seems to not want to embrace its importance. (Just an observation on my part.)&nbsp;</p><p>I refer to the 'Vis-Viva equation'. Google it.&nbsp;</p><p>Getting back to your original question, and our orbiting object, Vis-Viva shows you what happens to your energy as you move around up there, whether within a given orbit or from orbit to orbit. It isn't obvious, but in a certain form the equation will show you this upon inspection. </p><p>When raising an orbit, you are modifying the bound energy. This is way different than achieving the bound energy state in the first place, where you had to add a lot more KE than PE to your previous condition. </p><p>Here is the key factoid I want y'all to remember:</p><p>When raising an orbit (circular to circular), you add energy. The allocation of this energy between PE and KE is not what you might expect: if you change the value of bound energy by 1 unit, 2 units of energy goes to PE and 1 unit is deducted from your KE. This goes against intuition, and mathematically speaking is why a slower velocity at a higher orbit is counter-intuitive. You have to invest a whole bunch of energy to shift that bound energy to a higher level of energy storage, and the level of energy stored is given by the radius of the orbit, not the velocity of the craft.</p><p>Likewise, if you lower an orbit by 1 energy unit, your PE is reduced by two units and your KE is increased by one unit.&nbsp;</p><p>There is an explanation for this strange accounting of energy. At the risk of straying into confusing territory, it makes more sense if you look at it from the Universe's standpoint. From deep space, Earth is a gravity well. What we call the Potential Energy invested in climbing the well is to the universe energy recovered from having fallen into the well. But enough for now, I won't plumb those depths here. ;) </p> <div class="Discussion_UserSignature"> </div>
 
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drwayne

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<p>I think we were answering different questions.&nbsp; I was referring to:</p><p>"Where does the energy come for centripetal acceleration?"</p><p>and some other things from a later post that suggested that the poster was concerned about the notion of an acceleration changing a velocity vector&nbsp;without changing its magnitude.</p><p>Clearly, if we want to discuss changing orbits, then dealing with the problem from an energy standpoint as your did&nbsp;is clearer (to me).&nbsp; That should of course be caveated with the fact that I sometimes use non-inertial reference frames that are not always popular on oral exams.&nbsp; ;)</p><p>Wayne</p> <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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spacester

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think we were answering different questions.&nbsp; I was referring to:"Where does the energy come for centripetal acceleration?"and some other things from a later post that suggested that the poster was concerned about the notion of an acceleration changing a velocity vector&nbsp;without changing its magnitude.Clearly, if we want to discuss changing orbits, then dealing with the problem from an energy standpoint as your did&nbsp;is clearer (to me).&nbsp; That should of course be caveated with the fact that I sometimes use non-inertial reference frames that are not always popular on oral exams.&nbsp; ;)Wayne <br /> Posted by drwayne</DIV></p><p>You're welcome. Yes, I'm glad to make a contribution and it's always nice to know that my efforts will be accepted in the spirit in which they are intended. :rolleyes:</p><p>Next time, I'll make sure to narrowly confine my remarks to pre-approved areas that do not in any way attempt to out-do a vaunted moderator. Is there a section in this new and improved community where I can find out which things I am allowed to post on without feeling like I'm being drawn into a competitive pissing match?</p><p>Either that or I guess I won't effing bother anymore. </p> <div class="Discussion_UserSignature"> </div>
 
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spacester

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think we were answering different questions.&nbsp; I was referring to:"Where does the energy come for centripetal acceleration?"and some other things from a later post that suggested that the poster was concerned about the notion of an acceleration changing a velocity vector&nbsp;without changing its magnitude.Clearly, if we want to discuss changing orbits, then dealing with the problem from an energy standpoint as your did&nbsp;is clearer (to me).&nbsp; That should of course be caveated with the fact that I sometimes use non-inertial reference frames that are not always popular on oral exams.&nbsp; ;)Wayne <br /> Posted by drwayne</DIV></p><p>Tell me, Mr Physics Teacher, are you past the stage of your life where you learn new things?</p><p>Did you know the thing about 2 units of PE for every -1 units of KE?&nbsp; Most don't, and I've read you for years here, and you've never mentioned it.</p><p>There's some insight there if you want to access it. Or you can continue to align yourself with those here who seem determined to drive off every last long-term contributor except the chosen few such as yourself.&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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drwayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>You're welcome. Yes, I'm glad to make a contribution and it's always nice to know that my efforts will be accepted in the spirit in which they are intended. :rolleyes:Next time, I'll make sure to narrowly confine my remarks to pre-approved areas that do not in any way attempt to out-do a vaunted moderator. Is there a section in this new and improved community where I can find out which things I am allowed to post on without feeling like I'm being drawn into a competitive pissing match?Either that or I guess I won't effing bother anymore. <br />Posted by Spacester</DIV></p><p>Pardon me?&nbsp;&nbsp;&nbsp;</p><p>You replied to an earlier post, but referred to my statement of mine&nbsp;which referred to a later post.&nbsp; (Which was probably not clear, as I did not quote the question it was intended to answer)&nbsp; I thought that might be confusing for the original poster, so I clarified the regime that I was suggesting the solution for.&nbsp; </p><p>I did *not* say your solution was wrong, or inferior in the process.&nbsp; In fact, I said it was *clearer* for the regime you were looking at.&nbsp; (And I made a self-depricating remark about *my* methods for solving some of these problems - which were not very popular with my committee during orals)</p><p>How exactly do you think that a remark in which I tell the poster that your approach is clear, and made fun of the way that I solve problems is being competitive with you?&nbsp; </p><p>Wayne</p> <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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spacester

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Pardon me?&nbsp;&nbsp;&nbsp;You replied to an earlier post, but referred to my statement of mine&nbsp;which referred to a later post.&nbsp; (Which was probably not clear, as I did not quote the question it was intended to answer)&nbsp; I thought that might be confusing for the original poster, so I clarified the regime that I was suggesting the solution for.&nbsp; I did *not* say your solution was wrong, or inferior in the process.&nbsp; In fact, I said it was *clearer* for the regime you were looking at.&nbsp; (And I made a self-depricating remark about *my* methods for solving some of these problems - which were not very popular with my committee during orals)How exactly do you think that a remark in which I tell the poster that your approach is clear, and made fun of the way that I solve problems is being competitive with you?&nbsp; Wayne <br /> Posted by drwayne</DIV><br /></p><p>Criminy, dockeen, er, DrWayne, don't you see how you turned it into a competition? Maybe my post wasn't all about you.</p><p>Not every post has to be about 'I said, you said', does it? Even you have fallen into that trap?</p><p>Scroll up the thread. What was the original question? That's the question I was answering. I gave your approach a nod and proceeded to show mine. </p><p>I spent some energy on it. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> Those words do not come without some effort. It's called 'making a contribution to the community'. How the eff do you Moderators get off positioning yourselves as so high and mighty that you can't even be bothered to acknowledge that a contribution has positive value, that it is appreciated? </p><p>Even you DrWayne - my last, best hope for this place to pull its head out - go straight to competition mode. My exposition was not about you and your answer, OK? They were about the subject at hand fer crying out loud. Maybe I don't actually care what your answer was; maybe I do, but the thread does not have to revolve around what you said, does it? Can you not share the stage even a little bit?</p><p>If you guys want these forums to grow, you'll climb off your high horses and show some appreciation for contributions to the community and not turn everything into a competition and/or threat to your eminence.</p><p>You could have said 'That's interesting, spacester, thanks for sharing that. It is indeed interesting to compare and contrast different ways of getting a grasp of this counter-intuitive subject.' See, that puts us on an equal footing. How many years of posting on space math does it take before one of you ivory tower overlords can deign to show some respect for this long-term contributor? Even if you feel the need to codemn my attitude, you could try showing some respect when I try to return to my old habits of being Mr. Positive. It sure wouldn't hurt. </p><p>The incompetence here has just about killed this community (the My Lai of the internet - 'we had to destroy the village in order to save it') - are you trying to finish the job or what?</p> <div class="Discussion_UserSignature"> </div>
 
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aphh

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<p>Thanks for all of the info from all who have participated,</p><p>I've checked Braeunig's excellent course on OM: http://www.braeunig.us/space/orbmech.htm</p><p>Even with my limited math and physics I should be able to start calculating basic orbits and trajectories pretty soon. Atleast I think so. </p><p>However, I'm past the point where I was in school, where knowing the formula to calculate something to finish the course is no longer enough (I wish I had reached that point already back in school).</p><p>You've all provided valuable insight into how the system actually works, "what makes it tick", if you will. </p><p>In fact, there's so much information that digesting all of it is going to take a while. Thanks again!&nbsp;</p>
 
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aphh

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<p>Today I read an interesting concept or notion about the effects of gravity. The spacecraft on orbit actually flies on a straight line.</p><p>It's because the space around a large body is bent is what makes the craft go around the body.</p><p>This made me think about what actually is pulled or bent by gravity. Is radiation or light affected by gravity? Surely you need mass, but do charged particles or photons count? I read about the gravitational lens used to magnify distant objects, so it appears photons are affected.&nbsp; </p>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Today I read an interesting concept or notion about the effects of gravity. The spacecraft on orbit actually flies on a straight line.It's because the space around a large body is bent is what makes the craft go around the body.This made me think about what actually is pulled or bent by gravity. Is radiation or light affected by gravity? Surely you need mass, but do charged particles or photons count? I read about the gravitational lens used to magnify distant objects, so it appears photons are affected.&nbsp; <br /> Posted by aphh</DIV></p><p>Yes, they do.&nbsp; Because gravity is not a 'force' in the classical sense, but rather a geometry of spacetime.&nbsp; Objects/particles (even photons) follow a geodesic when within the gravity well of a sufficiently massive object.</p><p>An example of photons travelling along a geodesic is gravitational lensing.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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aphh

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yes, they do.&nbsp; Because gravity is not a 'force' in the classical sense, but rather a geometry of spacetime.&nbsp; Objects/particles (even photons) follow a geodesic when within the gravity well of a sufficiently massive object.An example of photons travelling along a geodesic is gravitational lensing.&nbsp; <br /> Posted by derekmcd</DIV></p><p>This is starting to make sense to me atleast on some basic level.</p><p>If the space is bent due to gravitational wells everywhere, does this mean that the stars we see aren't actually quite there, but in reality reside someplace else? The path of the light being curved by gravities along the way? </p>
 
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