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Elliptical orbit question
Reposting with corrections ("centrifugal force" is what was meant):
=================================== It might help to think of two-body orbit dynamics in a way that most people don't think of it: A satellite going around a planet acts like a mass hanging on the end of a spring. Basic diagram: https://i.ytimg.com/vi/lZPtFDXYQRU/maxresdefault.jpg Gravity pulls down on the mass, but the mass can move down and up in an oscillation. The spring is pulling up on the mass, and this is how the centrifugal force works, pulling the satellite up and away from the Earth. The centrifugal force is a manifestation of the inertial property of the satellite's mass. When gravity and the centrifugal force are in equilibrium, the mass remains at a constant altitude from the Earth. Circular orbits are static in this respect, in a reference frame that rotates at the same rate as the satellite is orbiting. And this is why you can bolt your DirecTV dish pointing to one point in the sky and the geometry does not change. The satellite is as still as the mass hanging on the end of the spring. Elliptical orbits are not still. They have a continual tradeoff of Potential Energy & Kinetic Energy. This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. Hopefully this makes it clear exactly what is causing the altitude changes with the satellite. When the mass moves down past what would be the static equilibrium point, it has plenty of kinetic energy. And that is getting packed into "the spring" of inertia. It bottoms out at perigee when the spring force finally overcomes the motion from the gravitational force, and the direction reverses. So yes, it is the inertia of the velocity that has built up during this downward part of the cycle that causes the altitude reversal. You can think of it as a spring that has been pulling on this satellite. You stretch the spring all the way down to perigee, and then its force will finally reverse the direction that the force of gravity was pulling in. But from every moment that the mass was below the point of equilibrium - the altitude of the circular orbit - that spring force was greater than the force of gravity. The satellite's velocity toward the Earth was decelerating the entire time since it had passed that equilibrium point. That equates to a flight path angle, gamma, with respect to the horizon, that was continually shallowing ever since passing that circular altitude equilibrium point. And upon passing perigee, the satellite's flight path angle goes through zero and turns from negative to positive and it starts climbing again. Just like the mass on the spring. ~ CT =================================== (Originally posted at https://groups.google.com/d/msg/sci....k/HxefrAHiBgAJ) |
#2
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Elliptical orbit question
On Sept/9/2018 at 04:25, Stuf4 wrote :
Reposting with corrections ("centrifugal force" is what was meant): =================================== It might help to think of two-body orbit dynamics in a way that most people don't think of it: A satellite going around a planet acts like a mass hanging on the end of a spring. Basic diagram: https://i.ytimg.com/vi/lZPtFDXYQRU/maxresdefault.jpg Gravity pulls down on the mass, but the mass can move down and up in an oscillation. The spring is pulling up on the mass, and this is how the centrifugal force works, pulling the satellite up and away from the Earth. The centrifugal force is a manifestation of the inertial property of the satellite's mass. When gravity and the centrifugal force are in equilibrium, the mass remains at a constant altitude from the Earth. Circular orbits are static in this respect, in a reference frame that rotates at the same rate as the satellite is orbiting. And this is why you can bolt your DirecTV dish pointing to one point in the sky and the geometry does not change. The satellite is as still as the mass hanging on the end of the spring. Elliptical orbits are not still. They have a continual tradeoff of Potential Energy & Kinetic Energy. This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. Hopefully this makes it clear exactly what is causing the altitude changes with the satellite. When the mass moves down past what would be the static equilibrium point, it has plenty of kinetic energy. And that is getting packed into "the spring" of inertia. It bottoms out at perigee when the spring force finally overcomes the motion from the gravitational force, and the direction reverses. So yes, it is the inertia of the velocity that has built up during this downward part of the cycle that causes the altitude reversal. You can think of it as a spring that has been pulling on this satellite. You stretch the spring all the way down to perigee, and then its force will finally reverse the direction that the force of gravity was pulling in. But from every moment that the mass was below the point of equilibrium - the altitude of the circular orbit - that spring force was greater than the force of gravity. The satellite's velocity toward the Earth was decelerating the entire time since it had passed that equilibrium point. It isn't quite at the moment where the mass passes the altitude of the circular orbit that the satellite's velocity towards Earth starts to decelerates. It starts decelerating when the centrifugal force becomes stronger than the force of gravity. When it crosses the altitude of the circular orbit corresponding to its energy level, it has the same potential energy as a satellite in the circular orbit since it is at the same height. It also has the same speed as that satellite since its total energy (potential energy + kinetic energy) is the same. But that speed isn't in the right direction and therefore doesn't give as much centrifugal force and the satellite's vertical speed is therefore still increasing when it is on the way down or decreasing when it is on the way up. Else than that minor detail, your explanation is in my opinion correct. Alain Fournier |
#3
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Elliptical orbit question
On 18-09-09 20:40 , JF Mezei wrote:
On 2018-09-09 04:25, Stuf4 wrote: This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. If I am 10km behind the ISS in circular orbit, and I turn on the impulse engines to try to catch up to ISS, my increased speed will also result in my gaining altitude becase I am going faster than speed needed to remain at that altitude. Right ? Yes, eventually; your altitude will peak after a half-orbit, and then start to fall again. However, if you have only 10 km to go, you may reach the ISS, and brake to match its speed, before the gain in altitude is very noticeable. When a satellite dropping from 10,000 to 400 gets to 400, isn't it correct to state that its speed is WAY higher than what is needed to remain at 400km altitude? And like the paragraph above, with it going faster than needed, it starts to gain altitude again. Yes indeed, assuming that 400 km is the perigee, where the satellite's velocity has no vertical (altitude component). What I don't understand is that the point where the satellite starts to go faster than needed for that altitude happens before perigee. How come it continues to drop even if it is going faster than needed to remain in that orbital altitude? Because the satellite's velocity has a downward component -- the satellite is moving in a direction that decreases altitude. The satellite starts to gain altitude only when the satellite's velocity vector has turned enough (relative to the local vertical, which is also turning as the satellite orbits) to bring the downward component to zero, and then to a positive value, in other words, when the satellite passes its perigee. So what is magical about perigee that causes the satellite who is already going way faster than necessary to finally stop losing altitude/accelerating and starts to behave normally for a satellite that is going faster than needed at that altitude? (gain altitude, lose speed) Perigee is the point where the satellite's vector changes from pointing down to pointing up. -- Niklas Holsti Tidorum Ltd niklas holsti tidorum fi . @ . |
#4
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Elliptical orbit question
JF Mezei wrote on Sun, 9 Sep 2018
13:40:09 -0400: On 2018-09-09 04:25, Stuf4 wrote: This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. If I am 10km behind the ISS in circular orbit, and I turn on the impulse engines to try to catch up to ISS, my increased speed will also result in my gaining altitude becase I am going faster than speed needed to remain at that altitude. Right ? Correct. If you do a single burn, you will have perigee at the ISS orbit. When a satellite dropping from 10,000 to 400 gets to 400, isn't it correct to state that its speed is WAY higher than what is needed to remain at 400km altitude? And like the paragraph above, with it going faster than needed, it starts to gain altitude again. True. What I don't understand is that the point where the satellite starts to go faster than needed for that altitude happens before perigee. How come it continues to drop even if it is going faster than needed to remain in that orbital altitude? Because it is not going faster IN THE RIGHT DIRECTION. So it continues to drop and gain speed until its velocity IN THE RIGHT DIRECTION is too high, at which point it starts going back up and slowing down. So what is magical about perigee that causes the satellite who is already going way faster than necessary to finally stop losing altitude/accelerating and starts to behave normally for a satellite that is going faster than needed at that altitude? (gain altitude, lose speed) Resolve the velocity into two components, one tangential to a circular orbit and one normal to that. When your tangential velocity exceeds orbital speed you start going back up. -- "If it's the fool who likes to rush in. And if it's the angel who never does try. And if it's me who will lose or win Then I'll make my best guess and I won't care why. Come on and get me, you twist of fate. I'm standing right here, Mr Destiny. If you want to talk, well then I'll relate. If you don't, so what? 'Cuz you don't scare me. -- "Gunfighter", Blues Traveler |
#5
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Elliptical orbit question
From Alain Fournier:
On Sept/9/2018 at 04:25, Stuf4 wrote : Reposting with corrections ("centrifugal force" is what was meant): =================================== It might help to think of two-body orbit dynamics in a way that most people don't think of it: A satellite going around a planet acts like a mass hanging on the end of a spring. Basic diagram: https://i.ytimg.com/vi/lZPtFDXYQRU/maxresdefault.jpg Gravity pulls down on the mass, but the mass can move down and up in an oscillation. The spring is pulling up on the mass, and this is how the centrifugal force works, pulling the satellite up and away from the Earth. The centrifugal force is a manifestation of the inertial property of the satellite's mass. When gravity and the centrifugal force are in equilibrium, the mass remains at a constant altitude from the Earth. Circular orbits are static in this respect, in a reference frame that rotates at the same rate as the satellite is orbiting. And this is why you can bolt your DirecTV dish pointing to one point in the sky and the geometry does not change. The satellite is as still as the mass hanging on the end of the spring. Elliptical orbits are not still. They have a continual tradeoff of Potential Energy & Kinetic Energy. This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. Hopefully this makes it clear exactly what is causing the altitude changes with the satellite. When the mass moves down past what would be the static equilibrium point, it has plenty of kinetic energy. And that is getting packed into "the spring" of inertia. It bottoms out at perigee when the spring force finally overcomes the motion from the gravitational force, and the direction reverses. So yes, it is the inertia of the velocity that has built up during this downward part of the cycle that causes the altitude reversal. You can think of it as a spring that has been pulling on this satellite. You stretch the spring all the way down to perigee, and then its force will finally reverse the direction that the force of gravity was pulling in. But from every moment that the mass was below the point of equilibrium - the altitude of the circular orbit - that spring force was greater than the force of gravity. The satellite's velocity toward the Earth was decelerating the entire time since it had passed that equilibrium point. It isn't quite at the moment where the mass passes the altitude of the circular orbit that the satellite's velocity towards Earth starts to decelerates. It starts decelerating when the centrifugal force becomes stronger than the force of gravity. When it crosses the altitude of the circular orbit corresponding to its energy level, it has the same potential energy as a satellite in the circular orbit since it is at the same height. It also has the same speed as that satellite since its total energy (potential energy + kinetic energy) is the same. But that speed isn't in the right direction and therefore doesn't give as much centrifugal force and the satellite's vertical speed is therefore still increasing when it is on the way down or decreasing when it is on the way up. Else than that minor detail, your explanation is in my opinion correct. Your reasoning looks sound to me. Thank you for highlighting my apparent error. I had never seen anyone explain a two-body orbit in terms of a spring-mass system. It's clear that I needed to put more thought into my reply before posting. Maybe some day someone will write a paper on this and nail it down. Or maybe such a paper was published long ago and I'm just not aware of it. I googled around and this was the closest I could find: ========================= Reactive centrifugal force https://www.revolvy.com/page/Reactive-centrifugal-force ---- Gravitational two-body case In a two-body rotation, such as a planet and moon rotating about their common center of mass or barycentre, the forces on both bodies are centripetal. In that case, the reaction to the centripetal force of the planet on the moon is the centripetal force of the moon on the planet.[6] ---- https://d1k5w7mbrh6vq5.cloudfront.ne...67f1ccca57.PNG ========================= The title of that page focuses on centrifugal force, but then describes the orbit case in terms of centripetal force. ~ CT |
#6
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Elliptical orbit question
On Sep/9/2018 at 17:33, Fred J. McCall wrote :
JF Mezei wrote on Sun, 9 Sep 2018 13:40:09 -0400: On 2018-09-09 04:25, Stuf4 wrote: This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring. If I am 10km behind the ISS in circular orbit, and I turn on the impulse engines to try to catch up to ISS, my increased speed will also result in my gaining altitude becase I am going faster than speed needed to remain at that altitude. Right ? Correct. If you do a single burn, you will have perigee at the ISS orbit. When a satellite dropping from 10,000 to 400 gets to 400, isn't it correct to state that its speed is WAY higher than what is needed to remain at 400km altitude? And like the paragraph above, with it going faster than needed, it starts to gain altitude again. True. What I don't understand is that the point where the satellite starts to go faster than needed for that altitude happens before perigee. How come it continues to drop even if it is going faster than needed to remain in that orbital altitude? Because it is not going faster IN THE RIGHT DIRECTION. So it continues to drop and gain speed until its velocity IN THE RIGHT DIRECTION is too high, at which point it starts going back up and slowing down. So what is magical about perigee that causes the satellite who is already going way faster than necessary to finally stop losing altitude/accelerating and starts to behave normally for a satellite that is going faster than needed at that altitude? (gain altitude, lose speed) Resolve the velocity into two components, one tangential to a circular orbit and one normal to that. When your tangential velocity exceeds orbital speed you start going back up. Not exactly, or poorly formulated. When your tangential velocity exceeds circular orbital speed you start accelerating upwards in the normal component of your velocity vector. But because at that point you still have some downward speed in that vector, you keep on going down until your upward acceleration cancels off your downward motion. At which point you are at perigee and start going up. But your tangential velocity exceeds the circular orbital speed well before perigee. Alain Fournier |
#7
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Elliptical orbit question
JF Mezei wrote on Sun, 9 Sep 2018
23:06:49 -0400: On 2018-09-09 19:16, Alain Fournier wrote: component of your velocity vector. But because at that point you still have some downward speed in that vector, you keep on going down until your upward acceleration cancels off your downward motion. Thansk for this addition. And, as I (poorly) stated, it's when your tangential velocity exceeds the orbital velocity at your altitude that you start getting a net acceleration upward, which is what turns that velocity vector. -- "May God have mercy upon my enemies; they will need it." -- General George S Patton, Jr. |
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Elliptical orbit question
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#10
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Elliptical orbit question
I wrote:
From Jeff Findley: In article , says... What I don't understand is that the point where the satellite starts to go faster than needed for that altitude happens before perigee. How come it continues to drop even if it is going faster than needed to remain in that orbital altitude? Because it is not going faster IN THE RIGHT DIRECTION. So it continues to drop and gain speed until its velocity IN THE RIGHT DIRECTION is too high, at which point it starts going back up and slowing down. So what is magical about perigee that causes the satellite who is already going way faster than necessary to finally stop losing altitude/accelerating and starts to behave normally for a satellite that is going faster than needed at that altitude? (gain altitude, lose speed) Resolve the velocity into two components, one tangential to a circular orbit and one normal to that. When your tangential velocity exceeds orbital speed you start going back up. This. You have to use vector math to analyze orbital mechanics, not scalar math. For a two body problem, the motion is at least planar, which reduces the complexity to a 2D vector math problem. As was posted yesterday to this thread, it was offered that 2-body orbit dynamics can be approximated by using a 1D spring-mass model. So that says that vector math is *not necessary* in order to grasp the basics of what is happening in circular and elliptical orbits. And for the circular orbit case, the motion further reduces to Zero- dimensions (0D). The satellite (or moon, planet, star, what have you) just sits there absolutely still (in a rotating coordinate frame of reference). My words above are yet again in error, for at least the third time here on this thread. It was foolish of me to assert that just because there is no motion in that circular orbit case that it is reducible to zero dimension. There is still the orbit altitude. The distance is static, but it is still a distance. Ok, I have clearly been jumping the gun repeatedly here in this thread and I will need to be a lot more careful before hitting 'post' in the future. My apologies to anyone who may have been misled by anything I've misstated here on this topic. ~ CT |
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