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Elliptical orbit question
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JF Mezei
Sep 6, 2018, 7:23:12 PM9/6/18
to
I know this is likely a newbie question but...
Say I have a satellite in elliptical orbit of 10,000km at apogee and
400km perigee.
At 400km, the satellite is going way too fast to stay at that altitude
and goes up. At 10,000km the satellite doesn't have enough speed to
remain at that altitude and will drop back in altitude.
So far correct?
Is it correct to state that the satellite's energy level is simular to
one with a circular orbit somewhere between 400km and 10,000? ( lets
say 5000 for sake of disussion).
If, between 5000 and 400, the satellite has more speed than needed to be
in orbit, how come it continues to drop all the way down to 400km before
rising?
Or is this a case of the innertia gained falling from 10,000 to 5000
will make the satellite w3ant to continue in the same direction towards
earth until the "slinghot" cause it to change direction and turn around
and rise up in altitude again?
Say I have a satellite in elliptical orbit of 10,000km at apogee and
400km perigee.
At 400km, the satellite is going way too fast to stay at that altitude
and goes up. At 10,000km the satellite doesn't have enough speed to
remain at that altitude and will drop back in altitude.
So far correct?
Is it correct to state that the satellite's energy level is simular to
one with a circular orbit somewhere between 400km and 10,000? ( lets
say 5000 for sake of disussion).
If, between 5000 and 400, the satellite has more speed than needed to be
in orbit, how come it continues to drop all the way down to 400km before
rising?
Or is this a case of the innertia gained falling from 10,000 to 5000
will make the satellite w3ant to continue in the same direction towards
earth until the "slinghot" cause it to change direction and turn around
and rise up in altitude again?
Alain Fournier
Sep 6, 2018, 8:09:45 PM9/6/18
to
On Sep/6/2018 at 19:23, JF Mezei wrote :
> I know this is likely a newbie question but...
>
> Say I have a satellite in elliptical orbit of 10,000km at apogee and
> 400km perigee.
>
> At 400km, the satellite is going way too fast to stay at that altitude
> and goes up. At 10,000km the satellite doesn't have enough speed to
> remain at that altitude and will drop back in altitude.
>
> So far correct?
>
>
> Is it correct to state that the satellite's energy level is simular to
> one with a circular orbit somewhere between 400km and 10,000? ( lets
> say 5000 for sake of disussion).
Yes. The 5000km figure isn't right, but you have the right idea.
> I know this is likely a newbie question but...
>
> Say I have a satellite in elliptical orbit of 10,000km at apogee and
> 400km perigee.
>
> At 400km, the satellite is going way too fast to stay at that altitude
> and goes up. At 10,000km the satellite doesn't have enough speed to
> remain at that altitude and will drop back in altitude.
>
> So far correct?
>
>
> Is it correct to state that the satellite's energy level is simular to
> one with a circular orbit somewhere between 400km and 10,000? ( lets
> say 5000 for sake of disussion).
> If, between 5000 and 400, the satellite has more speed than needed to be
> in orbit, how come it continues to drop all the way down to 400km before
> rising?
>
> Or is this a case of the innertia gained falling from 10,000 to 5000
> will make the satellite w3ant to continue in the same direction towards
> earth until the "slinghot" cause it to change direction and turn around
> and rise up in altitude again?
have the right idea. Somewhere between 400km and 10,000km the satellite
will have enough speed to achieve a circular orbit at that altitude, but
that speed will not be in the right direction. Therefore as you said
because of inertia, it will continue in that direction. Earth will
slowly change its direction, Earth pulling it always towards Earth. But
because it is going too fast, Earth will not be pulling it fast enough
to compensate the fact that it isn't going straight towards Earth. So as
it gets closer to Earth it will be pointing less and less towards Earth
(I'm simplifying a little here) until it reaches perigee at which point
it will actually start going further away from Earth.
Alain Fournier
Stuf4
Sep 9, 2018, 3:40:49 AM9/9/18
to
From Alain Fournier:
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 centripetal force works, pulling the satellite up and away from the Earth. The centripetal force is a manifestation of the inertial property of the satellite's mass.
When gravity and the centripetal 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
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 centripetal force works, pulling the satellite up and away from the Earth. The centripetal force is a manifestation of the inertial property of the satellite's mass.
When gravity and the centripetal 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
Stuf4
Sep 9, 2018, 3:59:21 AM9/9/18
to
Oops. I misspoke. I had said "centripetal force", but I meant centrifugal force. The spring in the diagram represents the centripetal force acting on the satellite, again just a manifestation of inertia.
(And of course, the bigger picture of physics is that inertia itself is just a manifestation of a more fundamental property. As is gravity. And the biggest clue to that is in how inertial mass is indistinguishable from gravitational mass. So it is helpful to keep in mind when speaking about things like inertia, centripetal force, centrifugal force, and gravity, that these are constructs. Useful and consistent. Like in how everybody uses the terms "sunrise" & "sunset" while when we stop to think about it, we know that the phenomenon we are actually referring to is "Earth spin". The Sun is only rising from the perspective of our non-inertial reference frame of standing on the surface of the spinning planet.)
~ CT
(And of course, the bigger picture of physics is that inertia itself is just a manifestation of a more fundamental property. As is gravity. And the biggest clue to that is in how inertial mass is indistinguishable from gravitational mass. So it is helpful to keep in mind when speaking about things like inertia, centripetal force, centrifugal force, and gravity, that these are constructs. Useful and consistent. Like in how everybody uses the terms "sunrise" & "sunset" while when we stop to think about it, we know that the phenomenon we are actually referring to is "Earth spin". The Sun is only rising from the perspective of our non-inertial reference frame of standing on the surface of the spinning planet.)
~ CT
Stuf4
Sep 9, 2018, 4:04:49 AM9/9/18
to
I wrote:
> Oops. I misspoke. I had said "centripetal force", but I meant centrifugal
> force. The spring in the diagram represents the centripetal force acting on
> the satellite, again just a manifestation of inertia.
DAMN. I misspoke two posts in a row.
> Oops. I misspoke. I had said "centripetal force", but I meant centrifugal
> force. The spring in the diagram represents the centripetal force acting on
> the satellite, again just a manifestation of inertia.
Correction to the correction:
The spring in the diagram represents the _centrifugal force_ acting on the satellite, again just a manifestation of inertia.
~ CT
.
.
Stuf4
Sep 9, 2018, 4:25:38 AM9/9/18
to
Reposting with corrections ("centrifugal force" is what was meant):
===================================
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.space.policy/jkZuEFhvBtk/HxefrAHiBgAJ)
===================================
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.
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
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.space.policy/jkZuEFhvBtk/HxefrAHiBgAJ)
Alain Fournier
Sep 9, 2018, 12:29:28 PM9/9/18
to
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
> 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.
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
JF Mezei
Sep 9, 2018, 1:40:11 PM9/9/18
to
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
> This is the situation you have in the lab, with the mass bobbing up and down on the end of the spring.
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 ?
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.
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?
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)
Niklas Holsti
Sep 9, 2018, 4:06:27 PM9/9/18
to
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
> 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 ?
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.
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?
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)
down to pointing up.
--
Niklas Holsti
Tidorum Ltd
niklas holsti tidorum fi
. @ .
Fred J. McCall
Sep 9, 2018, 5:33:49 PM9/9/18
to
JF Mezei <jfmezei...@vaxination.ca> 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
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 ?
>
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.
>
>
>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?
>
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)
>
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
Stuf4
Sep 9, 2018, 7:10:44 PM9/9/18
to
From Alain Fournier:
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.net/images/cache/72/8a/c0/728ac0b7ec97db9210c2f667f1ccca57.PNG
=========================
The title of that page focuses on centrifugal force, but then describes the orbit case in terms of centripetal force.
~ CT
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.net/images/cache/72/8a/c0/728ac0b7ec97db9210c2f667f1ccca57.PNG
=========================
The title of that page focuses on centrifugal force, but then describes the orbit case in terms of centripetal force.
~ CT
Alain Fournier
Sep 9, 2018, 7:16:49 PM9/9/18
to
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
JF Mezei
Sep 9, 2018, 11:03:01 PM9/9/18
to
On 2018-09-09 16:06, Niklas Holsti wrote:
> 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
Ok, thanks for finding the wording that makes me understand.
> 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
JF Mezei
Sep 9, 2018, 11:06:50 PM9/9/18
to
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.
> 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.
Fred J. McCall
Sep 9, 2018, 11:51:27 PM9/9/18
to
JF Mezei <jfmezei...@vaxination.ca> wrote on Sun, 9 Sep 2018
23:06:49 -0400:
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.
Jeff Findley
Sep 10, 2018, 6:47:01 AM9/10/18
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In article <e34bpdti7p9fvcvuv...@4ax.com>,
fjmc...@gmail.com says...
scalar math. For a two body problem, the motion is at least planar,
which reduces the complexity to a 2D vector math problem.
Jeff
--
All opinions posted by me on Usenet News are mine, and mine alone.
These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.
fjmc...@gmail.com 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
> >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.
>
scalar math. For a two body problem, the motion is at least planar,
which reduces the complexity to a 2D vector math problem.
Jeff
--
All opinions posted by me on Usenet News are mine, and mine alone.
These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.
Stuf4
Sep 10, 2018, 5:26:16 PM9/10/18
to
From Jeff Findley:
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).
~ CT
> In article <e34bpdti7p9fvcvuv...@4ax.com>,
> fjmc...@gmail.com 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.
> fjmc...@gmail.com 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.
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).
~ CT
Stuf4
Sep 11, 2018, 6:57:21 PM9/11/18
to
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
tetian...@gmail.com
Oct 5, 2018, 9:48:04 AM10/5/18
to
> Is it correct to state that the satellite's energy level is simular to
> one with a circular orbit somewhere between 400km and 10,000? ( lets
> say 5000 for sake of disussion).
>
>
> If, between 5000 and 400, the satellite has more speed than needed to be
> in orbit, how come it continues to drop all the way down to 400km before
> rising?
>
> Or is this a case of the innertia gained falling from 10,000 to 5000
> will make the satellite w3ant to continue in the same direction towards
> earth until the "slinghot" cause it to change direction and turn around
> and rise up in altitude again?
As part of my almost daily exercise routine, I use an ellipical 2-4 times a week. I have AFG Sport 5.9AE Elliptical Machine. Works great for me. You can read a review on it on https://www.theolive.com/reviews/best-elliptical/ .
> one with a circular orbit somewhere between 400km and 10,000? ( lets
> say 5000 for sake of disussion).
>
>
> If, between 5000 and 400, the satellite has more speed than needed to be
> in orbit, how come it continues to drop all the way down to 400km before
> rising?
>
> Or is this a case of the innertia gained falling from 10,000 to 5000
> will make the satellite w3ant to continue in the same direction towards
> earth until the "slinghot" cause it to change direction and turn around
> and rise up in altitude again?
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