A Space & astronomy forum. SpaceBanter.com

Go Back   Home » SpaceBanter.com forum » Space Science » Policy
Site Map Home Authors List Search Today's Posts Mark Forums Read Web Partners

Elliptical orbit question



 
 
Thread Tools Display Modes
  #1  
Old September 9th 18, 09:25 AM posted to sci.space.policy
Stuf4
external usenet poster
 
Posts: 554
Default 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  
Old September 9th 18, 05:29 PM posted to sci.space.policy
Alain Fournier[_3_]
external usenet poster
 
Posts: 548
Default 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  
Old September 9th 18, 09:06 PM posted to sci.space.policy
Niklas Holsti
external usenet poster
 
Posts: 168
Default 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  
Old September 9th 18, 10:33 PM posted to sci.space.policy
Fred J. McCall[_3_]
external usenet poster
 
Posts: 10,018
Default 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  
Old September 10th 18, 12:10 AM posted to sci.space.policy
Stuf4
external usenet poster
 
Posts: 554
Default 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  
Old September 10th 18, 12:16 AM posted to sci.space.policy
Alain Fournier[_3_]
external usenet poster
 
Posts: 548
Default 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  
Old September 10th 18, 04:51 AM posted to sci.space.policy
Fred J. McCall[_3_]
external usenet poster
 
Posts: 10,018
Default 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.
  #8  
Old September 10th 18, 11:46 AM posted to sci.space.policy
Jeff Findley[_6_]
external usenet poster
 
Posts: 2,307
Default Elliptical orbit question

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.

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.
  #9  
Old September 10th 18, 10:26 PM posted to sci.space.policy
Stuf4
external usenet poster
 
Posts: 554
Default Elliptical orbit question

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).

~ CT
  #10  
Old September 11th 18, 11:57 PM posted to sci.space.policy
Stuf4
external usenet poster
 
Posts: 554
Default 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
 




Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Elliptical orbit question Stuf4 Policy 2 September 9th 18 09:04 AM
But why an elliptical orbit tt40 Amateur Astronomy 73 October 14th 05 11:00 PM
De-orbit question George R. Kasica Space Shuttle 2 August 12th 05 01:20 AM
Elliptical Orbit Starman Astronomy Misc 3 February 3rd 05 01:05 AM
orbit question Jan Philips History 7 September 29th 03 06:16 PM


All times are GMT +1. The time now is 11:19 PM.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 SpaceBanter.com.
The comments are property of their posters.