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Old September 9th 18, 10:33 PM posted to sci.space.policy
Fred J. McCall[_3_]
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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.


--
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And if it's the angel who never does try.
And if it's me who will lose or win
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I'm standing right here, Mr Destiny.
If you want to talk, well then I'll relate.
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