How does the ISS maintain orbit across a range of altitudes?

I am a physics student preparing a presentation on the ISS, one of my
topics will cover orbital information. My question is that for a stable
orbit to be maintained a satellite must maintain a fixed relationship
between altitude and velocity, otherwise due to atmospheric drag it
will decrease in velocity and be overcome by the earth's
gravitational field and begin re-entry. After examining the altitude
information on the ISS released by NASA, I have seen that the ISS
operates over a wide range of altitudes, and as it slowly decreases in
height it re-boosts back up to maintain its height, from what I have
been taught so far for a satellite to maintain stable orbit it must
conform to V = SquareRoot GM/R, if for example the satellite falls
10km, does it therefore increase it's speed in order to stop it losing
more altitude, and when re-boosted does it decrease its speed in order
to stop it exiting orbit? Or can it operate comfortably across a range
of altitudes without significant effect? Anyone who can shed light on
this for me your help is much appreciated, thanks.

Jarrod.

"Rueffy" wrote in message
oups.com...
I am a physics student preparing a presentation on the ISS, one of my
topics will cover orbital information. My question is that for a stable
orbit to be maintained a satellite must maintain a fixed relationship
between altitude and velocity, otherwise due to atmospheric drag it
will decrease in velocity and be overcome by the earth's
gravitational field and begin re-entry. After examining the altitude
information on the ISS released by NASA, I have seen that the ISS
operates over a wide range of altitudes, and as it slowly decreases in
height it re-boosts back up to maintain its height, from what I have
been taught so far for a satellite to maintain stable orbit it must
conform to V = SquareRoot GM/R, if for example the satellite falls
10km, does it therefore increase it's speed in order to stop it losing
more altitude, and when re-boosted does it decrease its speed in order
to stop it exiting orbit? Or can it operate comfortably across a range
of altitudes without significant effect? Anyone who can shed light on
this for me your help is much appreciated, thanks.

You seem to be confusing velocity and energy in your explanations. You also
seem to assume that orbits must always be circular, which is absolutely not
the case.

If a spacecraft is in a circular orbit and is acted upon by a constant force
in a constant direction opposite the velocity vector (e.g. air drag), you
can think of this as a decrease in the kinetic energy of the object and can
solve for velocity and altitude, assuming that the orbit will remain
circular.

However, if a spacecraft is in a circular orbit and fires its engines in the
direction which increases the velocity of the spacecraft for a duration
that's very short compared to the time it takes to complete one orbit, then
we can model the burn as instantaneous (not that I'm saying you can make
this assumption with ISS reboost burns). Since this is modeled as an
instantaneous change in velocity (delta-V), the altitude at the point in the
orbit at which the burn occurred does not change. The orbit will continue
to pass through that point on subsequent orbits. What does change is the
*shape* of the orbit and the altitude of the orbit at the location on the
opposite side of the orbit. Since our burn was made in the direction which
increased the velocity at the burn point, this increases the altitude at the
point opposite the burn point. Note that at this opposite point the
altitude is now higher than it was before. This is where terms like apogee
and perigee come into play.

It's really hard to describe these things with words, so I suggest you
consult a good orbital mechanics book that has both equations and pictures.
;-)

Jeff
--
"They that can give up essential liberty to obtain a
little temporary safety deserve neither liberty nor
safety"
- B. Franklin, Bartlett's Familiar Quotations (1919)

JRS: In article , dated Wed, 14
Jun 2006 10:44:28 remote, seen in news:sci.space.station, Jeff Findley
posted :

If a spacecraft is in a circular orbit and is acted upon by a constant force
in a constant direction opposite the velocity vector (e.g. air drag), you
can think of this as a decrease in the kinetic energy of the object and can
solve for velocity and altitude, assuming that the orbit will remain
circular.

But under those circumstances the kinetic energy actually increases.

See URL:http://www.merlyn.demon.co.uk/gravity2.htm#EDOB and
thereabouts.

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

"Dr John Stockton" wrote in message
...
JRS: In article , dated Wed, 14
Jun 2006 10:44:28 remote, seen in news:sci.space.station, Jeff Findley
posted :

If a spacecraft is in a circular orbit and is acted upon by a constant
force
in a constant direction opposite the velocity vector (e.g. air drag), you
can think of this as a decrease in the kinetic energy of the object and
can
solve for velocity and altitude, assuming that the orbit will remain
circular.

But under those circumstances the kinetic energy actually increases.

See URL:http://www.merlyn.demon.co.uk/gravity2.htm#EDOB and
thereabouts.

You're right. I was mixing up my terms too. It's been a few years since my
orbital dynamics class.

;-)

One of the interesting things to note is the relationship between the
kinetic energy and potential energy, the sum of which is the total energy of
an orbit.

http://www.go.ednet.ns.ca/~larry/orbits/kepler.html

Jeff
--
"They that can give up essential liberty to obtain a
little temporary safety deserve neither liberty nor
safety"
- B. Franklin, Bartlett's Familiar Quotations (1919)