Great view of dropped camera drifting away

In article ,
nmp wrote:
According to the nasa tv pao commentator, as of today ground tracking of
the lost camera showed it to be about 1/2 nm below and 68 nm in front of
the space station complex.

So, nm = nautical miles I guess?

Seems likely. :-)

How can it be in front of the ISS? Less drag = higher velocity? Then why
is it also *below* the station and not at same height? I may be missing
something very obvious here, I'm afraid. But I wouldn't mind being told
what it is

Air drag. The camera is much more affected by air drag than the station,
because being smaller, it has much more surface area per unit mass. The
details get a bit complicated, but the ultimate bottom line is that drag
shrinks the orbit, and since lower orbits move faster, it moves below and
ahead.
--
spsystems.net is temporarily off the air; | Henry Spencer
mail to henry at zoo.utoronto.ca instead. |

nmp wrote:
Op Tue, 19 Dec 2006 13:52:26 -0800, schreef columbiaaccidentinvestigation:


Jim Oberg wrote:
Great view of camera drifting away -- about 6:27 PM EST:
http://forum.nasaspaceflight.com/for...chmentid=16026

Tools have been dropped before -- one was dropped by Fuglesang on his
first EVA a few days ago. The last camera lost in space, AFAIK, was by
Mike Collins on his spacewalk from Gemini-10 in July 1966.

According to the nasa tv pao commentator, as of today ground tracking of
the lost camera showed it to be about 1/2 nm below and 68 nm in front of
the space station complex.

So, nm = nautical miles I guess?

How can it be in front of the ISS? Less drag = higher velocity? Then why
is it also *below* the station and not at same height? I may be missing
something very obvious here, I'm afraid. But I wouldn't mind being told
what it is

Oh, and BTW, what kind of camera was it?

In addition to the aerodynamic drag differences between the two objects
a couple of possible explanations for the orientation of the lost
camera being ½ nm (nautical mile) below and 68 nm in front of the
space station complex, may be the iss is under normal attitude control
and the camera is not, and or the fact the iss's orbit was going to be
re-boosted as listed in the sts-116 press kit, both of which would
cause a different separation between the two objects than if the iss
were in free drift like the camera.

"Henry Spencer" wrote in message
...
In article ,
nmp wrote:
According to the nasa tv pao commentator, as of today ground tracking of
the lost camera showed it to be about 1/2 nm below and 68 nm in front of
the space station complex.

So, nm = nautical miles I guess?

Seems likely. :-)

How can it be in front of the ISS? Less drag = higher velocity? Then why
is it also *below* the station and not at same height? I may be missing
something very obvious here, I'm afraid. But I wouldn't mind being told
what it is

Air drag. The camera is much more affected by air drag than the station,
because being smaller, it has much more surface area per unit mass. The
details get a bit complicated, but the ultimate bottom line is that drag
shrinks the orbit, and since lower orbits move faster, it moves below and
ahead.


Not 100% applicable but:

East takes you out. Out takes you west. West takes you in.
In takes you east. North and south brings you back.

(once you can translate that into planetary orbits, it sort of makes sense.
:-)


--
spsystems.net is temporarily off the air; | Henry Spencer
mail to henry at zoo.utoronto.ca instead. |

in article , Henry Spencer at
wrote on 12/19/06 4:08 PM:

In article ,
nmp wrote:

According to the nasa tv pao commentator, as of today ground tracking of the
lost camera showed it to be about 1/2 nm below and 68 nm in front of the
space station complex.

So, nm = nautical miles I guess?

Seems likely. :-)

How can it be in front of the ISS? Less drag = higher velocity? Then why is
it also *below* the station and not at same height? I may be missing
something very obvious here, I'm afraid. But I wouldn't mind being told what
it is

Air drag. The camera is much more affected by air drag than the station,
because being smaller, it has much more surface area per unit mass. The
details get a bit complicated, but the ultimate bottom line is that drag
shrinks the orbit, and since lower orbits move faster, it moves below and
ahead.

I would be surprised if drag differential accounts for much of that 68
miles. My guess is that the main component of the camera's velocity was
backward and that has produced most of the separation.

George Evans

This stuff seems to happen a lot... you would think it is easy to keep
track of stuff. But then again, I have not been in zero-g, and I'm sure
it makes stuff a lot harder.

We should make some sort of magnet ray gun to attract objects that are
floating away. Maybe I'll get on it. I'll patent it and make a little
money :P

[Disclaimer: I am not a rocket scientist.]

nmp wrote:

How come that "drag shrinks orbit"? Isn't it because drag slows you
down, and lower speed makes you fall out of orbit?

Yes.

I understand that to maintain a relatively low orbit, higher speed is
necessary (compared to higher orbits).

No. A higher orbital velocity gives a higher orbital altitude. However,
a higher orbit is longer, and it works out that the time taken to
complete one orbit (the period) is longer, despite the higher velocity.

I wasn't aware though that a lower orbit automatically causes a
higher speed. Or does it? I'm confused. Probably has to do with me
not paying enough attention in physics class, twenty years ago

A lower orbit is shorter. As described above, this means an object in a
lower orbit has a shorter period, so will move ahead of an object in a
higher orbit.

Hope this helps.

--Chris

On Thu, 21 Dec 2006 11:32:55 +0100, nmp wrote:


I understand that to maintain a relatively low orbit, higher speed is
necessary (compared to higher orbits). I wasn't aware though that a lower
orbit automatically causes a higher speed. Or does it? I'm confused.
Probably has to do with me not paying enough attention in physics class,
twenty years ago

Yes, it automatically causes higher speeds. It all has to do with gravity,
and that it's a function of the inverse square of the distance between two
object. Lower altitude, the gravitational force increases, faster orbital
speed.

http://hyperphysics.phy-astr.gsu.edu.../isq.html#isqg

--
Craig Fink
Courtesy E-Mail Welcome @

On Thu, 21 Dec 2006 11:32:55 +0100, in a place far, far away, nmp
made the phosphor on my monitor glow in such a
way as to indicate that:

Op Wed, 20 Dec 2006 00:08:31 +0000, schreef Henry Spencer:

In article ,
nmp wrote:
According to the nasa tv pao commentator, as of today ground tracking of
the lost camera showed it to be about 1/2 nm below and 68 nm in front of
the space station complex.

So, nm = nautical miles I guess?

Seems likely. :-)

Why still use such a weird traditional unit when it is officially defined
as being 1852 METRES anyway? A wholly different can of worms, I'm aware of
that.

It's a natural unit (for this planet, anyway), representing the length
of one minute of arc on the earth's surface.

On Thu, 21 Dec 2006 21:43:28 +1100, in a place far, far away, Chris
Bennetts made the phosphor on my monitor
glow in such a way as to indicate that:

[Disclaimer: I am not a rocket scientist.]

nmp wrote:

How come that "drag shrinks orbit"? Isn't it because drag slows you
down, and lower speed makes you fall out of orbit?

Yes.

I understand that to maintain a relatively low orbit, higher speed is
necessary (compared to higher orbits).

No. A higher orbital velocity gives a higher orbital altitude. However,
a higher orbit is longer, and it works out that the time taken to
complete one orbit (the period) is longer, despite the higher velocity.

No, higher-altitude circular orbits have lower velocity than
lower-altitude ones. But when you add velocity to an orbit, it raises
the apogee. As you reach the top, you slow down, due to conservation
of energy. If you add more velocity at the top to raise the perigee
up to your current altitude (i.e., circularizing the orbit), you've
added velocity in both cases, but your new velocity will be lower than
the one you had at the original lower orbit, due to the higher
potential energy represented by the higher altitude.

For example, LEO is about 25,000 fps. GEO is about 10,000 fps.

On Thu, 21 Dec 2006 14:33:32 +0100, in a place far, far away, nmp
made the phosphor on my monitor glow in such a
way as to indicate that:

So I had that right? But that would mean I am *again* lost on how all the
other stuff works. With the camera ahead of station and all - sorry for
the fuzziness.

As the camera slows down, it falls closer to earth. But as it falls,
it accelerates and picks up speed. There are probably programs you
can get that simulate this. The mechanics of the interactions between
two orbiting objects are quite complex, and depend on what direction
you add velocity (radial, in-track, cross-track).

Isn't there some plain English step-by-step "Orbit Science 101" on the
WWW for me to start with? I do find the subject intriguing, but I'm
afraid I'm lacking the very basics. As evidenced by my confusion...

Probably, but I don't know off hand. If you find one via a web
search, come back and let us know.

But when you add velocity to an orbit, it raises
the apogee. As you reach the top, you slow down, due to conservation
of energy. If you add more velocity at the top to raise the perigee
up to your current altitude (i.e., circularizing the orbit), you've
added velocity in both cases, but your new velocity will be lower than
the one you had at the original lower orbit, due to the higher
potential energy represented by the higher altitude.

For example, LEO is about 25,000 fps. GEO is about 10,000 fps.

See, that does make sense to my untrained brain, since I have always
understood that you need the higher speed to "escape" the stronger
gravity pull on a lower orbit.

One step at a time.

You're in LEO, at constant altitude, and a constant velocity of 25,000
fps. You add several thousand feet per second in the direction of
motion. Now you're at perigee of a new elliptical orbit going (say)
35,000 fps. Your apogee is now much higher, and you start to climb
toward it. Half an orbit later, you're at apogee, and going much more
slowly, slower even than you were in a circular LEO orbit, because the
total energy is conserved, and your velocity has been converted to
altitude. Perhaps 5000 fps (these aren't real numbers, but they're
ballpark). If you continue another half orbit, you'll fall back down
to perigee in LEO, and be going 35,000 fps again. But if instead, you
add another 5000 fps, you've raised your perigee to the same altitude
as you're at. You're now in a high circular orbit, with a constant
altitude, and a constant velocity of 10,000 fps. And a much longer
orbital period, both because you are going more slowly, and because
you have a much bigger circle to make.