You are such a poor shot that you couldn't throw an AA battery from orbit and have a hope of even hitting the earth.

And so am I.

I noticed recently that a batch of 91 AA sized batteries was delivered
to the ISS.

http://www.usatoday.com/tech/news/20...s-at-iss_x.htm

I thought that it might be fun to throw the used ones at Earth during
a sojourn outside, but despite the size of the target, I reckon you'd
have to be a pretty good throw to hit it, unless you could get some
serious atmospheric braking happening.

But it's a while since I've done the kind of thinking required to
prove it. Given that you are in geostationary orbit (for example) what
direction should you throw the battery to get the closest approach to
earth? A hunch tells me that it'd be back along your orbital path, in
the opposite direction of the earths spin.

Anyone care to enlighten me (us)?

Thanks

Denys Williams

Your hunch is right. your pitching it retrograde will give you the
lowest perigee for its new orbit. Its apogee would be the orbit you
started from. Pitching it toward the earth, or away, would have the
battery crossing your path twice an orbit, however its perigee would not
be as low as the retrograde pitch.

Mike



In article ,
(Denys Williams) wrote:

And so am I.

I noticed recently that a batch of 91 AA sized batteries was delivered
to the ISS.

http://www.usatoday.com/tech/news/20...s-at-iss_x.htm

I thought that it might be fun to throw the used ones at Earth during
a sojourn outside, but despite the size of the target, I reckon you'd
have to be a pretty good throw to hit it, unless you could get some
serious atmospheric braking happening.

But it's a while since I've done the kind of thinking required to
prove it. Given that you are in geostationary orbit (for example) what
direction should you throw the battery to get the closest approach to
earth? A hunch tells me that it'd be back along your orbital path, in
the opposite direction of the earths spin.

Anyone care to enlighten me (us)?

Thanks

Denys Williams

Denys Williams wrote:

And so am I.

I noticed recently that a batch of 91 AA sized batteries was delivered
to the ISS.

http://www.usatoday.com/tech/news/20...s-at-iss_x.htm

I thought that it might be fun to throw the used ones at Earth during
a sojourn outside, but despite the size of the target, I reckon you'd
have to be a pretty good throw to hit it, unless you could get some
serious atmospheric braking happening.

But it's a while since I've done the kind of thinking required to
prove it. Given that you are in geostationary orbit (for example) what
direction should you throw the battery to get the closest approach to
earth? A hunch tells me that it'd be back along your orbital path, in
the opposite direction of the earths spin.

Anyone care to enlighten me (us)?

Your intuition is correct. Whichever direction you throw it, the change
in velocity due to your puny human arm's effort will be quite small
compared to the initial velocity. Therefore the battery will assume an
orbit only slightly different from the one you are in. It will come
closest to earth if you throw it rearwards, as you suggest, since that
will reduce its kinetic energy the most.

Mitchell Timin

--
"Many are stubborn in pursuit of the path they have chosen, few in
pursuit of the goal." - Friedrich Nietzsche

http://annevolve.sourceforge.net is what I'm into nowadays.
Humans may write to me at this address: zenguy at shaw dot ca

Jack Crenshaw wrote:
Say you could throw the battery 100 mph -- the equivalent of a
first-rate baseball pitch. Because the orbital velocity is roughly
18,000 mph, you would change the perigee of the orbit only 1/180 of its
original value. That works out to be 22 miles; not bad for a sandlot
pitcher, but not nearly enough to hit the earth.

It is enough to quite significantly increase drag, and they'd decay
quite a lot faster than one just "dropped".

Ian Stirling wrote in message .. .
Jack Crenshaw wrote:
Say you could throw the battery 100 mph -- the equivalent of a
first-rate baseball pitch. Because the orbital velocity is roughly
18,000 mph, you would change the perigee of the orbit only 1/180 of its
original value. That works out to be 22 miles; not bad for a sandlot
pitcher, but not nearly enough to hit the earth.

It is enough to quite significantly increase drag, and they'd decay
quite a lot faster than one just "dropped".

On a tangent:

If you had a space station at a 600-mile altitude circular orbit and
wanted to drop cannisters from orbit, what's the minimum retrograde
velocity that needs to be imparted to the cannisters to have them
re-enter fairly quickly [1]? Would 300mph be adequate?

[1] Assume the cannisters are as simple as possible and thus lack
navigation and maneuvering abilities of their own, so they need to be
deorbited with one impulse, ideally before their apogee intersects the
station.

Mike Miller, Materials Engineer

Mike Miller wrote:
Ian Stirling wrote in message .. .
Jack Crenshaw wrote:
Say you could throw the battery 100 mph -- the equivalent of a
first-rate baseball pitch. Because the orbital velocity is roughly
18,000 mph, you would change the perigee of the orbit only 1/180 of its
original value. That works out to be 22 miles; not bad for a sandlot
pitcher, but not nearly enough to hit the earth.

It is enough to quite significantly increase drag, and they'd decay
quite a lot faster than one just "dropped".

On a tangent:

If you had a space station at a 600-mile altitude circular orbit and
wanted to drop cannisters from orbit, what's the minimum retrograde
velocity that needs to be imparted to the cannisters to have them
re-enter fairly quickly [1]? Would 300mph be adequate?

I'd think not, I think the perigee only goes down to around 450 miles.

As a rough guide, you lower the orbit by 1/4 of the velocity.
So, you might need 2000mph or so.

[1] Assume the cannisters are as simple as possible and thus lack
navigation and maneuvering abilities of their own, so they need to be
deorbited with one impulse, ideally before their apogee intersects the
station.

You don't generally get that happening if the orbit is inclined.

[1] Assume the cannisters are as simple as possible and thus lack
navigation and maneuvering abilities of their own, so they need to be
deorbited with one impulse, ideally before their apogee intersects the


I heard such cannisters existed and were more or less tested from Mir
Space Station..

I think it was a common way to get back films from early times spy
satellites ..

--
Julie
"please save Yuri"
a href="http://membres.lycos.fr/andromedanews/"Andromeda et Julie/a

On a tangent:

If you had a space station at a 600-mile altitude circular orbit and
wanted to drop cannisters from orbit, what's the minimum retrograde
velocity that needs to be imparted to the cannisters to have them
re-enter fairly quickly [1]? Would 300mph be adequate?

[1] Assume the cannisters are as simple as possible and thus lack
navigation and maneuvering abilities of their own, so they need to be
deorbited with one impulse, ideally before their apogee intersects the
station.

Mike Miller, Materials Engineer

I'd say that it is possible to deorbit them rather quickly by using
the atmosphere at our advantage. If you can make your canisters graze
the upper atmosphere, let's say 100 km, they'd loose a good deal of
velocity at perigee, thus lowering their apogee...

So if one does not care where they fall, because they'll disintegrate
in the lower atmosphere, one might safely use aerobraking passes to
save on fuel... I doubt stuff coming this low would last more than one
or two atmospheric passes before splitting in smaller pieces, thus
being more susceptible to the atmospheric drag, and providing a nice
light show for folks somewhere in the Pacific or South America. ;-)

(Mike Miller) wrote in
om:

Ian Stirling wrote in message
...
Mike Miller wrote:

If you had a space station at a 600-mile altitude circular orbit
and wanted to drop cannisters from orbit, what's the minimum
retrograde velocity that needs to be imparted to the cannisters to
have them re-enter fairly quickly [1]? Would 300mph be adequate?

I'd think not, I think the perigee only goes down to around 450
miles.

As a rough guide, you lower the orbit by 1/4 of the velocity.
So, you might need 2000mph or so.

Whoa...the orbital velocity of a 600-mile orbit is 8000mph?

So, could you launch the cannister in any way from the station and
have it deorbit fairly quickly (over several days) with less than
1000mph delta-V?

Nope, see my other post on the subject. Working out all the insane unit
conversions, the delta-V needed to drop from a 600 statute-mile circular
orbit to an elliptical one that just grazes the atmosphere (400,000 ft
perigee, which is what NASA considers "entry interface"), is about 510 mph.


--
JRF

Reply-to address spam-proofed - to reply by E-mail,
check "Organization" (I am not assimilated) and
think one step ahead of IBM.

"Jorge R. Frank" wrote in message ...

600 miles isn't a terribly good altitude for a manned space station. It's
within the inner Van Allen belts.

...oops, unit mistake. I was mis-remembering an earlier thread
(November 2003: " Drag at Orbital Altitudes") where I was thinking of
a 600km (not 600-mile) orbit.

So, at 600km, it would take ~140m/s to kiss that "re-entry perigee"?
Cool.

Mike Miller, Materials Engineer