Space Forum / Space Flight / December 2009

There has to be some more information for the problem to be meaningful:
the satellite isn't going to fall to a height of 800 km without
something happening to cause it.  You've made one good assumption as to
what the question may have meant (without checking your arithmetic, your
approach looks correct to me), and the "right" approach isn't immediately
clear to me.

(FWIW, I tried just recalculating the orbital velocity at a height of
800 meters and got 7460, which is closer but also clearly not what they
had in mind).

It's ok, some specific info is lacking.
How does a disabled satellite retro fire?
((BS baffles brains))

It sounds like you are expected to find the orbital
velocity at 800km when an impulse was applied to
the sat at 2000 km, to retro fire into an elliptical
orbit with perigee at 800 then retro fire again to
circularize. Anyway assuming you had help from
pink fairies, the standard circular orbit equation is,

Vo = sqrt( GM/r )  .

Ken

OK , I think I have the answer now.  it was to do with the folly of
simply adding velocities.  if we convert teh original speed to kinetic
energy , add the kinetic energy that is gained by the loss of
potential energy as the ht reduced from 2000 to 800 km then convert
the total kinetic energy back to speed we get the "right?" answer.   I
guess what the question meant was that even though the satellite had
the correct speed for a circular orbit at 2000km , if that speed was
not in the right direction , ie perpendicular to a line towards the
center of the earth, then in fact it would be an elliptical orbit that
may come down to 800km ?  I think!

In sci.space.tech message <be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000v
ba.googlegroups.com>, Sun, 7 Jun 2009 21:56:02, dotcom
<tfmann@iprimus.com.au> posted:

No, because we don't know why it fell.

Three extreme cases can be modelled as

(1) Gentle Atmospheric drag; its new speed is that for the lower
circular orbit.

(2) It suddenly lost some speed, so that it is in an elliptical orbit
apogee 2000 km perigee 800 km.

(3) It suddenly lost all speed, is coming straight down, and is passing
800 km now.

Also,

(A) It bounced off something elastically, so that its elliptical orbit
has perigee 800 km and speed crossing 200 km is circular speed for that
height.

(B) It bounced off something elastically, is coming straight down, and
is passing 800 km now.

Probably one of the first two is intended; the first for a mid-range
student or the second for an advanced one.

The first thing is to get the exact question as asked.  I was once
requested by a young teenager to give the formula for a tree.  First
question to ask : is this chemistry/biology, is it engineering, is it
topology (carbohydrate; Euler's Strut, Euler's polyhedrons).  She seemed
satisfied to count branches, leaves, and vertices.

== == ==

The moderators' system is not working as it claims that it will.  I got
messages; I sent the replies which, the message said, would stop the
system mailing me again; but it did not have that effect.  It has an
obvious design fault.

I think you got the right answer but were probably misled by the
coincidence in that extra calculation you did.  The question is
unclear, but nothing you state suggests the satellite is actually in
a circular orbit.

I think here they are suggesting -- but not stating -- that no forces
other than gravity act on the satellite.  So the orbit is NOT
circular.

This is the right approach, but as you noticed in a later post, you
can't add speeds.  Instead add the potential energy change to the
original kinetic energy, and work out what speed corresponds to that
kinetic energy.  The mass of the satellite is irrelevant, as you will
probably notice.

As others noted, if non-gravitational forces act, the problem as
stated is incomplete.

In article
<be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000vba.googlegroups.com>,

Presumably it doesn't have any propulsion systems so it is in free-fall.

I haven't done the calculation, but assuming that you did it correctly:

If the orbital speed for a circular orbit is 6900 m/s at 2000 km
height, that does not mean that an object with that height and that
speed is in a circular orbit.

Suppose you climb a 2000 km tall ladder (from Antarctica so you can
ignore Earth's rotation and are initially motionless), and then fire a
rifle that has a muzzle speed of 6900 m/s.  Every bullet you fire will
have a 6900 m/s speed at 2000 km height.  If you fire it horizontally,
then you will have a circular orbit.  If you fire it straight up then
it will have a long, skinny orbit that intersects Earth before it gets
all the way around.  If you fire it straight down it will have the same
orbit, only it will complete even less of an orbit.  If you fire it at
an angle, then it will have a perigee below 2000 km and a apogee above
that height.  It will, of course, travel faster at perigee and slower
at apogee.

If the apogee is below 800 km, then you can complete the question.  It
turns out to have the same answer for all orbits with an apogee below
800 km (including the shoot-straight-down case).

You can't take E
.9e10 J -> 3970 m/s and add that to 6900 m/s to get
the new speed.

You have to take
Ekinetic_start .7e10 J
+ Epotential_change
.9e10 J
kinetic_end .6e10 J
And 7.6e10 J -> the 7900 m/s that's the textbook answer.

Basically, you have to add the energies, not the velocities.

Which means that you have to add the square of the velocity and the
square of the KE-equivalent velocity and take the square root of the
sum.  (The phrase used is 'add in quadrature')

Yes, the stated problem is a conservation of energy problem, and
solvable.

In article
<be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000vba.googlegroups.com>,

Sadly I dont have the maths to work this out, but since this is only
high school work, perhaps you just need to calculate the speed at 800km

You guys are very cluey, but you wouldn't expect a school student to do
all other stuff you are considering

Perhaps someone could just calculate the speed required to keep it in
orbit at 800km

David who is trying to learn maths, but...