Let's simplify this a bit. Let's say that you're way out in deep
space, and the space shuttle is approaching you at 17,500 m/h.
Suddenly the spacecraft fires a bullet rearwards. Subsequently, how
fast are both the shuttle and the bullet traveling relative to you?
Prior to ejecting the bullet both it & the shuttle are traveling
inertially at some velocity. (17,500 m/h in this case.) The two bodies
together have a mutual center of mass, and it is the mass center which
is, in fact, traveling at the mentioned velocity. By the conservation
of momentum, the velocity of the mass center must remain constant. If
a bullet is fired out the rear, then to balance the momentum the
shuttle must aquire a quantity of momentum equal to that of the fired
bullet but in the opposite direction. So propelling the bullet
backwards also propells the shuttle forewards. This is of course the
whole principle of rocketry; a rocket motor expends energy to impart
momentum to trillions of tiny bullets, the molecules of the exhaust
gases, which mandates that the rocket then aquire momentum equal to
that of the ejecting gas.
So then the bullet's velocity is a composition of the velocity of
its original mass center and the rate at which it is radiating from
that mass center. An astronaut onboard the shuttle does the honors and
pulls the trigger, then observes the bullet receding at a relative
velocity of 1,500 m/h. That's how fast the bullet & shuttle are
getting farther apart from each other. How fast then are each of them
moving with respect to you, hanging out in space? We must figure out
how much delta-v (change in velocity) each has aquired relative to
their mutual mass center, then add these to the mass center's velocity
relative to you.
The classical momentum is given by p = mv.
p = momentum
m = mass
v = velocity
The bullet/shuttle system has a mass of (m[b] + m[s]), with m[b] =
bullet, m[s] = shuttle. But the ratio of the system mass to that of
each component will determine its relative change in momentum, and
velocity.
So (m[b] + m[s])/m[b] gives the portion of 1,500 m/h which belongs
to the bullet, and similarly for the shuttle by just plugging in its
mass. Let's plug in some masses. I don't offhand know how the mass of
a bullet, but let's guess it's perhaps an ounce. A real space shuttle
orbiter is something like 80 tons, which is 2,560,000 ounces, giving
the system a mass of 2,560,001 ounces. The bullet will recede from the
system mass center over 2.5 million times faster than will the
shuttle. In fact, the shuttle will be creeping ahead of the system
center barely at all for most practical reasons. As long as we're not
too picky, the shuttle is still traveling towards you at very nearly
17,500 m/h, and the bullet is traveling towards you at (17,500 -
1,500) m/h = 16,000 m/h.
Now of course we guessed ages ago that this was pretty much the
case, because we know that the shuttle is heavy and a bullet is almost
nothing in comparison. So why go through all this? Because it's not
always two such distinctly different masses. Instead of a space
shuttle it might just as easily have been a firecracker whizzing by.
In such a case the changes in velocity would've been more like equal
for the exploding pieces. Then we'd be saying something like, "The
near half is approaching at (17,500 + 8,000) = 25,500 and the rear
half at (17,500 - 8,000) = 9,500. So it can make a big difference
depending on the variables.
Now, let's put the shuttle back on orbit with Earth. You're on the
ground directly beneath the satellite's perigee, the point of lowest
altitude on its orbit. As it passes directly overhead an astronaut
fires the bullet. The shuttle aquires a tiny bit of forward velocity.
What happens to it? The greater orbital energy puts it into a new,
more eccentric orbit. As it climbs towards its new, higher apogee (the
point of highest altitude), it actually slows *slows down*, to
conserve angular momentum. Conversely, the bullet drops into an orbit
of lower perigee and *speeds up*. The bullet's ground track will
actually *overtake* the shuttle's. So for a period of time their
ground tracks will get closer together, even though the bullet was
propelled definitely out the rear of the spacecraft. Upon crossing
their groundtracks the two will then again get father apart, but only
for a short while. If the shuttle is on orbit at 17,500 m/h, then any
negative delta-v will put an object at less than orbital velocity. The
bullet, having lost 1,500 m/h at perigee, will enter an orbit with a
new perigee which is somewhere under Earth's surface. It'll very soon
plow through the atmosphere and be vaporised. |
