Gautam Mukunda wrote:
>
>> India has recently launched a satellite; and an ICBM is much
>> simpler than a satellite.
>
> Are you sure?
>
No, but your question did motivate me to do some calculations.
But now I am convinced that it's easier to put a satellite in orbit
than launch an ICBM - because a satellite may be of any
size, and a nuclear warhead has a minimum size.
> I'm not a hundred percent certain that is the case. Didn't
> the Soviets launch Sputnik before they had ICBMs?
>
Yes, but as soon as Sputnik was launched the world was
aware that they had the power to send a nuke anywhere they
wanted.
> For that mattter,
> there's the question of miniaturization. I'm not sure what the status of
> Indian nuclear technology is, but it's at least possible that they haven't
> miniaturized enough for practical use in ICBMs.
>
But why would an ICBM have to be miniaturized? An ICBM must carry,
at least, the critical mass of the nuclear reaction, so they
have a _minimum_ mass.
> Certainly the Indian
> nuclear tests that caused such a stir a few years ago were not entirely
> successful - most were fizzles, I believe. Finally there's the question
> of reentry, right, which is something that satellites don't have to deal
> with.
>
Mathematically, the reentry is a similar problem to the exit,
because you must cross the athmosphere with a huge speed.
> I always had the impression that satellites launch capacity was a
> step on the road to ICBMs, but not necessarily the whole way there.
>
I guess it's time to...
<shut up and calculate>
I don't _know_ how to make ICBMs, but there are some standard
formulas that must be used to calculate some stuff.
For example, let's assume two extreme cases of putting a cargo
into a low-altitude orbit [say, 200 km above sea-level].
In the first case, let's use (1.1) a gun to raise the cargo to
200 km, and then (1.2) use a kick rocket to put the satellite
into a circular orbit.
(1.1) can be solved by the conservation of Energy, as this:
-GM / r(earth) + 1/2 v^2 = -GM / (r(earth) + 200 km)
where GM = 393000 km^3/s^2 and r(earth) = 6378.160 km, so
v = 1.9 km / sec
(1.2) is just the circular speed at that height, and we
get it by equating the centrifugal force to the gravitational
force:
v^2 / (r + 200) = GM / (r + 200)^2
so v = 7.7 km/sec, for a total Delta-V of 9.7 km/sec.
In an ideal case [no atmosphere], we might be able to
use a transfer orbit to get to the 200 km; the intermediary orbit
[Hohmann Transfer Orbit] would be an orbit with perigee
touching the Earth at the departure point, and the apogee would
be at the antipode's point 200 km above the surface. The
conservation of Energy [again], plus the fact that the orbit
is elliptic and _a_ [semi-major axis] is the mean between
the radius of perigee, and the conservation of angular
momentum, and the radius of apogee gives us the so-called
Vis-Viva equation:
v^2 = GM (2/r - 1/a)
Applying this equation to the perigee, we get a Delta-V
of sqrt(GM (2/r(earth) - 1/(r(earth) + 100 km)) = 7.9 km/sec,
but we can reduce this by 0.46 km/s if we launch at the
equator, so this Delta-V is 7.5 km/sec. In the apogee,
we must increase from initial speed of
sqrt(GM (2/(r(earth) + 200 km) - 1/(r(earth) + 100 km))
to final speed of
sqrt(GM / (r(earth) + 200 km))
for a Delta-V of 0.06 Km/s, and a total Delta-V of
7.5 km/sec.
Truth lies somewhere between these two extremes, as the
rocket raises in a curved path.
Now, let's analyse the launching of an ICBM, also under
the simplistic assumption that we give it a boost
and then let it raise and fall.
Let's imagine that we want the ICBM to cross 90 degrees
on the surface of the Earth. This means that [imagine
a figure here] the orbit is an ellipse that crosses
the surface of the Earth on two points, and each of
these points is 180 - 90 / 2 deg away from a central
meridian. In each of these points [launch site and
target site], the speed will be:
v^2 = GM (2/r - 1/a)
Not every velocity will enable us to come back where
we want, because when v approaches zero, the orbital
eccentricity approaches 1.0.
So, we must add a restriction based on the equation
of the ellipse:
r = a (1 - ecc^2) / (1 + ecc cos truan)
[where truan - the true anomaly - is 135 deg, and
ecc is the orbit's eccentricity]
I am lazy and I have a computer, so I get this
minimum Delta-V using brute force and ignorance,
in a loop where I input the semi-major axis _a_
[starting with r(earth) / 2] and increment it until
the equation of the ellipse [now in the form
a ecc^2 + r(earth) cos(135 deg) ecc + r(earth) - a = 0
] has real roots for ecc, and one of them is in the
0 - 1 range.
The solution comes when a = 4790 km,
with two possible eccentricities (0.546773 or 0.606390;
as _a_ goes up, the two solutions diverge, I imagine
that the optimum solution comes when both solutions
converge), and an apogee height of 1000 km, and
an initial speed of 6.4 km/sec.
Conclusion: even though the Delta-V required to send
an ICBM is lower than the Delta-V required to put a
satellite in orbit, the "minimum size" requirement for a
nuke compensates that, so that a satellite program is
slightly more complex than an ICBM program.
Alberto Monteiro
PS: I can draw some pictures, if anyone is still reading.
