What is the necessary amount of energy to propel a 100g from ground to orbit ? (Slingatron) I would like to know what is the energy needed considering the air resistance to shoot from ground to the orbit a 100g object (even if it vaporizes in the process).
Variables to make approx calculations:
Lets assume it has a pointy bullet shape (best know shape) with 5cm diameter x 10cm length and the orbit is the lower possible, lets assume the ISS orbit 150km above ground, we are at equator latitude shooting to the east (Earth spining thrust, 1500km/h)
The first thing that crossed my mind was the russian meteor entry and its tremendous bright and shock wave (dissipated energy)
According to a news on physicsworld.com
 A: This class of problem is very simple or very complicated, depending on the ratio of the momentum of the bullet to the momentum lost by air drag.  You have chosen the unfortunate latter case.  Basically, if the bullet's velocity decreases significantly you'll need some calculus for it.
In the "simple" case where the velocity doesn't drop much, we'll wind up reducing a lot out and using the mass-thickness of the atmosphere, $\mu \approx \rho h$.
$$ F_d = \frac{dE}{dh} = \frac{1}{2 } \rho A C_d v^2$$
With a little magic...
$$ \Delta E = \frac{1}{2 } \mu A C_d v^2$$
To get the numbers:

Lets assume it has a pointy bullet shape (best know shape) with 5cm diameter and the orbit is the lower possible out of atmosferic resistance, lets assume the ISS (150km).

$$C_d = 0.42$$
$$ A = \pi \left( 2.5 cm \right)^2 $$
$$ \mu =  10 \frac{ tons }{ m^2 } $$
$$ \frac{1}{2 } \mu A C_d = 3.75 kg $$
This is very bad for you, because your bullet weighs $100 g$.  What's the formula for kinetic energy of the bullet at launch?  Why that's $\frac{1}{2} m v^2$ of course.
Let me use a qualitative picture.  Imagine that the atmosphere is a coherent ceiling with no significant depth.  You bullet has to punch through it, but it makes a "perfect hole" when doing this.  In short, your bullet's momentum is decreased because it accelerates the mass in that hole to its same velocity.  It picks up that hole and keeps moving with it.  You should look at this as basic momentum balance.  If the ceiling's area within that hole is twice the mass of the bullet, the bullet's velocity will be decreased by half.  Here, the mass of the hole it has to punch is 37.5 times the mass of the bullet.  I hope you see where this is going.
With this mental model, the answer is quite obvious.  The velocity of the bullet must be 38.5 times orbital velocity.  That's because the combined mass of the bullet and the air that it "picks up" in its trip is 38.5x the mass of the bullet.
But this would only work if you shot the bullet straight up at orbital velocity.  This would be quite useless.  In actuality, you want to shoot it at a shallow angle - as shallow as possible so that the boosters that are used to circularize the orbit don't have to be as big.  This correction is just dividing by a sin function.  If you shoot 90 degrees straight up, then the above ratio holds.  If you shoot at an angle lower than this, then divide by sin of the angle (or multiply cosec), and that's the ratio for the extra mass of air you pick up.

lets assume the ISS orbit 150km above ground, we are at equator latitude

This establishes that the apogee should be at the 150 km elevation, if you wanted to do this with only one rocket burn.  That actually doesn't fully specify the problem.  The perigee is unknown, but we know that it can not be higher than the surface of Earth.
Still, that doesn't eliminate the possibility of shooting it straight up and doing the rocket burn for the full orbital velocity.  Even at a 45 degree shooting angle, the orbit would be extremely elliptical, and thus very far from full orbital velocity.
Nonetheless, even if you did shoot it straight up, you would require $1.2 km/s$.  Then using the air resistance multiplier, that would be $46 km/s$, which is just plain impossible to produce (except with nuclear bombs).
In short, you would send something heavier to get around this problem.
