Could one fire a bullet with sufficient speed to leave the Earth?

Consider a gun or rifle fired directly upwards. My original question was what speed would be required to escape the Earth.

The escape velocity from the surface of the Earth is the classic $$v_e = \sqrt{ 2GM \over r } \approx 11,000 \text{ m/s}$$ and bullets typically (see for example) leave the muzzle with a maximum speed one order of magnitude lower, ~$1,000$ m/s. Terminal velocity of bullets in STP is another magnitude lower, ~$100$ m/s.

Even if a bullet were fired with speed $v_e$ that of course would not be sufficient due to drag which would slow the bullet down. So a theoretical required speed $v_T > v_e$.

• If the bullet were fired with anything close to $v_e$ or $v_T$, would it would burn up very rapidly in STP? I understand that rockets typically don't achieve anything like $v_e$ until they are high in the atmosphere at least in part for this reason
• And hence is there no speed possible in realistic conditions with which a bullet could be fire and escape the Earth?
• If it is possible, what model of drag should one use to calculate $v_T$ and concretely, does anyone have an estimate of its value?
• – John Rennie Jun 16 '15 at 17:14
• Yes, you could shoot a bullet trough the atmosphere without it burning up. The total mass that needs to be penetrated is equivalent to about 12m of water, which requires a sufficiently dense projectile with a multiple of that mass per area. – CuriousOne Jun 16 '15 at 17:14
• @CuriousOne - 12 meters? Reference, please? – WhatRoughBeast Jun 16 '15 at 18:31
• @WhatRoughBeast: It's just a very rough estimate for the back of the envelope. The average height of the atmosphere is about 8km and the density is about 1.5kg/m^3. I just didn't care to look up the exact numbers. – CuriousOne Jun 16 '15 at 18:33
• 12 m of water is noting but with $K_E=\ m\cdot\ v^2/2$, that squared v will blow everything apart. – Helder Velez Jun 16 '15 at 20:13

As far as I know, there is no data to calculate the drag effects of the atmosphere at these speeds.

The Pascal-B shot of Operation Plumbob did, apparently, launch a 1-ton steel plate at 6 times escape velocity. https://en.wikipedia.org/wiki/Operation_Plumbbob

Nobody has the faintest idea of whether or not it actually made it out of the atmosphere, although the most likely result is that it vaporized.

• You are basically in a deep shock regime where nothing else but the material right in front of you matters. A cone will simply shove those molecules out of the way in a purely kinetic way. There is an additional loss term due to ionization, but I thought those things were rather well understood. Since the exposure to the atmosphere would be very short, less than a second, there is no significant heat transfer in the nose cone, so one does not even need an ablator. The forces on the vehicle would be very strong, though, so in order to keep it intact, it would have to have a high density. – CuriousOne Jun 16 '15 at 18:47
• @CuriousOne - Sorry, I meant "experimental data". 60 kps would be a pretty ambitious wind tunnel, and depending on altitude that's in the Mach 50 regime. And shock tunnels aren't much for durations over a few milliseconds. Not that the atmospheric flight duration would be all that much greater. – WhatRoughBeast Jun 16 '15 at 18:53
• We have seen meteors enter at up to 26km/s. Large enough iron meteors survive that just fine and the heating only melts off a thin layer of their surface. A tungsten nose cone will survive just fine, although in reality one would probably use a pusher design which would create a near-vacuum tunnel in front of the actual vehicle and the exit of the electromagnetic accelerator would be far above sea level, but these are minor engineering considerations. – CuriousOne Jun 16 '15 at 19:05