# 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