# Assuming I could clip through the earth, what is the delta-v requirement to reach any orbit?

Nevermind how I intend to clip through the surface. You could say I intended to drill a tunnel kind of like the drop through the earth problem.

I know how to do this calculation assuming all of earth's mass is at its center, and this yields the ridiculously low delta v requirement of 460m/s (velocity of earth's rotation) to drop PE to zero relative to the core, an infinitesimal amount to raise AP to 1.5 million km, and 515.7m/s (Vorbit = sqrt(G * M / R)) to establish orbit. If I ignore Oberth and other stupid tricks altogether I get the simple answer of 7207 m/s (velocity required for LEO - velocity of earth's rotation) to increase my eastward velocity adequately.

But the earth isn't a point mass so that calculation is stupid wrong. I can't get infinite Oberth from being at the center; however the effective Oberth value is still very large, but I can't figure out how to estimate what it is.

[Deliberately ignoring moon assist and other such maneuvers; that's not the point of this question.]

So I've been working on this and realized I made a mistake; I accidentally assumed circular orbits where I had intended only orbits completely clear of the ground. Thanks to a tip from Cosmas Zachos, I now think I asked a question I am unlikely to recall enough calculus to understand the answer to.

• What you mean by "clip" is unclear. If you are thinking of a tunnel though the center of the earth then you can never reach orbit because you need a tangential velocity to the radius of the earth to get into an orbit you will need rocket propulsion for tangential velocity Otherwise you will just be oscillating through the center of the earth., after reaching 0 kinetic energy at the extremes.i f you are thinking of a tunnel not passing through the center you will need rocket propulsion not to hit the walls – anna v Jul 25 '17 at 5:57
• According to the shell theorem (en.wikipedia.org/wiki/Shell_theorem), since the Earth is very nearly spherically symmetric, its gravity is the same as a point mass for anything outside the normal Earth radius. So no, the point-mass approximation is not "stupid wrong," it's actually very accurate. – probably_someone Jul 25 '17 at 7:59
• Can you add more detail as to how you calculated each number (i.e. which formulas were used)? – probably_someone Jul 25 '17 at 7:59
• @probably_someone: "clip through the earth" is important here. It's stupid wrong when computing for 100m from the center of the earth. – Joshua Jul 25 '17 at 15:33
• It might be conceivable you are thinking of the obverse of this one, 285689, but one cannot tell... too many unspecified terms. – Cosmas Zachos Jul 25 '17 at 19:29