If we were to build a high speed rail up the side of a mountain like in some Science Fiction movies, what is the velocity needed at the point of leaving the mountain excluding angular momentum from Earth’s rotation to achieve escape velocity?

  • $\begingroup$ Is this actually a homework question? If so, you really should expand on it - I don't mean providing specific numbers (which I edited out because they're irrelevant for our purposes), but include what you've tried and what exactly you are having trouble. If it's not a homework question, it would still help to say what you're having trouble with, and perhaps mention why you're asking it. In any case, it's very close to being closeable as too localized, but I'll let it stay for now. $\endgroup$ – David Z Jul 14 '11 at 17:26
  • $\begingroup$ I was just trying to have some cooperative work. My intent was to ask a series of questions related to launching of a mountain. But you edited out the specifics. Am 60 I don't do homework. Please just close the question it has no relevance now. $\endgroup$ – Fortunato Jul 14 '11 at 20:44
  • $\begingroup$ This is actually not a bad question as it stands now. The key is that it has to invite answers which show how to do the calculation rather than just giving a result. You don't need the numbers for that. I'd also add that this is a place to get questions answered, not a site to get people working together. $\endgroup$ – David Z Jul 14 '11 at 22:34
  • $\begingroup$ Whiteout the location of the mountain the answer is incorrect. The diameter relative to the rotation at that latitude is not that same as on the equator. The question posted is not the question I asked. $\endgroup$ – Fortunato Jul 15 '11 at 1:27
  • $\begingroup$ Right, the question as it's posted now is not the question you asked, it's a more general one. What you originally asked is just a special case; any correct answer to the question in its current form will also answer the original version. So I don't see what you're saying about something being incorrect... if you'd like to continue discussing it we should probably move to the chat room. And remember you are free to continue to edit your own question. $\endgroup$ – David Z Jul 15 '11 at 3:07

The total escape velocity is about 11,200 m/s (approximately 7 miles/sec.) in a direction roughly tangential to the earth's surface. At 30 degrees north latitude (e.g., somewhere in southern Texas) the eath's spin would contribute about 400 m/s to the tangential velocity, so actual speed, relative to the earth's surface at said launch point would be about 10,800 m/s, if the load was launched in a roughly tangential direction, eastward; closer to the 11,200 m/s if you went straight up with it.

The height of the mountain will serve a good purpose in getting you up to where the air is thinner when you leave the track; but, otherwise has negligible on how fast you need to go.

The main problem with this is that 11,200 m/s, at 14000 feet, is about Mach 24; roughly 13 times the muzzle velocity of a 30-06 high-powered rifle. Heating, shredding, melting, ablation, and all sorts of bad stuff occur at these speeds, even in thin air.

Studies have been done on this, to engineer a ground-power-assisted launch. The idea is to get the load up to a very high speed --but not high enough to shred or melt it-- and let the rocket engines take over from there. I believe I read about a track hypothesized for somewhere in the vast wastelands of Texas.

  • $\begingroup$ The height of the mountain becomes more critical once your realize the amount of speed you need at the relise point so I was going to ask next is the propulsion needed to live the ramp at just under mach 1. But the sight controllers don't to. $\endgroup$ – Fortunato Jul 14 '11 at 21:02

Escape velocity has got nothing to do with mass of the rocket.

$v_e = \sqrt{\frac{2GM}{(r_0+h)}}$

$r_0 -$ Radius of the earth
$h -$ Height of the mountain

  • $\begingroup$ $G$ - Gravitational constant; $M$ - Mass of Earth. $\endgroup$ – Nic Jul 14 '11 at 11:35
  • $\begingroup$ Escape speed, not Escape Velocity :) $\endgroup$ – Bernhard Heijstek Jul 14 '11 at 13:41
  • $\begingroup$ This is true at the equator where do you account for the 39 N $\endgroup$ – Fortunato Jul 14 '11 at 20:48

An intuitive way to think about it. Escaping an inverse square law to infinity requires energy (kinetic energy is this example) equivalent to the force of gravity times the radius. So the number od gees you would need is the ratio of track length divided by the radius of the earth. You have the additional complication that you must curve from horizontal velocity to velocity at and angle, and this means you are also dealing with centripetal forces during the transition from plain to mountain.

  • $\begingroup$ I WAS going to deal whit this in subsequent questions but our leaders don't like the question. $\endgroup$ – Fortunato Jul 14 '11 at 20:53

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