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The escape velocity of earth is roughly $11 kms^{-1}$. However, what if a long ladder was built extending out of Earth's atmosphere and considerably more. Then if something was to climb up at much less than the escape velocity, what would happen when it reached the end?

And what if the object that climbed the "ladder" then fired some kind of thruster/rocket and was going fast enough so that it orbited the Earth. It would mean less energy required to get into orbit?

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"Escape velocity" is really just a measure of the kinetic energy an object near the surface of the Earth would need to have to start with in order to just run out of energy at the point where it was infinitely far from Earth, having converted all of its initial kinetic energy to gravitational potential energy.

Even if you built a giant ladder or a space elevator or whatever, the total energy required to get to orbit is exactly the same, it just comes in a less spectacular form. Rather than burning a rocket the whole way, you would be doing a slower conversion of energy into gravitational potential energy-- electricity running a motor to winch the rocket up to the top of a space elevator, or chemical energy from food as you climbed a bazillion stairs to get there, or whatever.

A space elevator would be an attractive way to get a rocket into orbit, or away from the Earth because it reduces the amount of rocket fuel you need to use to get there, replacing it with some other source that is more convenient (and less explosive) to work with. But you still need the same total amount of energy to get your payload into orbit.

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  • $\begingroup$ Then how about balloons that everybody attached a camera to and sends to the edge of space there is energy to inflate the balloon on the ground but this can not equal the amount of energy for the balloon to rise? $\endgroup$ – Jonathan. Nov 20 '10 at 0:22
  • $\begingroup$ I'm not quite sure what you're asking. In the case of a balloon, the lifting force comes because the balloon displaces a large volume of air, but has a mass less than the mass of the displaced air. This generates a buoyant force that pushes the balloon up. It won't get you all the way to orbit because it can only work as long as there is air of sufficient density to displace. $\endgroup$ – Chad Orzel Nov 20 '10 at 1:22
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    $\begingroup$ Your answer is correct, but I think it has some important caveats. Imagine turning gravity off and again comparing a rocket to a ladder. If you want to add energy to the rocket, you fire its rockets and burn fuel. But if you want to gain energy while climbing the pole, you do nothing at all. The Earth is spinning, and this spinning will hurl you out away from the Earth like a lacrosse ball. All you have to do it sit there and go along for the ride. The same amount of energy is used to get you to a given velocity, but in one case it is stolen from the Earth's kinetic energy. $\endgroup$ – Mark Eichenlaub Nov 20 '10 at 5:00
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    $\begingroup$ On those grounds, I disagree with "you need the same total amount of energy to get your payload into orbit", since surely we're not interested in counting energy taken from the Earth. Further, with a rocket the problem isn't explosiveness. It really does take a huge amount of energy to get a rocket into orbit because you waste lots of energy accelerating the fuel and carrying the fuel up higher into the gravitational field with you. These are fundamental limitations. For a fixed exhaust velocity, rockets can only be so efficient, and that efficiency is far less than for a space elevator. $\endgroup$ – Mark Eichenlaub Nov 20 '10 at 5:02
  • $\begingroup$ Well, my response would be that the 11 km/s escape velocity figure doesn't take any of that into account, either. You get that number by just equating the kinetic energy to the gravitational potential at the Earth's surface. So I think it is true that you use the same amount of energy raising a given amount of mass out of the Earth's gravity well, regardless of whether you do it with a rocket or an elevator, in the very simple approximation that goes into the escape velocity calculation. $\endgroup$ – Chad Orzel Nov 20 '10 at 14:50
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If your ladder were the height of a geostationary orbit, which is about 6 times the radius of the Earth, then when you got to the end you could step off and be in orbit. If it were lower, you would need to add energy to obtain a circular orbit.

The idea you're talking about is essentially a space elevator. If it were feasible, it would be a far more efficient means of getting to orbit than chemical rockets. Unfortunately, it is not feasible at this time for a variety of engineering reasons.

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  • $\begingroup$ But how could this work in theory because the object climbing is not going faster than the escape velocity $\endgroup$ – Jonathan. Nov 20 '10 at 0:17
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    $\begingroup$ @Jonathan: as explained in Chad's answer, escape velocity is simply the velocity needed to escape the Earth's (or any body's) gravitational field with no further propulsion. It does not mean that anything going slower than the escape velocity can't get into orbit. $\endgroup$ – David Z Nov 20 '10 at 0:25
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    $\begingroup$ There's a guy who often shows up in my university town who uses the escape velocity argument to demonstrate that the moon landings and all space travel are a big government conspiracy. That's not the only flaw in his argument(s). Funny dude, but totally off his rocker. $\endgroup$ – Raskolnikov Nov 20 '10 at 10:29
  • $\begingroup$ Could the down vote please explain? $\endgroup$ – Mark Eichenlaub Dec 5 '10 at 6:53
  • $\begingroup$ @Jonathan. the space elevator concept is not about achieving or circumventing escape velocity; it is about achieving orbital velocity... not quite the same thing. The exit point of the typical space elevator concept is at geostationary orbital altitude - some 22,000 miles above the equator. An object in orbit, and the exit point of the space elevator (at that altitude) would have a velocity of something like 3,400 miles per hour - far less than escape velocity. Maintaining a lower orbit (like that of ISS) requires much higher velocity (~17,000 mph); perhaps that is the cause of confusion. $\endgroup$ – Anthony X Dec 9 '13 at 3:16
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Energy required to escape from the earth's gravity is

$\frac{GMm}{R} = mgR$

Now you may pay it in installments, you may pay it slowly but you may not pay less.

Escape velocity from point at height $h$ above the earth's surface is $V_{eh} = \sqrt{2g(R-h)}$.

Case I: Launching rocket from the surface of the earth i.e. h = 0

Total energy = Energy required to launch = $\frac{mv_{e0}^2}{2} = \frac{2mg(R+0)}{2} = mgR $

Case II: Launching rocket from height $h$ above the surface of the earth

Total energy = Energy required to climb upto height $h$ + Energy required to launch

$mgh + \frac{mv_{eh}^2}{2} = mgh + \frac{2mg(R-h)}{2} = mgR$

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    $\begingroup$ The energy required to climb to a given height along a space elevator can come partially from the kinetic energy of Earth's rotation. Hence, you do not actually have to "pay" all the energy. See comment on Chad Orzel's post. $\endgroup$ – Mark Eichenlaub Nov 20 '10 at 9:31
  • $\begingroup$ Yes. I read that answer. So earth helps. I neglected it for simplicity. Thanks! :-) $\endgroup$ – Pratik Deoghare Nov 20 '10 at 10:22
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Don't know how tall the ladder is, but it would get easier and easier to climb the ladder as you went up, because the gravititational force goes down as distance from the earth's center of mass increases.

If you're already up at some orbit height, you need only to thrust along your intended orbit until you reach the orbital velocity. No need to thrust to get up to altitude since you're already there.

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    $\begingroup$ It would get easier to climb the higher you go for two reasons. One is that gravity gets weaker. $g = GM/R^2$ so $dg = -2g/R dR$. The other is the centrifugal acceleration $f = \omega^2 R$ so $df = \omega^2 dR$. Their ratio is $2g/R\omega^2$, so the change in the centrifugal term compared to the change in gravity is weaker by about the same factor as the size of the centrifugal force compared to gravity - a factor of about 1000. $\endgroup$ – Mark Eichenlaub Nov 20 '10 at 0:16
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When you reach the end of the ladder you will fall back to earth, unless you have reached geo-synchronous orbit. Just compare the energy expended climbing with that required to reach orbit (which is more that the energy to just escape gravity vertically). If the ladder is longer than geo-synchronous orbit then you are no longer expending energy climbing, but on the contrary the earth is giving you energy. If you let go of the ladder you will fly out in space overcoming gravity.

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    $\begingroup$ You will not necessarily fall back to Earth if you're below geosynchronous orbit. You will fall in an elliptical path, which, if you're high enough that it your path does intercept the atmosphere, will simply become and elliptical orbit. $\endgroup$ – Mark Eichenlaub Nov 20 '10 at 5:35
  • $\begingroup$ Correct, remember the space elevator project... $\endgroup$ – John Alexiou Nov 22 '10 at 4:08
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Short answer: climbing such a ladder is acceleration. This is because the ladder is stuck into spinning earth, and you get higher speed by going up to higher radius.

Regardless of spinning, escape velocity does depend on height. If you are static at infinity, you already have escaped.

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  • $\begingroup$ surely the closer to the earths center the faster you will be going? $\endgroup$ – Jonathan. Nov 30 '10 at 21:33
  • $\begingroup$ I beg your pardon? $\endgroup$ – Pavel Radzivilovsky Dec 1 '10 at 6:53

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