Can a "free launch" from a space elevator really be free?

Question

The question Benefit of sling shot effect with a space elevator implies that you can get free $\Delta V$ from a space elevator launch. As highlighted in this answer on SciFi quoting Sheffield book "The Web Between the Worlds", and this answer on Space The launch is not really free, it is stealing rotational energy from Earth.

So two questions, given a space elevator launch from 100,000 kilometers, ignoring the energy cost getting mass to and from the launch point, and assuming all mass involved goes up and down the space elevator:

1. How much mass would need to be launched for the Earth to lose 1 second of rotational speed?

2. How much mass would need to be lowered to the Earth to gain 1 second of rotational speed?

For prospective, on potential launch mass.

• The Space Shuttle Orbiter has a fully loaded weight of about 120 tons

• The USS Enterprise (CVN-65) (an American aircraft carrier) has a fully loaded weight of about 94,781 tons

Answer 1

This is a simple calculation in the conservation of angular momentum.

The angular momentum of uniform sphere is ${2\over 5}MR^2\omega$. We will assume that Earth is a uniform sphere (close enough for this question), so $M=5.972\times 10^{24}\,\mathrm{kg}$, $R=6371\,\mathrm{km}$ (mean), and $\omega=7.292\times 10^{-5}\,\mathrm{s}$. So the angular momentum of the Earth is $L_e=7.071\times 10^{33}\mathrm{kg\,m^2\over s}$.

The angular momentum of a small object rotating about a point is $mr^2\omega$. $\omega$ does not change when climbing the elevator, just $r$. So the change in angular momentum is $\Delta L=m\omega(r_f^2-r_i^2)$. $\omega$ is that of Earth above.

To change the Earth's rotation rate by $1\,\mathrm{s}$, we need to change its angular momentum by $1\over 86164$ of what it was, so $\Delta L=8.206\times 10^{28}\mathrm{kg\,m^2\over s}$. Assuming a space elevator at the equator, and using the equatorial radius of the Earth, $6378\,\mathrm{km}$, and $100000\,\mathrm{km}$ above that, we get $\Delta L=8.206\times 10^{28}\mathrm{kg\,m^2\over s}=m\omega\left(\left(106378\,\mathrm{km}\right)^2-\left(6378\,\mathrm{km}\right)^2\right)$. To the extent that $m$ reduces (or increases) $M$, there is also a small change in the rotation rate for the given angular momentum when it is released. However that is negligible here.

This can be simplified and approximated, in order to better see the sensitivities, as: $m={2\over 5}M{\Delta T\over T}{R^2\over H^2}$, where $T$ is the rotation period of the Earth, $\Delta T$ is the small amount by which you want to change it, and $H$ is the radius of the release point from the elevator. $m$ is the amount of mass to let go of at the release point to effect a change in rotation period of $\Delta T$.

That gives $m\approx 10^{17}\mathrm{kg}$. About a billion (US-style) of your aircraft carriers.

If each aircraft-carrier size spaceship only carried a handful of people, then we could evacuate the entire human population from the Earth, only slowing Earth down by one second. So I would say, for all practical purposes, yes, a space elevator is free. You just have to pay the energy to climb to that altitude, which is quite small compared to the energy from the velocity that you get.

Answer 2

FWIW, I'm with Arthur C. Clarke on this whole "space elevator" business. There is no "free lunch" (or is it "free launch"?) to be had with any "space elevator" that is supposedly capable of taking a payload with non-negligible mass and with non-negligible air resistance in the lower atmosphere to the stratosphere and beyond. It doesn't really matter then where we put this alleged "space elevator", so I'll take a few shortcuts to answer this as easy as possible. Take it as it is, I have no interests in discussing "space elevators".

For your first question, there is no upper limit, short of removing the total of the Earth's mass. The remaining mass would still have radial velocity equal to the total mass from before, only it's kinetic potential will be lowered as the mass spinning decreases, so it would be more prone to changes in its rotation velocity, if we added some of that mass back later on.

The second question is also relatively easy, if we assume the kinetic potential of the added mass in the direction of the Earth's rotation is equal to zero, i.e. it's lowered directly perpendicular to the Earth's surface and dropped at either of the poles (another answer by @MarkAdler already includes calculations for dropping the mass at the equator):

$$\frac{duration\ of\ Earth\ day\ in\ seconds}{Earth's\ total\ mass} = x \left(\frac{duration\ of\ Earth\ day\ in\ seconds + 1\ }{Earth's\ total\ mass}\right)$$

So this comes out as $1/86400$  of the Earth's mass ($5.972 * 10^{23} kg$), which is $6.9123 * 10^{18} kg$, if my data input was correct. But please do check again. ;)

But assuming my pocket calculator doesn't lie, this would come out as:

• 72,929,329,875 (nearly 73 billion) of USS Enterprise (CVN-65) aircraft carriers
• 57,602,623,456,791 (a good 57 and a half trillion) of fully loaded Space Shuttle Orbiters