Yesterday, I read the NASA post The Tyranny of the Rocket Equation. Very interesting read.

In there, it states:

The rocket equation contains three variables. Given any two of these, the third becomes cast in stone.
They are

  • the energy expenditure against gravity (often called delta V or the change in rocket velocity)
  • the energy available in your rocket propellant (often called exhaust velocity or specific impulse) and
  • the propellant mass fraction (how much propellant you need compared to the total rocket mass).

This led me to think: the first factor, $\Delta$V, is the main obstacle in being able to leave Earth. So, if we had a means to alter the rate of spin of the Earth using magic, could we defeat it? Or would we just get thrown out in a chaotic manner, to our eventual death due to loss of control?

Note: Assume the magic is specific to the spin of the Earth and nothing else; i.e. we don't have generalized energy/magic of levels required to alter planets.


closed as off-topic by Aify, Hohmannfan, a CVn, T3 H40, Gianluca Apr 16 '16 at 19:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

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    $\begingroup$ Umm.... the Earth's spin doesn't affect the gravitational pull of the Earth on your rocket... In fact, the Earth's spin has nothing to do with that equation at all. $\endgroup$ – Aify Apr 16 '16 at 8:27
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    $\begingroup$ It may help us answer this question if you explain what you think the connection between the Earth's spin and escape velocity is. $\endgroup$ – 2012rcampion Apr 16 '16 at 10:44
  • $\begingroup$ @2012rcampion centrifugal force at the equator, of course. $\endgroup$ – cst1992 Apr 16 '16 at 11:03
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    $\begingroup$ Since you simply posit "magic", the magic clause applies. Specifically: "If your question involves magical elements, you must provide enough details about the magic system in question so that answers may be given that conform to the particular magic system." Specifically in this case, if we posit the ability to arbitrarily speed up or down Earth's rotation, why are we not able to use that to equally arbitrarily speed up a spacecraft? Seems to me like with that, we wouldn't need rockets at all, so the rocket equation doesn't apply! $\endgroup$ – a CVn Apr 16 '16 at 12:01
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    $\begingroup$ Centripetal acceleration gives the illusion of making the rocket lighter. What it does is gives the rocket a portion of the velocity needed to achieve orbit. Why you can't use this to help launch your rockets is if you rotate the Earth fast enough to launch rockets without fuel, you are also rotating fast enough to launch everything else too (air, oceans, cars, people, etc.). $\endgroup$ – Jim2B Apr 16 '16 at 12:47

If the Earth were a rigid sphere, and we spun it up to:

$$ a_c = r \omega^2 \\ a_c \approx g \\ \omega = \sqrt{\frac{1~g}{R_\text{E}}} = 1.2\times 10^{-3}~\text{rad}\cdot\text{s}^{-1}= \frac{1~\text{revolution}}{1.4~\text{hours}} $$

Then the centrifugal acceleration would roughly balance out gravity, and objects on the equator would appear weightless: essentially, they would be in orbit!

This is a problem.

The Earth contains a lot of things that are not nailed down (like the atmosphere) that are normally held down by gravity[citation needed]. On the super-spinning Earth, centrifugal force balances gravity and these things are no longer held down. You can probably already see where this is going.

However, it turns out that it's even worse! Rocks seem pretty rigid to us, so we might be tempted to think of the Earth like a rigid object. However, on a planetary scale materials like rock behave like fluids (not to mention that parts of the inside are actually molten). This means that centrifugal force causes the equator to bulge outward. To first order, we can approximate the flattening of a rotating, self-gravitating body composed of incompressible fluid by:

$$ \frac{15\pi}{4GT^2\rho} $$

In real life this formula gives a value of about $1/230$ for the Earth (the true value is closer to $1/300$ due to the fact that the core of the Earth is denser than the surface). For the super-spinning Earth this formula gives a value of $1.25$. For such an extreme value this formula no longer applies and the actual value is much worse. The Earth would probably look more like a pancake.

A useful comparison is to look at the amount of energy we've added to the Earth by spinning it up this fast.

$$ E = \frac{1}{2}I\omega^2 = \frac{2MR^2 \omega^2}{2\cdot 5} = \frac{g M R}{5} = \frac{G M^2}{5R} < \frac{3GM^2}{5R} $$

The term on the right is the gravitational binding energy. The fact that we have only added about a third of the gravitational binding energy means that the pieces of the Earth are likely to stay gravitationally bound: that is, they won't all fly off into space. Basically, we've turned the Earth back into a protoplanetary disk. Eventually, after they've dissipated enough energy (probably only 100 million years or so), the pieces will collapse back into a planet a little smaller than the Earth. Of course, it will take another one or two billion years for the molten mass to cool enough for oceans to form.

So in summary, if the Earth was spinning fast enough, then yes, we would all be flung into space, and this is pretty much as bad an idea as it sounds.

  • $\begingroup$ Very nice answer. Another reminder of how delicately balanced Earth really is. $\endgroup$ – cst1992 Apr 16 '16 at 14:42
  • $\begingroup$ Facts + funny a winning combo +1 $\endgroup$ – Jim2B Apr 16 '16 at 23:02
  • $\begingroup$ Another point to consider is how long it takes for the Earth to magically spin up and down. 'Cause if it's near-instantaneous, everything not nailed down would (appear to) fly off to the west as the solid Earth accelerates out from under it. So if you happened to be inside a stoutly constructed building, you'd splatter against the wall. $\endgroup$ – jamesqf Apr 17 '16 at 5:22

If we had energy on that scale and control over planetary sized masses, then yes, launching a sattelite would be trivial.

Looking at the Earth's rotation in isolation, let's see:

If you launch at the equator towards the east, you get 1000 miles per hour velocity for free, towards your orbital speed requirement of 17,500 miles per hour.

If you turn off the spin, you lose that. Why would you think that would be helpful? Note that you can launch towards the west, reversing the benefit; or launch at an angle near north where it doesn't help or hurt, and get a polar orbit.

Also, look at other Answers concerning what happens to the air, water, and stresses in the crust and mantle when the Earth's rotation is stopped. That's a lot of wreckage to put up with to just launch a package.

  • $\begingroup$ You are broadening the scope too much. We can control only the spin of the Earth, nothing else. $\endgroup$ – cst1992 Apr 16 '16 at 8:31
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    $\begingroup$ That sounds like a classic way to make a lame story. You don't get just one isolated extroninary thing; it's part of an infrastructure, and it finds other uses, etc. $\endgroup$ – JDługosz Apr 16 '16 at 9:47
  • $\begingroup$ We aren't just turning it on or off. We're able to speed it up or slow it down. $\endgroup$ – cst1992 Apr 16 '16 at 11:06
  • $\begingroup$ "Note that you can launch towards the west, reversing the benefit" Of course, if you want your satellite in a retrograde orbit, that's just about the only realistic way to do it with modern technology. Thankfully, most satellites are equally happy (or happier) in a prograde orbit, so this isn't needed very often. $\endgroup$ – a CVn Apr 16 '16 at 12:16

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