There is a lot of debris left in space from our activities up there. It makes it very difficult to put satellites or astronauts into orbit. There is also the problem that there isn’t nearly enough time in the day to get everything done that we want to get done. What if I were to tell you that we can fix both of these seemingly disparate problems with one simple solution?

The angular momentum of a spinning system is invariant such that when its rotational velocity decreases, the distance increases, and vise versa. You may have seen this effect with ice skaters.

Thus, if we slow down the rotation of the earth, all of the debris orbiting the earth will move further away, like the ice skaters arms when she slows her rotation. And like the moon.

So, if we slow the rotation down enough, it will send all that debris flying off into deep space, making it some extraterrestrial’s problem. Here’s how we do it:

We calculate the slowdown of earth thus:

and using the ratio of mass density of ocean water to earth:

k = water/rock = 1 g/cm^{3} vs. 5.5 g/cm^{3} = 1.8

and use the moon’s bulge angle:

α = 1/13

we plug in the debris’ mass for m, debris distance from earth’s rotational axis for A, earth’s mass for M, and earth’s radius for r, we get

(45/8)(k){[Gm^{2}A^{3}]/[Mr^{6}]}⋅sin(2α)

=(45/8)(0.18){[(6.7×10^{-11} m^{3}/kg⋅s^{2})(8.4×10^{6} kg)^{2}(7.2×10^{6 }m)^{3}]/[(6.0×10^{24 }kg)(6.4×10^{6 }m)^{6}]}⋅sin(2/13)

= (0.0027)⋅{[3.4×10^{10}]/[4.1×10^{65}]}

= (2.7×10^{-3})⋅(8.3×10^{-56}) = (2.2×10^{-58})s^{-2}

which means that the earth’s rotation about its axis is decelerating by 2.2×10^{-58} radians per second every second due to tidal forces caused by the space debris. At that rate, it will take

(7.3×10^{-5 }rad/s) / (2.2×10^{-58} rad/s^{2}) = 3.3×10^{53 }seconds or 10^{46} years for the earth to slow to a stop and all the debris to go flying off.

But if we want to slow the earth’s 7.3×10^{-5 }rad/sec rotation down to 0 rad/sec, we’ll need to use some energy. Why not use petroleum gas energy?

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We can get the rotational kinetic energy of the earth rotating on its axis with:

KE_{rot} = (Iω^{2})/2

where

I = moment of inertia: I = (2mr^{2})/5 for a uniform sphere (earth approximation)

m = the mass of the earth, which is 6.0×10^{24 }kg

r = the distance from its pivot, which is the radius of the earth: 6.4×10^{6 }meters

ω = rotational velocity at the earth’s surface in radians/time, which is:

2π/24hr = 2π/86,400s = 7.3×10^{-5 }rad/sec

So:

KE_{rot} = (Iω^{2})/2 = ([(2⋅(6.0×10^{24 }kg)⋅(6.4×10^{6 }m)^{2})/5]⋅(7.3×10^{-5 }rad/sec)^{2})/2

= [(1.5×10^{31})⋅(5.3×10^{-9 }rad/s^{2})]/2 = [8.1×10^{22} kg⋅m^{2}/s^{2}]/2

= 4.0×10^{22} Joules

We get 1.25×10^{8} Joules/gallon from petroleum gasoline

So we would need

4.0×10^{22} Joules / 1.25×10^{8} Joules/gallon

= 3.3×10^{14} gallons of petroleum gas

And there are an estimated 1.5×10^{12} gallons of petroleum left on earth…which isn’t nearly enough. Oh well. I guess we’re stuck with the junk… in SPACE!