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As a follow-up to the two previous questions

What is the scientific reason for the inclination of earth's rotation axis?

and

How much effort would be required to fix the Earth's rotation?

What would be the energy needed to change Earth current inclination of an amount of 1 degree toward the vertical?

I recall a classic novel by Jules Verne where a group try to make the axis completely perpendicular to the ecliptic plane with a huge canon but discover they are off by many order of magnitudes.

Edit:

I looked at a summaries of Verne novels and found the one I was referring too. In fact there are the same characters than in the "From the Earth to the Moon".

A description of the novel I read when I was young: "The purchase of the North Pole"

The Wikipedia article for "From the Earth to the Moon" were computations and assessment of the scientific details are presented

A calculation by John Ptak of the exit velocity of Jules Verne's cannon bullet in "From the Earth to the Moon"

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ogerard
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1 Answers1

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To move the axis of rotation of the earth, you need to change its angular momentum; this in turn means you need to transfer the angular momentum to an object that is not (no longer) attached to the earth.

The "energy" needed for that can literally be anything, since the relationship between angular momentum and energy depends on the moment of inertia (or if you like, the velocity of the cannon ball).

But let's assume we do this by shooting a cannon ball at the surface of the earth, and let's give the cannon an evacuated barrel that reaches to the outer atmosphere, so there is no atmospheric drag (which would cost a lot of energy but it would keep the angular momentum on earth).

Using round numbers:

The mass of the earth is $6\cdot 10^{24}\ \mathrm{kg}$

The radius of earth is $6.3 \cdot 10^6\ \mathrm{m}$

Moment of inertia of sphere = $\frac25 m r^2 \approx 10^{38}\mathrm{\ kg\ m^2}$

Rate of rotation: $2\pi/86400\mathrm{\ s^{-1}}$, so angular momentum is $7\cdot 10^{33} \mathrm{\ kg\ m^2 /s}$.

We need a 1° change in the tilt - that's about 1.8% of the angular momentum, or $2.5\cdot 10^{32} \mathrm{\ kg\ m^2 / s}$.

If we have a cannon at the surface of the earth, pointing in the right direction, the distance is $6.4\cdot 10^6 \mathrm{m}$, so we need a linear momentum of

$$p = \frac{2.5\cdot10^{32}}{6.4\cdot10^6}\mathrm{kg\ m/s}\approx 4\cdot 10^{25} \mathrm{\ kg\ m/s}$$

The energy needed depends on the mass of the cannon ball, since

$$E = \frac{p^2}{2m}$$

With $p$ fixed, the heavier the cannon ball the better. But clearly we need it to reach escape velocity, so there is a lower limit on the velocity. There's also the practical consideration that you don't want to dig such a big hole in the earth to make your cannon (and cannonball) that you generate earth quakes.

If your cannon ball had the mass of the earth, you'd be firing it at 6 m/s. Using just the entire mass of Mount Everest, (considered a 10 km tall cone with base equal to 10 km, volume ~$300\ \mathrm{km^3}$ and density $5000 \mathrm{\ kg/m^3}$ I get a mass of about $10^{20} \mathrm{\ kg}$), you would have to shoot it at a velocity of 400 km/s. The kinetic energy would be about $8\cdot 10^{30}\mathrm{J}$.

You can play around with these assumptions - but the short answer is "forget about it".

Floris
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