# Energy required to tilt Earth’s axis by 90 degrees? [closed]

How much energy would be required to tilt Earth’s rotation axis in such a way that what is now poles would be where to equators are relative to Sun and vice versa?

• The poles are points; the equator is a circle. You can rotate about 2 perpendicular axes to achieve two different final configurations. You'll also need to read up on "precession of a gyroscope". Further, there's a big difference between rotating the axis and changing the axis of rotation. Which do you want? Feb 9, 2021 at 15:05

Assuming you are preserving the magnitude $$L$$ of the spin angular momentum, a $$\pi/2$$ flip of the earth will result in an angular momentum change of $$\sqrt{2} L$$. The magnitude of the torque necessary to achieve this is time $$\Delta t$$ is $$\Delta \tau = \frac{\sqrt{2} L}{\Delta t}.$$ The work necessary is $$\Delta W = \Delta \tau \Delta \theta = \frac{\sqrt{2} L}{\Delta t} \cdot \frac{\pi}{2} = \frac{L\pi}{\sqrt{2}\Delta t}.$$ Further assuming the earth is a perfect sphere rotating around an axis connecting its poles, $$L \approx 7 \times 10^{33} \, \mathrm{kg m/s}$$ (see this). This means the work done as a function of the total time taken to flip the earth is $$\Delta W = \frac{1.55 \times 10^{34}}{\Delta t}.$$ You can put whatever value of $$\Delta t$$ in there; in particular assume you can rotate the earth in $$1$$ second. In that case, $$\Delta W = 1.55 \times 10^{34} \, \mathrm{J},$$ which is a lot of energy.
• Note that Earth's gravitational binding energy is $2.49 \times 10^{32}\,\mathrm J$, so you need to be careful how you twist it. ;) Feb 9, 2021 at 16:04