Can One Harness Somehow Energy From Earth

Can't we harness energy from the rotation of The Earth?

Hmmmm.Try this. Get in a car,don’t start the engine, try to move it by pushing it forward. No matter how hard you push the car from inside, it won’t move an inch. To move the car, you need to get out and push from outside.You might be wondering what does that have anything do with the question.well, A lot.When you push the car from inside, you get the reaction force from the car, but from outside, you get the reaction force from the ground.When the things changes their state of motion, they requires Force and Force comes from energy. So if you want to transfer the energy (which is Kinetic energy in this case), relative state of motion of two bodies must be different from each other.If you have fuel for 1 car up to 100 km, you cant drive two cars with the same fuel up to 100. Other car can proceed from where the first one stops, but in order to drive second car, first one needs to stop.Image Source:- Youtube.When the first ball hit the other three, it transfer the energy to 4th ball but itself comes to rest. These two balls can not have same displacement simultaneously provided energy is constant. If first one moves, fourth comes at rest and vice-versa.Right now on earth, we are moving with the same speed as the Earth and we are carrying the same momentum. Earth is rotating for sure but from our prospective, it is stationary and you can’t harness kinetic energy from stationary things. can you?. Therefore we cannot harness the energy from Earth’s rotation. We need to get out of the car.

How much energy could we harness by slowing the Earth's rotation period to 1/360 year?

Let's assume circular orbits. The total energy of the Earth in its current orbit is given by[math] E_{0} = -\frac{GMm}{2r} [/math] where M is the mass of the sun, m the mass of the Earth and r the current distance between the sun and the Earth. You want to decrease the current period of rotation by a factor of 0.9856 to get 360 days. One of Kepler's Laws (can't remember which) says that [math] T^2 \propto R^3 [/math]. So this means that you have to decrease the radius by a factor of (0.9856)^3/2 = 0.9785. Therefore, when you slow the Earth's rotation down to 360 days, you get [math] E = -\frac{1}{0.9785}E_{0} [/math]If all the difference in energy gets converted to useful energy, you would have taken[math] \Delta E = -(\frac{1}{0.9785} - 1)E_{0} [/math]which works out to about 3 x 10E31 J if I did all the math correctly, enough to power the Earth at its current power consumption for 100 billion years.ORyou can slow down the revolution speed of the Earth so that 1 day coincides exactly with one rotation around the Sun. To do this, you need to decrease the angular speed by a factor of 72/73. Approximating the Earth to be a sphere, the loss in rotational energy is then[math] \Delta E = \frac{1}{2} \cdot \frac{2}{5}mR^{2}_{E} ( 1 - ( \frac{72}{73} )^{2} )\omega^{2} [/math]where [math] \omega [/math] is the current angular speed of the Earth. If you put all of that together, and assuming I've done the math right again, that gives 7 x 10E27 J, enough to power the Earth at its current power consumption for 20 million years. (This assumes the sidereal day is 24 hours long.)