Classical Mechanics

Proofs of Kepler’s Laws

Kepler’s law is what concerning the movement of a planet. If you can master this rule, you can easily think about the movement of the planet. And in order to think about this, it is the quickest to introduce the motion equation of two-dimensional polar coordinates. In this post, I’ll introduce Kepler’s law and prove it.

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Centrifugal Force and Centripetal Force

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In order to understand the centrifugal force and the centripetal force, it is first necessary to understand the imaginary force that is related with observer’s condition.

The centripetal force is the real force required for the circular movement of the object. On the other hand, the centrifugal force is the imaginary force that works when the observer moves with the object.

The direction of the centripetal force and the centrifugal force are diametrically opposed, and their magnitudes are the same. Therefore, when the observer moves with the object, it seems that the centrifugal force and the centripetal force are balanced.

The magnitude of the centripetal force or the centrifugal force \(F\) is

$$F=mrω^2=m\frac{v^{2}}{r}.$$

\(m\):mass   \(r\):radius of the orbit of the object   \(ω\):angular velocity[rad/s] \(v\):object speed[m/s]

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Moment of Inertia

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What is the moment of inertia?

Energy is necessary for the moving of stationary objects. Likewise, when a stationary object rotates, it also needs energy. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. This difficulty of rotation is called the moment of inertia. While the physical quantity representing the difficulty of moving the object with respect to the force is “mass”, the rotating version thereof is “moment of inertia”.

This moment of inertia \(I_ {j}\) is to be written

$$I_j=m_{j}r_{j}^{2}$$

where the distance from the mass point \(m_ {j}\) to the rotation axis stands for \(r_ {j}\).

All objects can be regarded as a mass of mass. Therefore, we can obtain the moment of inertia \(I\) of any object by integrating the above expression as the integral range of the whole object.

Furthermore, by using this moment of inertia \(I\) and the angular velocity \(ω\) of the rigid body, it is possible to express the kinetic energy \(K\) of rotation of the object.

$$K=\frac{1}{2}I\omega^{2}$$

You can remember this formula easily by comparing the above equation about \(K\) with the below equation about the kinetic energy E that you know.

$$E=\frac{1}{2}mv^{2}$$

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