The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition

Applying Newton's second law, this expression can be rewritten as

$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$

Using the definition of angular velocity

we get

$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$

The difference in angular velocities is

$\Delta\omega=\omega_2-\omega_1$

On the other hand, angular velocity itself can be approximated with the average angular velocity

$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$

Using both expressions, we obtain the equation

$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$

Thus, the change in energy is given by

$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$

This allows us to define kinetic energy as

(ID 3244)

The work variance ($\Delta W$) required for an object to change from the initial Angular Speed ($\omega_0$) to the angular Speed ($\omega$) is obtained by applying a the torque ($T$) that produces an angular displacement the difference of Angles ($\Delta\theta$), according to:

Applying Newton's second law for rotation, in terms of the moment of inertia for axis that does not pass through the CM ($I$) and the mean Angular Acceleration ($\bar{\alpha}$):

this expression can be rewritten as:

$\Delta W = I \alpha \Delta\theta$

or, using the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$):

we get:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta$

Using the definition of the mean angular velocity ($\bar{\omega}$) and the time elapsed ($\Delta t$):

results in:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta = I\omega \Delta\omega$

where the difference in Angular Speeds ($\Delta\omega$) is expressed as:

On the other hand, the angular velocity can be approximated by the average angular velocity:

$\bar{\omega}=\displaystyle\frac{\omega_1 + \oemga_2}{2}$

By combining both expressions, we obtain the equation:

$\Delta W = I \omega \Delta\omega = I(\omega_2 - \omega_1) \displaystyle\frac{(\omega_1 + \omega_2)}{2} = \displaystyle\frac{I}{2}(\omega_2^2 - \omega_1^2)$

Therefore, the change in energy is expressed as:

$\Delta W = \displaystyle\frac{I}{2}\omega_2^2 - \displaystyle\frac{I}{2}\omega_1^2$

This allows us to define the rotational kinetic energy as:

(ID 3255)

The relationship between the angular Momentum ($L$) and the moment ($p$) is expressed as:

Using the radius ($r$), this expression can be equated with the moment of Inertia ($I$) and the angular Speed ($\omega$) as follows:

Then, substituting with the inertial Mass ($m_i$) and the speed ($v$):

and

it can be concluded that the moment of inertia of a particle rotating in an orbit is:

(ID 3602)

(ID 3686)

(ID 3687)

Just as the relationship between the speed ($v$) and the angular Speed ($\omega$) with the radius ($r$) is expressed by the equation:

we can establish a relationship between the angular Momentum ($L$) and the moment ($p$) in the context of translation. However, in this case, the multiplicative factor is not the arm ($r$), but rather the moment ($p$). This relationship is expressed as:

(ID 9874)

The total Kinetic Energy ($K$) corresponds to the sum of the translational Kinetic Energy ($K_t$) and the kinetic energy of rotation ($K_r$):

Since the translational Kinetic Energy ($K_t$) is expressed as a function of the inertial Mass ($m_i$) and the speed ($v$):

and the kinetic energy of rotation ($K_r$), as a function of the moment of inertia for axis that does not pass through the CM ($I$) and the angular Speed ($\omega$), is defined as:

the final expression is:

(ID 9944)

As the gravitational force is

To move a mass $m$ from a distance $r_1$ to a distance $r_2$ from the center of the planet, a potential energy is required

resulting in the gravitational potential energy being

$W_2-W_1=\displaystyle\int_{r_1}^{r_2}\displaystyle\frac{GmM}{r^2}dr=\displaystyle\frac{GmM}{r_1}-\displaystyle\frac{GmM}{r_2}$

thus yielding

(ID 12551)

(ID 12552)

The total Energy ($E$) depends on the total Kinetic Energy ($K$) and the general gravitational potential energy ($V$), according to:

When the object is in orbit, the total Kinetic Energy ($K$) consists of a translational part and a rotational part. Considering the inertial Mass ($m_i$), the speed ($v$), the moment of Inertia ($I$) and the angular Speed ($\omega$), we have:

Since the angular Momentum ($L$) is:

and using the distance to the center of the celestial body ($r$), we obtain:

On the other hand, the gravitational potential, expressed in terms of the mass of the celestial body ($M$), the gravitational mass ($m_g$) and the gravitational constant ($G$), is:

Therefore, it follows that:

(ID 16251)