Why is angular momentum conserved in a satellite?

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Why is angular momentum conserved in a satellite?

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Answer 1 · 2018-03-08T16:14:50

The principle of conservation of momentum, both linear (also known as translational) and angular, is a universal principle of Physics.

Explanation:

Angular momentum, $L$ $L$ , is defined, in mathematical terms as

$L = I \cdot ω$ $L = I*omega$

Take the time derivative of that

$\frac{d L}{d t} = I \cdot \frac{d ω}{d t}$ $(dL)/dt = I*(domega)/dt$

We give the term "torque" to $I \cdot \frac{d ω}{d t}$ $I*(domega)/dt$ and we give torque the variable name $τ$ $tau$ .
We give $\frac{d ω}{d t}$ $(domega)/dt$ the variable name $α$ $alpha$ .

Expanding that last equation with the variable names $α and τ$ $alpha and tau$

$\frac{d L}{d t} = I \cdot \frac{d ω}{d t} = I \cdot α = τ$ $(dL)/dt = I*(domega)/dt = I*alpha = tau$ ... Equation1

The equation $τ = I \cdot α$ $tau = I*alpha$ is the angular equivalent of $F = m \cdot a$ $F = m*a$ which Newton made famous in Newton's 2nd Law.

$τ, I, and α$ $tau, I, and alpha$ are the angular equivalents of $F, m, and a$ $F, m, and a$ respectively.

If there are no external torques acting on a body, $τ = 0$ $tau = 0$ , and that there is no angular acceleration follows. That is shown in the repeat of Equation 1 below.

$\frac{d L}{d t} = I \cdot \frac{d ω}{d t} = I \cdot α = 0$ $(dL)/dt = I*(domega)/dt = I*alpha = 0$

Therefore $\frac{d L}{d t} = 0$ $(dL)/dt = 0$ . And therefore, the angular momentum is constant if $τ = 0$ $tau = 0$ .

Therefore, we have Conservation of Angular Momentum .

I hope this helps,
Steve

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Why is angular momentum conserved in a satellite?

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