- #1

Benjamin_harsh

- 211

- 5

- Homework Statement
- How magnitude of L is calculated here?

- Relevant Equations
- How magnitude of L is calculated here?

When the Sun disappears the Earth heads off in a straight line at constant velocity as shown by the horizontal dashed line, so after some time ##t## it has moved a distance ##x = vt## as I've marked on the diagram. The question is now how the angular momentum can be conserved.

The answer is that angular momentum is given by the vector equation:

## \mathbf L = \mathbf r \times m\mathbf v ##

where ##\mathbf r## is the position vector, ##\mathbf v## is the velocity vector and ##\times## is the cross product. We are going to end up with the vector ##\mathbf L## pointing out of the page and the magnitude of ##L## is given by:

## |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta \tag{1} ##

but looking at our diagram we see that:

## \sin\theta = \large\frac{d}{|\mathbf r|} ##

and if we substitute this into our equation (1) for the angular momentum we get:

## |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\large\frac{d}{|\mathbf r|}\normalsize = m\,|\mathbf v|\,d \tag{2} ##

And this equation tells us that the angular momentum is constant i.e. it depends only on the constant velocity ##\mathbf v## and the original orbital distance ##d##.

I didn't understand how ##|\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta## is written? Why ##sin## instead of ##cos## or ##tan##?

I only know one formula for angular momentum; ##L = mvr##.

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