- #1

embphysics

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- 0

## Homework Statement

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the xy-plane, with the sun at the origin, and label the planet's position by polar coordinates [itex](r, \theta)[/itex]. (a) Show that the planet's angular momentum has magnitude [itex]L = mr^2 \omega[/itex], where [itex]\omega = \dot{\theta}[/itex] is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" is [itex]\frac{dA}{dt} = \frac{1}{2} r^2 \omega[/itex], and hence [itex]\frac{dA}{dt} = \frac{L}{2m}[/itex],

## Homework Equations

## The Attempt at a Solution

If we consider the sun and the planet as one system, then the gravitational force will be an internal force, which means there are no external force, and consequently [itex]\sum \vec{\tau} = \vec{0}[/itex], which means angular momentum is constant.

[itex]\vec{L} = \vec{r} \times \vec{p}[/itex]

[itex]L = rp \sin \theta[/itex], where the radial distance, r, and the angle, theta, both vary as a function of time.

Substituting in the expression for linear momentum,

[itex]L = rmv_{tan} \sin \theta[/itex], where [itex]v_{tan} = r \omega[/itex]

[itex]L = rm(r \omega) \sin \theta[/itex]

[itex]L=mr^2 \omega \sin \theta[/itex]. This is where I get stuck. What am I to do with the sin theta, which varies as time progresses? I would appreciate a hint, but not the answer.