Keplers laws and kinetic energy

[sheetman]

Homework Statement

As part of my coursework I've had to show that kinetic energy can be written in a certain way. As a cheat I used this as a starting point and worked my way back to the following equation

Kinetic energy = \frac{1}{2}m\dot{r^{2}} + \frac{1}{2}mr^{2}\dot{\theta}^{2}

r dot is the derivative of radius with respect to theta, and theta dot is the derivative of theta with respect to time.

Homework Equations

(edit: This is for the kinetic energy of a planet in an elliptical orbit around the sun)

The Attempt at a Solution

the first part of the equation for kinetic energy is angular velocity, but I can't for the life of me figure out what r\frac{d\theta}{dt} is. Is it another way of calculating angular velocity or have I probably messed up the algebra somewhere?

Thanks

 

[vela]

You may recognize it as ωr since ω=dθ/dt. The other way you can view that term is as 1/2 (mr²)ω².

 

[sheetman]

So the second term is angular velocity, so what kind of velocity is dr/d(theta)?

 

[vela]

So the second term is angular velocity

Not exactly. First, the second term is an energy, not a velocity. Second, angular velocity is dθ/dt while r dθ/dt is the tangential velocity, a component of the linear velocity. Depending on how you choose to interpret the second term, you can say it's a function of either the angular velocity or the tangential velocity.

so what kind of velocity is dr/d(theta)?

It's the component of velocity perpendicular to the tangential velocity. It's the radial velocity of the body.

 

[sheetman]

Hero, thanks very much!

 

[vela]

I just noticed you said dr/dθ, not dr/dt. It's dr/dt which is the radial velocity. I don't know of a name for dr/dθ. It's not a velocity, though.