# Help understanding Angular V as it applies to Kepler's 2nd law

1. Mar 24, 2013

### masteriti

Robert Braeunig has an great blog on rocket science here: http://www.braeunig.us/space/index.htm, and I've been trying to follow the math his piece on Orbital Mechanics.

I'm having trouble in particular following his description on Kepler's 2nd law, here's how he puts it:

Figure 4.5 shows a particle revolving around C along some arbitrary path. The area swept out by the radius vector in a short time interval t is shown shaded. This area, neglecting the small triangular region at the end, is one-half the base times the height or approximately r(rt)/2. This expression becomes more exact as t approaches zero, i.e. the small triangle goes to zero more rapidly than the large one. The rate at which area is being swept out instantaneously is therefore

$$\lim_{t\rightarrow 0}{\Big[ \frac{r(rωΔt)}{2} \Big]} = \frac {ωr^2}{2}$$

I understand the area of the triangle is given by (base x height)/2. I understand r is the triangle base in this case, but how does the opposite side equal rt or rωΔt?

The way I'm reading this is that Angular speed is just change in angle given a change in time ω = Δθ / Δt, so as I understand it, ωΔt just leaves the angle Δθ. So shouldn't the height then be r tan(Δθ)? What am I missing?

2. Mar 24, 2013

### Simon Bridge

For small time periods, $\tan\Delta\theta \approx \Delta\theta$ ... it's an approximation.

The author is giving a description for people who don't have calculus yet.

3. Mar 25, 2013

### masteriti

I got hung up by the formula for a triangle in the figure and forgot that the distance covered on a circumference is s = rθ, so it would make sense for it to approximate the height of a very small triangle. Thank you for the reply!!

4. Mar 25, 2013

### Simon Bridge

Its a common approach - what they really wanted was $dA=\frac{1}{2}r^2(\theta)d\theta$ and you can integrate to get the area swept by motion between two angles.

However, it can take a bit of getting used to.