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

[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?

 

[Simon Bridge]

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?

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.

 

[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!

 

[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.

 

[pmacias]

Angular V, also known as angular velocity, is a measure of how quickly an object is rotating or revolving around a central point. In the context of Kepler's 2nd law, it is used to describe the rate at which area is being swept out by an object as it moves along its orbital path.

In the blog post, Braeunig is explaining how to calculate the instantaneous rate of area swept out by a particle in orbit. He uses the formula for the area of a triangle, (base x height)/2, to represent the area swept out in a short time interval t. The base of the triangle is given by r, which is the distance between the particle and the central point (C). The height of the triangle is represented by rt, which is the distance the particle has moved along its orbit in that short time interval t.

The opposite side of the triangle, rωΔt, represents the distance the particle has moved perpendicular to its radius vector (the line connecting the particle to the central point). This distance is equal to the change in angle, Δθ, multiplied by the radius, r.

So in this case, the height of the triangle is not just r tan(Δθ), but it is r multiplied by the change in angle, Δθ, which is represented by ωΔt. This is because the height is not just the distance along the orbit, but also takes into account the direction in which the particle is moving.

Overall, Braeunig's explanation is using mathematical concepts to describe how Kepler's 2nd law works in terms of angular velocity and area swept out by a particle in orbit. I hope this helps clarify the concept for you.