# Area Swept by particle under central forces is an approximation

From Kepler's second law, we infer, the conservation of angular momentum is equivalent to saying the areal velocity is constant,

And the proof goes like this $$mr^2{\dot\theta=L}$$ where $$L$$ is angular momentum. $$\frac{d(\frac{1}{2} r^2\dot\theta)}{dt}={0} \qquad ...(1)\ (as\ L\ and\ m \ are \ constant)$$

from the figure we can write $$dA= \frac{1}{2}r\ r d\theta \qquad..(2)$$ and from (1) and (2) we get $$\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{1}{2}r^2\frac{\mathrm{d}\theta}{\mathrm{d}t}$$

now since angular momentum is constant, we say, The radius vector sweeps out equal areas in equal times.

My problem is,

(1). To calculate the area element, we are taken an approximation, so how we are getting accurate results, (means we have equated it to angular momentum only after approximation and not in general)?

(2) Is Kepler's second Law true up to some approximation? Because at the end we have just cared about the approximated triangle, but there remains a smaller region (as depicted in the figure) that we have not included. And at the planetary level, our approximation will not work?

(3) Can I say that it is not an approximation and I have missed something?

End Note: The 3 questions are actually 1 question, just written in points for the convenience of reader. I know the policy.

This is just the standard trick of calculus. This looks like an approximation for finite $$\Delta \theta$$, but as you divide the area into smaller and smaller pieces, the error goes to zero. The same reasoning allows you to write the area under a curve in terms of integrals.

• this looks like a trick because I have taken a simple approach. the same when proved in other way, we don't take a $dr d\theta$ component, Also in the proof done in my question, in one step I have to assume sin x = x, then only you get perpendicular, – crabNebula Feb 15 at 7:25
• I have not written every step, but as you can see, to prove perpendicular height as $rd\theta$ you have to assume sinx=x – crabNebula Feb 15 at 7:26
• You assume $\sin(d \theta)/d\theta =1$, which is true in the limiting case where you divide into smaller and smaller pieces. This is exactly the same calculus trig for integrals. – Rd Basha Feb 15 at 9:08

Kepler's second law is not an approximation. Within Classical Mechanics it is an exact result provided two bodies interact with Newton's gravitational law.

The key point is that your equation $$(3)$$ is a controlled approximation for the area swept by the position vector. By "controlled" I mean that it is possible to show that the error is at least of the order $$d\theta^2$$. Therefore, since the evaluation of a finite area requires to integrate $$dA$$ between an initial and a final angle, the first-order approximation for $$dA$$ contains the information to get the exact result.

• And that's my question, after neglecting the second-order term, you get the desired answer. but at the planetary level, this second-order approximation can contribute to like a large inaccuracy. – crabNebula Feb 15 at 7:28
• @Kunalkumar, to evaluate the integral you have to take a limit $\Delta \theta \rightarrow 0$ whatever is the size of the orbit. So, it doesn't matter if you are dealing with distances on $nm$ or $parsec$. – GiorgioP Feb 15 at 7:32
• you are right, but what about the error ? – crabNebula Feb 15 at 7:34
• @Kunalkumar When you take the limit there is no error. It is exactly like for an integral: if you work out all the limits and you find the primitive the result is exact. If you evaluate it numerically there is a (controllable) error due to the discretization. However theoretical results correspond to the case where the limit is properly done and there is no final error. – GiorgioP Feb 15 at 11:32
• do we take these higher-order terms in practical projects (say at NASA)? – crabNebula Feb 15 at 12:53