In the context of an object moving under the influence of gravity, in Bate, Mueller, White it says, sect. 1.4.2

The expression $\bf{r} \times \bf{v}$ which must be a constant of the motion is simply the vector $\bf{h}$...

Why does it have to be a constant and from the text up to that section how is it explained?

It starts from the two-body equation of motion:

$$ \ddot{\bf{r}} + \frac{\mu}{r^3}{\bf{r}} = 0 \tag{1} $$

where $\mu = GM$ and $M$ is the solar mass and $G$ is graviational constant. $\bf{r}$ is the vector from the sun to the planet.

The text goes on to obtain an expression for angular momentum $\bf{h}$ by cross multiplying the above equation by $\bf{r}$ to yield

$$ {\bf{r}} \times \ddot{\bf{r}} + {\bf{r}}\times\frac{\mu}{\it{r}^3}{\bf{r}} = 0 \tag{2} $$

With ${\bf{a}}\times{\bf{a}} = 0$ in general, the second term vanishes and

$$ {\bf{r}}\times\ddot{\bf{r}} = 0 \tag{3} $$

Using $\frac{d}{dt}({\bf{r}}\times\dot{\bf{r}}) = \dot{\bf{r}}\times\dot{\bf{r}} + {\bf{r}}\times\ddot{\bf{r}}$ the above equation becomes

$$ \frac{d}{dt}({\bf{r}} \times \dot{\bf{r}}) = \frac{d}{dt}({\bf{r}}\times{\bf{v}}) = 0 \tag{4} $$

Then it says, what I wrote at the beginning, the expression..., hence my question.

The text makes two assumptions: the bodies are spherically symmetrical and there are no external or internal forces other than gravity.

  • $\begingroup$ What is the context…? There is no way to answer this question if we do not know the system for which the angular momentum is a constant. $\endgroup$ Commented Aug 11, 2023 at 16:53
  • $\begingroup$ Thank you, added to the question that this is within the context of gravity. $\endgroup$
    – TMOTTM
    Commented Aug 11, 2023 at 16:58
  • $\begingroup$ That is still nowhere near enough context. You need to explicitly outline the full nature of the problem you are attempting to solve. $\endgroup$ Commented Aug 11, 2023 at 17:12
  • $\begingroup$ Just to explain why the information you provides is not enough, a satellite in a polar orbit around Earth, does not see a central force (due to the equatorial bulge), and consequently angular momentum is not a constant of motion. $\endgroup$ Commented Aug 11, 2023 at 19:57
  • $\begingroup$ I added the two assumptions the text makes. $\endgroup$
    – TMOTTM
    Commented Aug 11, 2023 at 20:22

1 Answer 1


By the looks of it: that section in that textbook is a very weird section.

I am of the opinion: in that book the subject of conservation of angular momentum (in orbital motion) is approached in such a way that the student is prevented from understanding the concept.

As we know: the concept of angular momentum (of orbiting objects) had a precursor: Kepler's law of areas.

In his book 'Principia Mathematica' Newton gave a derivation of Kepler's law of areas from first principles, demonstrating that Kepler's second law is an instance of a more general property: if the force that is causing the (orbital) motion is a central force then the resulting motion has the property that in equal intervals of time equal areas are swept out.

For Newton's demonstration there is a 2021 answer by me to a question titled Intuition for angular momentum The key points are illustrated with diagrams.

Returning to the book that is your source:
The approach that you describe - taking a vector cross product multiple times - is one that would make sense if the goal is to present a very generic demonstration.

In that book: The initial expression is for the specific case of an inverse square force law

$$ \ddot{\bf{r}} + \frac{\mu}{r^3}{\bf{r}} = 0 \tag{1} $$

The thing is: two steps later 'the second term' vanishes. The reason that that second term vanishes: it describes a force that is acting in radial direction.

The fact that that term was describing an inverse square force does not affect the outcome. We have that for any force acting in radial direction that second term vanishes. The key property is: for a force that acts purely in radial direction that second term vanishes.

So the approach in that book is a weird mismatch of generic and specific-to-one-case.

I shouldn't judge the book that you are using based purely on that section. So I won't go on about that book.

I recommend that you read the story 'Richard Feynman on the education system in Brazil'

Feynman recounts that in the Brazil physics community he encountered a culture, deeply ingrained, of attempting to teach physics purely through memorization, without regard to understanding.

Needless to say: Feynman recounts that experience precisely because such a rote learning approach is the very opposite of proper physics education

  • $\begingroup$ Why is a section titled "Two-body orbital mechanics" that takes the shortest path to show conservation of angular momentum "weird"? And what has that to do with the Brazilian education system? $\endgroup$
    – Kurt G.
    Commented Aug 12, 2023 at 7:23
  • $\begingroup$ @KurtG. The shortest path isn't necessarily a transparent path; the shortest path isn't necessarily an informative path. The Feynman story is about opinion on what to strive for in a physics textbook. Newton's demonstration of conservation of angular momentum gives the student the means to think about what angular momentum is. Conversely, it would be useless to effectively say to the student: "Angular momentum is conserved because the vector cross product says so." $\endgroup$
    – Cleonis
    Commented Aug 12, 2023 at 8:25
  • $\begingroup$ As long as you don't find something from Feynman that better explains conservation of angular momentum that Brazil story is irrelevant. It also surprises me that nobody has mentioned the Noether theorem yet. The rotational symmetry of the Lagrangian leads to the conservation law in question. $\endgroup$
    – Kurt G.
    Commented Aug 12, 2023 at 9:53
  • $\begingroup$ @KurtG. There is the series of lectures by Feynman, for a general audience, on the theme of 'The character of physical law'. In the second lecture one of the subjects touched upon is Kepler's second law. Feynman walks over to the blackboard and he proceeds to draw the diagram for Newton's demonstration of the area law: conservation of angular momentum. I take it as evident that Feynman regarded Newton's demonstration as the most accessible demonstration. $\endgroup$
    – Cleonis
    Commented Aug 12, 2023 at 10:36
  • $\begingroup$ @KurtG. Here's the thing: I think it is necessary to be aware of the following difference. In his teaching Feynman was always keen to create an accessible presentation of the subject he was teaching. $\endgroup$
    – Cleonis
    Commented Aug 12, 2023 at 10:40

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