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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.

Cleonis
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1 Answers1

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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

Cleonis
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