In the book of Classical Mechanics by Goldstein, at page 88, it is given that:

$$ \frac{d^{2} u}{d t^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right) . $$ The preceding equation is such that the resulting orbit is symmetric about two adjacent turning points. To prove this statement, note that i= the orbit is symmet-

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

In the book of Classical Mechanics by Goldstein, at page 88, it is given that:

$$ \frac{d^{2} u}{d t^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right) . $$ The preceding equation is such that the resulting orbit is symmetric about two adjacent turning points. To prove this statement, note that i= the orbit is symmetrical.

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

In the book of Classical Mechanics by Goldstein, at page 88, it is given that

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

In the book of Classical Mechanics by Goldstein, at page 88, it is given that:

$$ \frac{d^{2} u}{d t^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right) . $$ The preceding equation is such that the resulting orbit is symmetric about two adjacent turning points. To prove this statement, note that i= the orbit is symmet-

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

In the book of Classical Mechanics by Goldstein, at page 88, it is given that

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes  ?

In the book of Classical Mechanics by Goldstein, at page 88, it is given that

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?