Deriving Newtonian law of gravitation

[TimeRip496]

Homework Statement

Homework Equations

$$\frac{\omega}{v}=\frac{s}{r}$$

The Attempt at a Solution

How to get the equation above? Cause
$$\frac{s}{r}=\omega$$ & $$\frac{\omega}{v}=\frac{1}{r}$$ since v=rw.

Source: http://www.relativitycalculator.com/Newton_Universal_Gravity_Law.shtml

 

[Andrew Mason]

Homework Statement

2. Homework Equations [/B]
$$\frac{\omega}{v}=\frac{s}{r}$$

This is not a correct equation. You can see that from the dimensions. LS = 1/distance and RS = distance / distance. Your source is not very well written and takes incomprehensible mathematical steps that appear to me to be wrong.

AM

 

[TimeRip496]

This is not a correct equation. You can see that from the dimensions. LS = 1/distance and RS = distance / distance. Your source is not very well written and takes incomprehensible mathematical steps that appear to me to be wrong.

AM

Thanks! That's what I thought so. But without that step, I cannot continue and the subsequent steps that the source provides are useless in helping me to derive the Newtonian law of gravitation. Is there a correct way to derive the Newtonian laws of gravitation?

 

[Andrew Mason]

The author is trying to show how Kepler's laws lead to the $F \propto 1/r^2$ relationship for circular orbits.

I am not sure what the author of your source is trying to show with Kepler's second law. You can find the derivation of centripetal force in any first year physics textbook: $a_c = v^2/r = \omega^2 r$.

From Kepler's third law, $r^3 \propto T^2 \text{ or } \frac{r^3}{T^2} = \text{ constant}$. Since $\omega = 2\pi/T$ this means that $r^3\omega^2 = K$ or $\omega^2 = \frac{K}{r^3}$

The gravitational force provides the centripetal force required for circular orbit, so $g=v^2/r = \omega^2 r$, this means that $g = Kr/r^3 = K/r^2$

AM

 

[mpresic]

The angular velocity omega is r x v. The angular momentum vector is not even in the plane of the orbit. The angular momentum is normal to the plane of orbit. This source is clearly incorrect.

 

[mpresic]

Sorry for the last post I was thinking of (specific) angular momentum not the angular velocity. I still think the source is unclear. A better explanation can be found in many textbooks.

 

[Andrew Mason]

Sorry for the last post I was thinking of (specific) angular momentum not the angular velocity. I still think the source is unclear. A better explanation can be found in many textbooks.

Your first post was correct. The direction of $\vec{\omega}$ is in the same direction as the angular momentum vector: perpendicular to the plane of rotation:

(1) $\vec{L} = I\vec{\omega}$

and for a point mass
(2) $\vec{L} = \vec{r} \times \vec{p}$

so:
(3) $\vec{\omega} = \frac{\vec{L}}{I} = \frac{\vec{r} \times \vec{p}}{I}$

Since, for a point mass: $I = m|\vec{r}|^2$, for a point mass (3) can be rewritten:
$\vec{\omega} =\frac{\vec{r} \times \vec{p}}{m|\vec{r}|^2} = \frac{\vec{r} \times \vec{v}}{|\vec{r}|^2}$

I agree that the OP's source is incorrect in several other ways. Better to stick to reliable texts or online sites.

AM