-3
$\begingroup$

The following exercise has made me question the properties of a conservative field:

enter image description here

Going through the first steps quickly so I can get to my point:

$$\vec a = \vec a_r = -\frac{\beta ^2}{r} \hat r$$ $$a_r = \ddot r - r\dot \theta ^2$$

$$r = r_0 $$ $$ \therefore \dot r = 0$$

$$\therefore -\frac{\beta ^2}{r_0} = -r_0\dot \theta ^2$$

Rearranging gives:

$$\dot \theta = \beta / r_0$$

Now that's all well and good, but part (d) confuses me.

I can agree with this mathematically:

$$r \dot \theta = \beta = const$$

$$ \vec v = \dot r \hat r + \dot \theta r \hat \theta = r\dot \theta $$

$$\therefore \vec v = \beta$$

But the statement as a whole? Hmm. For example, in a different conservative field with central force:

$$\vec F = -\frac{GMm}{r^2}\hat r$$

This shouldn't be the case, at least to my intuition. You need a far higher tangential speed to orbit Earth at a height of $10$ meters than you would have to at a height of $10,000$ meters.

My guess is, at some stable radius $r$:

$$r\dot \theta ^2 = \frac{GMm}{r^2}$$ $$\vec v_{\theta} = \sqrt{\frac{GMm}{r}}$$

Now I have the tangential velocity as a function of $r$, so due to the fact that the radial force is merely inversely proportional but not an inverse square law, the tangential velocity required to orbit a specific height is constant. This is intuitively hard top grapple with. Does anyone have any intuitive understanding behind this? And, in addition, did I make any mistakes in my thought processes?

$\endgroup$

closed as unclear what you're asking by John Rennie, Jon Custer, glS, sammy gerbil, Yashas Nov 19 '17 at 14:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ It isn't clear what you are asking. The central acceleration is $v_\theta^2/r$ so you can just equate this to the force per unit mass to show that $v_\theta$ is constant for any circular orbit. This is unintuitive because it is not a physically reasonable potential - for example try calculating the potential energy at infinity. $\endgroup$ – John Rennie Nov 13 '17 at 6:47
  • $\begingroup$ The statement "the tangential velocity required to orbit a specific height is constant." Is true for circular orbits. As value of r increases value of v will decrease, matching the statement "You need a far higher tangential speed to orbit Earth at a height of 10 meters than you would have to at a height of 10,000 meters." $\endgroup$ – See Jian Shin Nov 13 '17 at 12:05
2
$\begingroup$

Does anyone have any intuitive understanding behind this?

That depends on what you find intuitive. For a circular orbit, we need the central force to equal the centripetal force:

$$\beta_n^2\,m\,r^n = \frac{v_\circ^2}{r}m$$

and so

$$v_\circ(r) = \sqrt{\beta_n^2r^{n+1}}$$

Clearly, for $n=-2$ (inverse square law), the tangential velocity decreases with $r$ while for $n=1$ (Hooke's force law), the tangential velocity increases with $r$.

Intuitively, there must then be some force law 'in between' for which the tangential velocity neither decreases or increases with $r$ and, by inspection, that force law is $n = -1$ (inversely proportional law).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.