# Intuitivelty speaking, if a radial force is only inversely proportional to $r$, why is speed required for any orbit independent of $r$, intuitively? [closed]

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

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?

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

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• 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. – John Rennie Nov 13 '17 at 6:47
• 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." – See Jian Shin Nov 13 '17 at 12:05

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).