Minimum velocity to maintain orbit around a perfect sphere

Question

I'm looking at the concept of the Schueler period and wondering if there is a way to begin from this definition to calculate the lower bound on velocity required to orbit a perfectly spherical planet with no atmosphere.

Beginning with the definition of orbital period from Kepler's third law, and considering the case where the object is just above the surface, so $r$ is approximately equal to the radius of the planet, $r_p$:

$$ T = 2\pi\sqrt{\frac{r_p^3}{GM}} $$

We know that for an ideal orbit, the force of gravity the only force acting on the planet, and at the surface, the gravitational acceleration can be defined as g. $$ F = ma = \frac{GMm}{r^2} $$ $$ g = \frac{GM}{r_p^2} $$ Which makes the period at the surface: $$ T = 2\pi\sqrt{\frac{r_p}{g}} $$

Then the orbital velocity is found by dividing the distance per orbit, $2\pi r_p$, and dividing it by T, meaning the minimum velocity required to maintain orbit around this perfectly spherical planet is: $$ v = \sqrt{r_p g} $$

And at higher altitudes, larger $r$, the velocity required to maintain orbit will increase. Is this correct?

Answer 1

Your equation:

$$ v = \sqrt{r g} $$

is correct, but what you have forgotten is that the gravitational acceleration $g$ is a function of $r$. You should write:

$$ v = \sqrt{r g(r)} \tag{1} $$

where:

$$ g(r) = \frac{GM}{r^2} \tag{2} $$

As you go to increased $r$ the $\sqrt{r}$ term makes $v$ increase but the $\sqrt{g(r)}$ term makes $v$ decrease. The end result is that the orbital velocity decreases with increasing $r$. We can make this quantitative by using equation (2) to substitute for $g(r)$ in equation (1) to get:

$$ v = \sqrt{\frac{GM}{r}} $$