Question

Consider a hypothetical planet which is very long and cylindrical. The density of the planet is $\rho$, its radius is R.

If the possible orbital speed of the satellite moving around the planet in circular orbit in a plane which is perpendicular to the axis of planet is $\sqrt{xG\rho R^2}$ then find x.

Answer 1

2π

Answer 2

The gravitational field at a distance $r$ from the axis of a very long cylindrical planet of radius $R$ and uniform density $\rho$ can be found using Gauss's law for gravity.

For a point outside the cylinder ($r > R$), consider a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the planet. The gravitational field $\vec{g}$ is radial and its magnitude is constant on the surface of the Gaussian cylinder. The gravitational flux through the curved surface is $\oint \vec{g} \cdot d\vec{A} = |g| (2\pi r L)$. The flux through the flat ends is zero. The mass enclosed within the Gaussian surface is the mass of the cylinder of radius $R$ and length $L$, which is $M_{enclosed} = \rho (\pi R^2 L)$.

According to Gauss's law for gravity, $\oint \vec{g} \cdot d\vec{A} = -4\pi G M_{enclosed}$.

So, $|g| (2\pi r L) = 4\pi G (\rho \pi R^2 L)$.

$|g| = \frac{4\pi^2 G \rho R^2 L}{2\pi r L} = \frac{2\pi G \rho R^2}{r}$, for $r > R$.

For a satellite of mass $m$ in a circular orbit of radius $r$ in a plane perpendicular to the axis, the gravitational force provides the centripetal force.

Gravitational force $F_g = m |g| = m \frac{2\pi G \rho R^2}{r}$.

Centripetal force $F_c = \frac{mv^2}{r}$, where $v$ is the orbital speed.

For a circular orbit, $F_g = F_c$.

$m \frac{2\pi G \rho R^2}{r} = \frac{mv^2}{r}$.

$v^2 = 2\pi G \rho R^2$.

$v = \sqrt{2\pi G \rho R^2}$.

This orbital speed is independent of the orbital radius $r$ (for $r > R$). The question asks for "the possible orbital speed", which is consistent with this result. The smallest possible orbital radius for an orbit around the planet is $R$. If the orbit is at the surface ($r=R$), the speed is still given by the same formula.

The problem states that the possible orbital speed is $\sqrt{xG\rho R^2}$.

Comparing this with our derived speed, we have:

$\sqrt{xG\rho R^2} = \sqrt{2\pi G \rho R^2}$.

Squaring both sides, we get:

$xG\rho R^2 = 2\pi G \rho R^2$.

Assuming $G$, $\rho$, and $R$ are non-zero, we can cancel $G\rho R^2$ from both sides:

$x = 2\pi$.