2.6 Polar coordinates

For given a point $(x,y)\ne (0,0)$ in ${ℝ}^2$, there exist unique $r>0$, $\theta \in [0,2\pi )$ such that $x=r\cos \theta$, $y=r\sin \theta$. We say that $(r,\theta )$ are the polar coordinates of $(x,y)$; $r=\sqrt{x^2+y^2}$ and $\theta$ is the angle that the vector $(x,y)$ makes with the positive part of the $x$-axis.

Suppose a particle $P$ has position $(x(t),y(t))$ at time $t$. Then we can equally define time-dependent polar coordinates $(r(t),\theta (t))$. We will now decompose the velocity and acceleration of $P$ into radial and transverse components.

Since $(x,y)=(r\cos \theta ,r\sin \theta )$, we obtain by differentiating, using the chain rule:

Thus $(x^{\prime },y^{\prime })=r^{\prime }(\cos \theta ,\sin \theta )+r{\theta }^{\prime }(-\sin \theta ,\cos \theta )$. In other words, the radial component of velocity is $r^{\prime }.(\cos \theta ,\sin \theta )$, and the transverse component is $r{\theta }^{\prime }.(-\sin \theta ,\cos \theta )$. We say that ${\theta }^{\prime }$ is the angular velocity.

For acceleration, we differentiate a second time:

Rearranging terms, we obtain

So the radial component of acceleration is $(r^{\mathrm{\prime \prime }}-r{\theta }^{\mathrm{\prime }2})(\cos \theta ,\sin \theta )$, and the transverse component is $(2r^{\prime }{\theta }^{\prime }+r{\theta }^{\mathrm{\prime \prime }})(-\sin \theta ,\cos \theta )$.

Suppose a particle $P$ rotates on the edge of a circle of radius $r$. Then its position at time $t$ is $(x(t),y(t))=(r\cos \theta (t),r\sin \theta (t))$. Thus its angular velocity at time $t$ is ${\theta }^{\prime }(t)$.

Suppose further that it rotates at a constant angular velocity $\omega$. Then $\theta =\omega t+c$. So

A satellite moves in a geostationary orbit about the earth if it is above a fixed point on the equator. With the above description - and the following information: the mass $M$ of the earth, the gravitational constant $G$, and the radius $R$ of the earth - we can calculate the height of a geostationary orbit above the earth’s surface.

Recall that the gravitational force exerted on an object of mass $m$ at a distance $r$ from the centre of the earth is $GMm/r^2$; in other words, the acceleration it experiences is $-GM/r^2$ in the radial direction. Suppose the object rotates about the earth at a constant distance $r$ and a constant angular velocity $\omega$. Then the radial component of acceleration:

Thus $r^3{\omega }^2=GM$, whence $r^3=GM/{\omega }^2$.

What does it mean for the orbit to be geostationary? The angular velocity is the angle swept out per second. Since the orbit is geostationary, $2\pi$ radians is swept out in one day $=86,400$ seconds. So the angular velocity is $\omega =2\pi /86,400$. The mass of the earth is approximately $6\times {10}^{24}$ kilogrammes. The gravitational constant is approximately $6.7\times {10}^{-11}$ m${}^3$ kg${}^{-1}$ s${}^{-2}$. Thus, using a calculator, $r^3\approx 7.6\times {10}^{22}$. Then we take the cube root: $r\approx 42,400,000$m, i.e. 42,400 kilometres. But this is the distance from the centre of the earth. The distance from the surface of the earth is $r-R$. The (equatorial) radius of the earth is about $6,400$ kilometres. Thus a satellite on a geostationary orbit is approximately 36,000 kilometres above the equator.