2.6 Circular orbits in inverse square law.

In the previous section we talked about the gravitational field strength and the gravitational potential. Here we'll talk about the circular orbits in the inverse square law.

It is important to remember that the actual closed orbit of a planets around a host star is not a circular motion rather an ellipse. For simplicity we can approximate the orbits to be circular because this simplifies our calculations. Now that we have assumed the circular orbit, we can the use the formulas for the circular motion as reviewed in this note.

Lets consider the circular motion of earth-moon system,

Now according to the Newtons laws of motion we know that $F_{G}=m\times a_{centripetal}$. Substituting the force term by the Newtons universal gravitational formula and the acceleration by the centripetal acceleration, we then have,
$$G\frac{Mm}{r^{2}}=m\times r\omega^{2}$$
which gives,$$\omega=\sqrt\frac{GM}{r^{2}}$$
 where the symbols have their usual meaning refered in the previous notes. Now to find the time period of the revolution, we use the fact that $\omega =2\pi f$, where $f$ is the frequency of revolution. Rearranging and substituting we have,

Now let's talk a little on the geostationary orbit. Geostationary orbit is a circular orbit around the Earth on which a satellite would appear to be stationary to an observer on the Earth's surface.

A geostationary satellite must be placed vertically above the equator because the centripetal force must be directed towards the centre of the earth and so any circular orbits must have its centre at the centre of the earth and also if the orbit is not on the equator, the satellite must be over the northern hemisphere and sometimes over the southern hemisphere and so cannot be geostationary. A geostationary satellite must move from west to east. In order to appear stationary to an observer on the earth's surface, it must therefore move in the same direction.

Here are some applications of the Geostationary satellite are in the global communications, television broadcasting, weather forecasting and so on.

Ex. Find the time taken for moon to orbit the earth, given that earth-moon distance=$384,400$km, Mass of the earth= $5.972\times 10^{24}$kg

Soln: $27.46$ days

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