# (a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to...

## Question:

(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of the earth's rotation.

(b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = {eq}6 \times 10^{24} {/eq} kg.

## Angular Speed :

The angular position of a rotating or revolving body varies with time. The angular velocity is a measure of the rate change of the angular position. If the object covers an angle {eq}\displaystyle { \Delta \theta } {/eq} in time {eq}\displaystyle { \Delta t } {/eq}, then its angular velocity is by definition, {eq}\displaystyle { \omega =\frac{ \Delta \theta}{\Delta t }} {/eq}. The angular velocity is related to the time period of rotation or revolution, {eq}\displaystyle { T } {/eq}, by : {eq}\displaystyle {\omega = \frac{2 \pi }{T} } {/eq}

(a) A satellite moving with the same angular speed as the earth has an orbital period equal to the period of rotation of the earth.

Earth completes one rotation about its axis in 24 hrs.

Hence the satellite has an orbital period of {eq}\displaystyle { T=24 \ hrs=24\times 3600 \ s } {/eq}

The orbital period of a satellite in orbit around the earth is given by:

{eq}\displaystyle {\quad T= 2\pi \sqrt{ \frac{r^3}{G M}} } {/eq}.

Here r is the radius of the orbit as measured from the center of the earth, M is the mass of the earth and G is the gravitational constant.

Now, the mass of the earth, {eq}\displaystyle {M= 6\times 10^{24} \ Kg} {/eq}.

Also {eq}\displaystyle {G= 6.67 \times 10^{-11} \ SI \ Units.} {/eq}.

Then the radius of the satellite's orbit is given by :

{eq}\displaystyle {r = \left (\frac{GMT^2}{4 \pi^2} \right )^{ 1/ 3}\\ r = \left (\frac{ 6.67 \times 10^{-11} \times 6\times 10^{24} \times (24\times 3600)^2}{4 \pi^2} \right )^ {1/ 3}\\ \text{or} \quad \boxed{r=42311Km}} {/eq}

(b) Suppose the satellite is directly above the north pole at some instant.

Then to come over to the equatorial plane it will take a time equal to one-quarter of its orbital period.

This is : {eq}\displaystyle {t= \dfrac T 4 =\dfrac {24} 4 = 6 \ hrs } {/eq}