Question

Two artificial satellites ($A$ and $B$) of earth are orbiting in equatorial plane at such altitudes that their periods of revolution are 4 hours and 16 hours respectively. Sense of rotation of satellite $A$ is west to east while that of $B$ is opposite. After being nearest to each other and lying in a given longitude of earth at a given point of time, the minimum time interval after which both are again nearest to each other and lying in the same longitude is $x$ hours. Find 10x.

Answer 1

32

Answer 2

Let the angular speed of Satellite A be

$$\omega_A = \frac{360^\circ}{4 \text{ h}} = 90^\circ/\text{h},$$

and Satellite B (moving in the opposite direction) be

$$\omega_B = -\frac{360^\circ}{16 \text{ h}} = -22.5^\circ/\text{h}.$$

Their relative angular speed is

$$\omega_{\text{rel}} = \omega_A - \omega_B = 90 - (-22.5) = 112.5^\circ/\text{h}.$$

They will again be aligned (i.e. nearest to each other on the same longitude) when

$$112.5 \, t = 360^\circ \quad \Longrightarrow \quad t = \frac{360}{112.5} = \frac{16}{5} \text{ hours} = 3.2 \text{ hours}.$$

Thus, $x = 3.2$ hours and hence

$$10x = 10 \times 3.2 = 32.$$

Core Explanation:

Use relative angular speeds (90°/h and -22.5°/h) to find the synodic period $t = \frac{360}{112.5} = 3.2$ h, then multiply by 10.