Satellites in orbit

Once in space, the motion of the satellites is quite independent of the earth's daily rotation. There is no need to find their speeds relative to point $P$ .

Due to the symmetry of the situation, the satellites meet each other twice during each revolution, on opposite sides of their orbits. Let's call the original meeting point $X$ and the point at the opposite side, $Y$ . The satellites meet $\begin{cases} \text{at point}\ X & t = 2.4n\ \ \text{for even}\ n, \\ \text{at point}\ Y & t = 2.4n\ \ \text{for odd}\ n.\end{cases}$ The observer can only see the satellites overhead if he is directly below either point $X$ or point $Y$ . He is $\begin{cases} \text{below}\ X & t = 12m\ \ \text{for even}\ m, \\ \text{below}\ Y & t = 12m\ \ \text{for odd}\ m.\end{cases}$ Thus we must solve $2.4n = 12m\ \ \ \text{for}\ n,m > 0\ \text{both even or both odd}.$ This reduces to $n = 5m\ \ \ \text{for}\ n,m > 0\ \text{both even or both odd},$ which is obviously true for $m = 1, 2, 3, \dots$ and $n = 5, 10, 15, \dots$ .

The smallest solution is $m = 1$ , $n = 5$ , ant $t = \boxed{12}\ \text{hours}$ .

Satellites A and B crosses above point P and will cross again there and at the opposite side (180 degrees) of their circular orbits, indefinitely. They will cross every 2.4 hours with each satellite traveling half the orbit. It will take point P 12 hours to make it to the other side of the circle where the satellites pass one another for the 5th time since the first observation.

First, we have to convert the periods into speeds. To make things easy, we can use units of spins per hour.

Obviously the Earth spins one time in a day, so its angular velocity is $\omega_E = 1/24$ spins per hour counterclockwise. Similarly, the angular velocities of satellites are $\omega_A = 1/4.8$ spins per hour counterclockwise and $\omega_B = 1/4.8$ spins per hour clockwise.

We have to find out how fast these satellites move relative the person standing at $P.$ We can do this by simply addition and subtraction, a Galilean transformation into the frame of the person at $P.$ Relative to the person standing at point $P,$ satellite A has angular speed $\omega_A^\prime = \omega_A - \omega_E = 1/6$ spins per hour counterclockwise. Similarly, satellite B has angular speed $\omega_B^\prime = \omega_B + \omega_E = 1/4$ spins per hour clockwise.

Thus, in the reference frame of the person at $P,$ their periods are $T_A = 1/\omega_A = \SI{6}{\hour}$ and $T_B = 1/\omega_B = \SI{4}{\hour}.$ Obviously, it will take $\SI{12}{\hour}$ for these satellites to overlap again above the person standing at $P.$

• Earth completes one revolution in 23.92 hours (sidereal time). Which means that it is able to cover 15 degrees in an hour.
• Both satellites can cover 75 degrees in an hour.
• Sat A, as seen by an observer on Earth, is able to cover (75-15) degrees per hour.
• Sat B, as seen by observer, covers (75+15) degrees per hour.
• To the observer, the sat A will take 6 hours to complete one revolution and sat B will take 4 hours. (Hint: One revolution(=360 deg)/Degrees covered per hour will give the time taken for one revolution.)
• The LCM of 4 and 6 is 12. So there is your answer!

PS: No complex relations involved. Just pure logic and basic math required.

If the satellites are orbiting at the same speed, they cross paths every 2.4 hours - 0.5 of a complete orbit ( $O$ ).

Because the satellites are moving at the same speed, point $P$ can only observe a potential crossing every 0.5 of a complete rotation of Earth ( $R$ ) - every 12 hours.

We need to find a multiple of 2.4 that is a multiple of 12, but also the first such instance where the progress of the satellites within a given orbit ( $O$ ) matches the progress of $P$ within a given rotation of Earth ( $R$ ) - I.e. Where the satellites & $P$ are specifically on the same side of Earth.

$O\pmod{1} = R\pmod{1}$

Thanks goes to @Pranshu Gaba for pointing out that we should check that the crossing of the orbits and point $P$ are on the same side of Earth.

For example, with an orbit time of 9 hours, the first multiple of 4.5 that is a multiple of 12 is 36, but while the timing of a crossing and a 12 hour interval has been met, point P would be on the opposite side of Earth to the satellites (as Earth would be half way through its 2nd rotation, while the satellites complete their 4th full orbit).

Relevant wiki: Uniform Circular Motion

We have to also take into account rotation of the earth.

So for one satellite $ω_{relative}$ $=$ $ω_{Earth}$ $+$ $ω_{satellite}$

Whereas for the other satellite $ω_{relative}$ $=$ $ω_{satellite}$ $-$ $ω_{Earth}$

We know, $ω_{Earth}$ $=$ $2π / T$ $=$ $7.27 * 10^{-5} rad s^{-1}$

(where $T = 24 hours = 86400 s$ )

Also $ω_{satellite}$ $=$ $2π / T$ $=$ $3.63 * 10^{-4} rad s^{-1}$

(where $T = 4.8 hours = 17280 s$ )

Calculating,

$ω_{satellite A}$ $=$ $2.903 * 10^{-4} rad s^{-1}$

$ω_{satellite B}$ $=$ $4.357 * 10^{-4} rad s^{-1}$

Converting their angular velocities into Time periods,

$T_{A}$ $=$ $2.164 * 10^4 s$ $=$ $6.011 hours$ $=$ $6 hours$ (estimation)

$T_{B}$ $=$ $1.442 * 10^4 s$ $=$ $4.005 hours$ $=$ $4 hours$ . (estimation)

The time after which they meet again is just their LCM which is $12$ $hours$ .

Hence the satellites will cross each other at $00:00$ $hours$ .

A，B相遇一次所需时间：2.4h
A，P相遇一次所需时间：3h
B，P相遇一次所需时间：2h
取最小公倍数：12h

English translation:

A, B meet once required time: 2.4h

A, P meet time: 3h

B, P meet once required time: 2h

Take the least common multiple: 12h

Si marcamos la velocidad angular por Revoluciones/hora a partir del periodo de los satelites y de la tierra, y sabiendo que son movimientos no inerciales, podemos marcar una velocidad relativa de los satelites respecto la tierra restando la velocidad angular de esta a la de los satelites. Si metemos esto en ecuaciones de ángulo recorrido y sumamos una vuelta al satelite A, dado que este tiene que recorrer un periode antes de volverse a encontrar con el punto P, la variable de tiempo se despeja como 12 h.

English translation:

If we mark the angular velocity by Revolutions/hour from the period of the satellites and the Earth and knowing that they are non-inertial movements, we can mark a relative speed of the satellites with respect to the Earth subtracting the angular velocity of this one to one of the Satellites. If we put this in angle equations and add a return to satellite A, since it has to travel a period before returning to meet the point P, the time variable is cleared as 12 h.