A geostationary satellite above the equator is orbiting around the earth at a fixed distance r1 from

Q: Why do we add the angular velocities?

Because the satellites are moving in opposite directions. Just like two cars driving towards each other seem to be moving faster from the perspective of the drivers, the relative angular velocity is the sum of their individual speeds.

Q: How did we know the time period of the geostationary satellite?

By definition, a geostationary satellite remains fixed above a single point on Earth's equator. This means it must complete one orbit in exactly the same time it takes Earth to rotate once on its axis, which is 24 hours.

Q: Is it safe to approximate 1.331 as 4/3?

Yes, in this context it is a very safe and smart approximation. $4/3$ is $1.333...$, which is extremely close to $1.331$. It simplifies the algebra significantly, and the final answer ($2.33$) remains the same within the required precision.

Q: Why is $\omega \propto r^{-3/2}$?

Kepler's Third Law states $T^2 \propto r^3$, which means $T \propto r^{3/2}$. Since angular velocity $\omega = \frac{2\pi}{T}$, it is inversely proportional to $T$. Therefore, $\omega \propto \frac{1}{r^{3/2}} = r^{-3/2}$.

Q: What does 'time period as measured from the geostationary satellite' mean?

It refers to the synodic period—the time it takes for the second satellite to return to the same relative position with respect to the geostationary satellite. This is governed by their relative angular velocity.

Question

A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_2$ from the center of the earth, such that $r_1 = 1.21 r_2$. The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is _____

Sigma Percentile · Accepted Answer

### The Orbital Setup

Imagine you are standing on a geostationary satellite (GSS) orbiting the Earth. By definition, this satellite is locked in sync with Earth's rotation, meaning its time period $T_1$ is exactly $24 	ext{ hours}$.

Now, look down towards the Earth. There is a second satellite, let's call it $S$, orbiting at a closer distance $r_2$. The problem gives us a precise geometric constraint: the radius of the GSS orbit $r_1$ is $1.21$ times the radius of the inner orbit $r_2$.

Crucially, this second satellite is moving in the **opposite direction** to the Earth's rotation. Our mission is to find the time period of this second satellite *as observed from the GSS*.

### Invoking Kepler's Third Law

To understand how fast the second satellite is moving, we must consult Kepler's Third Law of Planetary Motion. This fundamental law states that the square of the orbital time period is directly proportional to the cube of the orbital radius:

$T^2 \propto r^3$

Taking the square root of both sides, we see that $T \propto r^{3/2}$.

However, we are interested in the angular velocity $\omega$. Since angular velocity is inversely proportional to the time period ($\omega = \frac{2\pi}{T}$), the relationship flips:

$\omega \propto r^{-3/2}$

### The Ratio of Angular Velocities

Using this inverse relationship, we can set up a ratio between the angular velocities of the two satellites. The angular velocity of the second satellite $\omega_2$ compared to the GSS $\omega_1$ is:

$\frac{\omega_2}{\omega_1} = \left(\frac{r_1}{r_2}ight)^{3/2}$

Notice how the radii indices are flipped ($r_1$ is on top) because of the negative exponent. We know from the problem that $\frac{r_1}{r_2} = 1.21$. Substituting this in:

$\frac{\omega_2}{\omega_1} = (1.21)^{3/2}$

### The Art of Smart Approximation

At first glance, $(1.21)^{3/2}$ looks like a nightmare to calculate without a calculator. But let's break it down. We know that $1.21$ is simply the square of $1.1$.

$(1.21)^{3/2} = ((1.1)^2)^{3/2}$

The powers of $2$ cancel out beautifully, leaving us with:

$(1.1)^3 = 1.331$

So, $\omega_2 = 1.331 \omega_1$.

Here is where we apply a brilliant physicist's trick. The number $1.331$ is incredibly close to $1.333...$, which is exactly the fraction $\frac{4}{3}$. By approximating $1.331 \approx \frac{4}{3}$, we make the subsequent algebra vastly more elegant without sacrificing meaningful precision.

$\omega_2 \approx \frac{4}{3}\omega_1$

### Calculating Relative Motion

Because the two satellites are orbiting in **opposite directions**, their relative angular velocity is the sum of their individual angular velocities. Think of two cars driving towards each other; their closing speed is the sum of their speeds.

$\omega_{	ext{rel}} = \omega_2 + \omega_1$

Substituting our approximated value:

$\omega_{	ext{rel}} = \frac{4}{3}\omega_1 + \omega_1 = \frac{7}{3}\omega_1$

### The Final Time Period

The time period of the second satellite as measured from the GSS is simply $2\pi$ divided by their relative angular velocity:

$t_0 = \frac{2\pi}{\omega_{	ext{rel}}}$

$t_0 = \frac{2\pi}{\frac{7}{3}\omega_1} = \frac{3}{7} \left(\frac{2\pi}{\omega_1}ight)$

But wait! The expression $\frac{2\pi}{\omega_1}$ is exactly the time period of the geostationary satellite, which we know is $24 	ext{ hours}$.

$t_0 = \frac{3}{7} 	imes 24 	ext{ hours}$

The problem states that this relative time period is equal to $\frac{24}{p}$ hours. Equating the two expressions:

$\frac{24}{p} = \frac{3 	imes 24}{7}$

The $24$s cancel out perfectly, leaving:

$\frac{1}{p} = \frac{3}{7} \implies p = \frac{7}{3}$

Calculating the final decimal value gives us $p \approx 2.33$.

*(Note: Even if we had used the exact value of $1.331$, we would have found $p = 2.331$, which still rounds to $2.33$. The approximation simply saved us from tedious decimal arithmetic!)*

Enter Numerical Value:

Visualized Solution

$r_{1} = 1.21 r_{2}$

$T^{2} \propto r^{3}$

$\frac{ω _{2}}{ω _{1}} = (\frac{r _{1}}{r _{2}})^{3/2}$

$\frac{ω _{2}}{ω _{1}} = (1.21)^{3/2}$

$ω_{2} \approx \frac{4}{3} ω_{1}$

$ω_{rel} = ω_{1} + ω_{2}$

$t_{0} = \frac{2 π}{ω _{rel}}$

$p = 2.33$

The Sigma Insight: Orbital Motion of a Satellite

The Orbital Setup

Invoking Kepler's Third Law

The Ratio of Angular Velocities

The Art of Smart Approximation

Calculating Relative Motion

The Final Time Period

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