Animated Solution for Physics - Gravitation: Consider two satellites $S_1$ and $S_2$ with periods of revolution 1 h and 8 h respectively, revolving around a planet in circular orbits. The ratio of angular velocity of satellite $S_1$ to the angular velocity of satellite $S_2$ is

Imagine a massive planet suspended in the vastness of space. Around this central body, two satellites, $S_1$ and $S_2$, are locked in a continuous cosmic dance, revolving in perfectly circular orbits. The problem provides us with a crucial piece of information: the time it takes for each satellite to complete one full revolution, known as their time period.

The first satellite, $S_1$, is a speedster, completing its orbit in just $T_1=1\text{ h}$. The second satellite, $S_2$, takes a much more leisurely pace, requiring $T_2=8\text{ h}$ to make its way around the planet. Our mission is to determine the ratio of their angular velocities, ${\omega }_1$ to ${\omega }_2$.

To solve this, we need to bridge the gap between the time period we are given and the angular velocity we need to find. But what exactly is angular velocity? In simple terms, it is the rate at which an object sweeps out an angle as it moves along a circular path.

For a complete circular orbit, the total angle swept is $2\pi$ radians. The time taken to sweep this angle is the time period, $T$. Therefore, the fundamental relationship connecting these two quantities is:

This elegant equation tells us exactly how fast the satellite is rotating in terms of angle per unit time.

Let's look closely at our master equation. The numerator, $2\pi$, is a constant value for any complete circle. Because the time period $T$ is in the denominator, we can deduce a direct mathematical proportionality:

This inverse relationship is highly intuitive. If a satellite takes a very long time to complete an orbit (a large $T$), it must be sweeping out angles very slowly (a small $\omega$). Conversely, a short time period implies a rapid angular sweep.

Because of this inverse proportionality, when we set up a ratio comparing the angular velocities of our two satellites, the time periods will flip. The ratio of ${\omega }_1$ to ${\omega }_2$ will be equal to the ratio of $T_2$ to $T_1$:

Now, we execute the final atomic compute. We substitute the raw values given in the problem statement into our derived ratio equation. We know that $T_2=8\text{ h}$ and $T_1=1\text{ h}$.

Plugging these in, we get:

This gives us our final answer. The ratio of the angular velocity of satellite $S_1$ to the angular velocity of satellite $S_2$ is $8:1$. This means the inner satellite is sweeping through its orbit eight times faster in terms of angle than the outer satellite. Looking at our options, this corresponds perfectly to option (a).

Kepler's Third Law

While this problem was straightforward, it's always good to anticipate how concepts can be stretched. What if the examiner had asked for the ratio of their orbital radii instead?

In that scenario, we would need to invoke Kepler's Third Law of Planetary Motion, which states that the square of the time period is directly proportional to the cube of the orbital radius:

By mastering these fundamental relationships, you build a robust toolkit that can dismantle any orbital mechanics problem thrown your way!