Animated Solution for Physics - Gravitation: A geostationary satellite is orbiting around an arbitrary planet $P$ at a height of $11R$ above the surface of $P$, $R$ being the radius of $P$. The time period of another satellite in hours at a height of $2R$ from the surface of $P$ is ............ . $P$ has the time period of 24 h.

Welcome to a fascinating journey into the mechanics of the cosmos! Today, we are going to tackle a brilliant problem involving satellites, planetary orbits, and the timeless elegance of Kepler's Laws. This question is a fantastic test of your conceptual clarity, specifically regarding how we measure distances in gravitational physics.

Let's dive right in and decode the universe!

Visualizing the Orbits

Imagine an arbitrary planet, let's call it $P$, with a radius $R$. We are given two satellites orbiting this planet.

The first satellite is a geostationary satellite. What does that mean? A geostationary satellite is one that appears stationary relative to a point on the planet's surface. For this to happen, the satellite must complete one full orbit in the exact same time it takes the planet to complete one full rotation on its axis. The problem tells us that planet $P$ has a time period of $24\text{ h}$. Therefore, the time period of our first satellite, $T_1$, is also exactly $24\text{ h}$.

Now, let's look at its position. We are told it orbits at a height of $11R$ above the surface of the planet.

Surface vs. Center

Here lies the most common pitfall in gravitation problems! Gravity acts as if all the mass of the planet is concentrated at its center. Therefore, all orbital distances must be measured from the center of the planet, not its surface.

If the height is $11R$, the true orbital radius $r_1$ is the radius of the planet plus the height:

Similarly, we have a second satellite orbiting at a height of $2R$. Applying the same logic, its true orbital radius $r_2$ is:

Kepler's Third Law

We now have the orbital radii of both satellites ($12R$ and $3R$) and the time period of the first satellite ($24\text{ h}$). We need to find the time period of the second satellite, $T_2$.

This is where Johannes Kepler comes to our rescue. Kepler's Third Law of Planetary Motion states that the square of the orbital time period is directly proportional to the cube of the semi-major axis of its orbit (which is simply the orbital radius for circular orbits).

Mathematically, this is expressed as:

Because this proportionality holds true for any satellite orbiting the same central body (planet $P$), we can set up a beautiful ratio comparing our two satellites:

Crunching the Numbers

Now, we simply substitute our known values into this master equation. Let's plug in the raw data:

Look at the right side of the equation. The variable $R$ elegantly cancels out, leaving us with a simple fraction:

We know that $4^3=4\times 4\times 4=64$. So, our equation simplifies to:

To solve for $T_2$, we take the square root of both sides. The square root of $64$ is $8$:

A quick rearrangement gives us our final answer:

The second satellite completes its orbit in just $3\text{ h}$! Notice the profound physical reality here: the satellite that is closer to the planet ($3R$ vs $12R$) orbits much, much faster ($3\text{ h}$ vs $24\text{ h}$). This is the essence of orbital mechanics—closer orbits require higher speeds to counteract the stronger gravitational pull.

Always remember the golden rule of gravitation: Measure from the center! Master this, and Kepler's laws will always guide you to the right answer.