Animated Solution for Physics - Gravitation: The time period of a satellite of earth is $5\text{ h}$. If the separation between the earth and the satellite is increased to $4$ times the previous value, the new time period will become

Welcome, future engineers and physicists! Today, we are going to embark on a cosmic journey. We are not just solving a problem; we are uncovering the very laws that govern the dance of the planets and the satellites in our universe.

Imagine you are standing in a control room, monitoring a satellite orbiting our beautiful blue planet, Earth. This satellite is cruising along at a steady pace, completing one full revolution every 5 hours. It's a perfect, predictable rhythm.

But what if we decide to push this satellite further out into the depths of space? What if we fire its thrusters and move it to an orbit that is exactly four times farther away from the Earth than its current position? How will this drastic change in distance affect its time period? Will it take 4 times longer? 16 times longer? Let's find out!

Kepler's Third Law

To solve this cosmic puzzle, we need to consult one of the greatest astronomers in history: Johannes Kepler. After years of meticulously analyzing the astronomical data collected by Tycho Brahe, Kepler discovered three fundamental laws of planetary motion.

The law that holds the key to our problem is Kepler's Third Law, also known as the Law of Harmonies. It states something profoundly beautiful: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

For a circular orbit, the semi-major axis is simply the radius of the orbit, $r$. Mathematically, we can write this as:

This isn't just a random mathematical coincidence. It is a direct consequence of Isaac Newton's Law of Universal Gravitation. If we equate the gravitational force to the required centripetal force, $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$, and substitute the orbital velocity $v=\frac{2\pi r}{T}$, we arrive exactly at this proportional relationship!

Now that we have our master equation, let's apply it to our specific scenario. We have two distinct states for our satellite: the initial orbit and the final, expanded orbit.

Because $T^2$ is proportional to $r^3$, the ratio of the squares of the time periods must equal the ratio of the cubes of their respective radii. We can set up this elegant equation:

Let's carefully substitute the values given in the problem. We know the initial time period $T_1$ is $5\text{ h}$. Let's call the initial radius $r_1=r$. The problem states that the new separation is 4 times the previous value, so $r_2=4r$.

Plugging these into our machinery, we get:

This is where the math starts to look beautiful. Let's focus on the right side of our equation. We have $r^3$ in the numerator and $(4r{)}^3$ in the denominator.

When we expand the denominator, we must remember to cube both the $4$ and the $r$. The cube of $4$ is $4\times 4\times 4=64$. So, the denominator becomes $64r^3$.

Our equation now looks like this:

Notice how the $r^3$ terms perfectly cancel each other out! The specific initial radius doesn't even matter; only the ratio of the radii dictates the change in the time period. We are left with a clean, pure numerical fraction:

We are now just a few algebraic steps away from our answer. To isolate $T_2^2$, we can simply cross-multiply.

Now, you could multiply $25$ by $64$ to get $1600$, and then take the square root. Or, you can be a bit more clever and take the square root of both sides immediately, since both $25$ and $64$ are perfect squares!

Let's do it the clever way:

And there we have it! The new time period of the satellite is exactly 40 hours.

Why Does It Slow Down?

Let's take a step back and appreciate what this result means physically. We increased the radius by a factor of 4, but the time period increased by a factor of 8 (from 5 hours to 40 hours). Why such a dramatic increase?

There are two compounding factors at play here. First, by moving the satellite 4 times further away, the circumference of its orbit (the total distance it must travel) has become 4 times larger.

Second, and more importantly, the gravitational pull of the Earth is significantly weaker at that distance. Because gravity follows an inverse-square law, moving 4 times further away means the gravitational force is 16 times weaker!

With a weaker gravitational force pulling on it, the satellite doesn't need to travel as fast to maintain its orbit. In fact, its orbital speed decreases. So, the satellite is traveling a much longer path, and it's doing so at a slower speed.

These two effects combine to produce the massive increase in the time period. This is the profound beauty of physics: a simple mathematical ratio perfectly captures the complex, dynamic reality of the cosmos. Keep exploring, keep questioning, and never lose your sense of wonder!