Animated Solution for Physics - Gravitation: The time period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $9R$ is

Imagine you are standing on a distant planet, looking up at the night sky. You see two satellites gracefully orbiting the planet. The first satellite is relatively close, tracing a circular path of radius $R$, and it takes a time $T$ to complete one full revolution. The second satellite is much farther out, orbiting at a massive distance of $9R$.

The question we need to answer is: How long does it take for the second, farther satellite to complete its orbit?

To solve this, we don't need to know the mass of the planet or the exact speed of the satellites. We just need to understand the fundamental rule that governs all orbital motion.

Kepler's Third Law

Whenever you see a problem relating the time period of an orbit to its radius, your brain should immediately jump to Kepler's Third Law of Planetary Motion.

Johannes Kepler discovered that the universe follows a beautiful, mathematical rhythm. His third law states that the square of the time period ($T$) of a satellite is directly proportional to the cube of its orbital radius ($R$). Mathematically, this is written as:

Because this proportionality holds true for any satellite orbiting the same central body, we can set up a powerful ratio to compare our two satellites:

This equation is our master key. It allows us to find the unknown time period $T_2$ simply by plugging in the known values.

Let's bring in the values given in the problem. For the first satellite, the radius is $R_1=R$ and the time period is $T_1=T$. For the second satellite, the radius is $R_2=9R$.

Substituting these into our ratio, we get:

Notice how elegantly the $R$ terms cancel out on the right side! This leaves us with a pure number:

Now, we need to isolate $T_2$. To do this, we take the square root of both sides. This gives us a fractional exponent on the right side:

I know fractional exponents can sometimes look intimidating, but let's break it down. The expression $(9{)}^{3/2}$ simply means we first take the square root of $9$, and then cube the result.

The square root of $9$ is $3$. And $3$ cubed ($3\times 3\times 3$) is $27$.

Multiplying both sides by $T$, we arrive at our final answer:

The second satellite takes exactly 27 times longer to complete its orbit compared to the first one!

This makes perfect physical sense. A satellite that is farther away not only has a much longer path to travel, but it also moves slower because the gravitational pull from the planet is weaker at that distance. Kepler's Third Law perfectly captures this dual effect, showing us the elegant clockwork of the cosmos.