Animated Solution for Physics - Gravitation: A satellite is launched into a circular orbit of radius $R$ around Earth, while a second satellite is launched into a circular orbit of radius $1.02R$. The percentage difference in the time periods of the two satellites is

The problem asks us to find the percentage difference in the time periods of two satellites orbiting the Earth, given their orbital radii. This is a classic application of Kepler's Third Law of Planetary Motion, combined with a neat calculus trick for small percentage changes.

Imagine the Earth at the center of our system. We have two satellites. The first satellite is launched into a circular orbit of radius $R$. The second satellite is launched into a slightly larger circular orbit with a radius of $1.02R$.

Because the second satellite is further away from the Earth, it will take longer to complete one full orbit. Our goal is to find out exactly how much longer, in terms of a percentage.

To connect the time period of a satellite to its orbital radius, we rely on Kepler's Third Law. This law states that the square of the time period ($T$) is directly proportional to the cube of the orbital radius ($R$). Mathematically, we write this as:

Taking the square root of both sides, we can express the time period directly in terms of the radius:

This is our master equation. It tells us exactly how $T$ scales with $R$.

Now, we could calculate the exact time periods $T_1$ and $T_2$ and find their percentage difference. However, notice that the change in radius is very small—it goes from $R$ to $1.02R$, which is just a $2\%$ increase.

When dealing with small percentage changes (typically less than $5\%$), we can use the error approximation method from calculus. By taking the natural logarithm of our master equation and differentiating it, we get a direct relationship between the fractional changes:

This equation is incredibly powerful. It tells us that the fractional change in the time period is simply $\frac{3}{2}$ times the fractional change in the radius.

Let's find the fractional change in the radius, $\frac{dR}{R}$. The change in radius $dR$ is:

So, the fractional change is:

Now, we substitute this value back into our differential equation:

This means the fractional change in the time period is $0.03$. To convert this to a percentage difference, we simply multiply by $100\%$:

The time period of the second satellite is exactly $3\%$ longer than the first one. This method not only saves time but also elegantly demonstrates the power of calculus in physics!