Question

If T is the time period of satellite in an orbit of radius r, then the time period in an orbit of radius 3r is:

Hint:

A useful shortcut for Kepler's 3rd law problems where radius increases by a factor $k$ ($r_2 = kr_1$): the new time period will be $k^{3/2} \times T_1$ or $k\sqrt{k} \times T_1$. Here $k=3$, so $T_{new} = 3\sqrt{3} T$.

Answer 1

Step 1: Understanding the Concept:
The relationship between the orbital time period of a satellite and its orbital radius is governed by Kepler's Third Law of Planetary Motion.

Step 2: Key Formula or Approach:
Kepler's Third Law states that the square of the time period ($T$) is directly proportional to the cube of the orbital radius ($r$):
\[ T^2 \propto r^3 \]
For two different orbits around the same central body:
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]

Step 3: Detailed Explanation:
Let the initial orbit be: radius $r_1 = r$, time period $T_1 = T$.
Let the new orbit be: radius $r_2 = 3r$, time period $T_2$.
Using the ratio formula:
\[ \frac{T_2^2}{T^2} = \left(\frac{3r}{r}\right)^3 \]
\[ \frac{T_2^2}{T^2} = (3)^3 \]
\[ \frac{T_2^2}{T^2} = 27 \]
Multiply both sides by $T^2$:
\[ T_2^2 = 27 T^2 \]
Take the square root of both sides:
\[ T_2 = \sqrt{27} \cdot T \]
Simplify the radical ($\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$):
\[ T_2 = 3\sqrt{3} T \]

Step 4: Final Answer:
The new time period is $3\sqrt{3} T$.