The time period of a satellite of earth is $5\text{ hours}$. If the separation between the earth and the satellite is increased to four times the previous value, the new time period of the satellite will be
Show Hint
When applying Kepler's third law fractional exponents mentally ($r \rightarrow r^{3/2}$), remember this simple sequence: take the square root of the distance scaling factor first, then multiply that result by the initial time period value. Here: $\sqrt{4} = 2 \rightarrow 2^3 = 8 \rightarrow 8 \times 5 = 40\text{ hours}$.
Step 1: State the problem. A satellite has a period of $5$ hours. Its orbital separation is made $4$ times larger. We want the new period. Step 2: Recall Kepler's third law. The square of the period is proportional to the cube of the orbital radius: $T^2 \propto r^3$, so $T \propto r^{3/2}$. Step 3: Write the ratio of periods. $\dfrac{T_2}{T_1} = \left(\dfrac{r_2}{r_1}\right)^{3/2}$. Step 4: Insert the radius factor. Here $r_2 = 4 r_1$, so $\dfrac{r_2}{r_1} = 4$, giving $\dfrac{T_2}{T_1} = 4^{3/2}$. Step 5: Evaluate $4^{3/2}$. Take the square root first then cube: $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$. Step 6: Find the new period. $T_2 = 8 \times T_1 = 8 \times 5 = 40$ hours. \[ \boxed{T_2 = 40\ \text{hours (option 2)}} \]