Question

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

• A. $20\text{ hours}$
• B. $40\text{ hours}$
• C. $80\text{ hours}$
• D. $10\text{ hours}$

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: $\sqrt{4} = 2 \rightarrow 2^3 = 8 \rightarrow 8 \times 5 = 40\text{ hours}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question explores how the orbital period ($T$) of an Earth satellite changes when its separation distance ($r$) from the center of the Earth is increased by a factor of 4.

Step 2: Key Formula or Approach:
According to Kepler's Third Law of Planetary Motion, the square of the orbital time period of a satellite is directly proportional to the cube of its orbital radius:
$$T^2 \propto r^3 \implies T \propto r^{3/2}$$ This allows us to set up a direct ratio comparison between the final and initial states:
$$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$$

Step 3: Detailed Explanation:
Let the initial time period be $T_1 = 5\text{ hours}$ and the initial separation be $r_1$.
The new separation is given as $r_2 = 4r_1$.
Substitute these relations into our ratio equation:
$$\frac{T_2}{5} = \left(\frac{4r_1}{r_1}\right)^{3/2}$$ $$\frac{T_2}{5} = (4)^{3/2}$$ To compute $(4)^{3/2}$ easily, take the square root of 4 first, and then cube the result:
$$(4)^{3/2} = (\sqrt{4})^3 = (2)^3 = 8$$ Now substitute this back into the equation:
$$\frac{T_2}{5} = 8$$ Isolating $T_2$ by multiplying both sides by 5:
$$T_2 = 8 \times 5 = 40\text{ hours}$$

Step 4: Final Answer:
The new time period of the satellite is $40\text{ hours}$, which corresponds to option (B).