Question

Two satellites A and B are orbiting the earth at heights of 2.5R and 7.5R from the centre. The ratio of their time periods is:

• A. $\sqrt{3}:1$
• B. $1:3\sqrt{3}$
• C. $1:\sqrt{3}$
• D. $1:2\sqrt{3}$
• E. $3\sqrt{3}:1$

Hint:

$x^{3/2}$ is calculated as $x \cdot \sqrt{x}$. Here, $3^{3/2} = 3\sqrt{3}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves Kepler's third law of planetary motion, which relates the orbital period of a satellite to its orbital radius.
Step 2: Key Formula or Approach:
Kepler's third law states that the square of the orbital period (T) of a satellite is directly proportional to the cube of the semi-major axis of its orbit. For circular orbits, this is the cube of the orbital radius (r). \[ T^2 \propto r^3 \] This means \( \frac{T^2}{r^3} = \text{constant} \). We can use this to form a ratio for two satellites: \[ \left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_A}{r_B}\right)^3 \] Taking the square root of both sides: \[ \frac{T_A}{T_B} = \left(\frac{r_A}{r_B}\right)^{3/2} \] Step 3: Detailed Explanation:
We are given the orbital radii (distances from the center of the Earth): - For satellite A, \( r_A = 2.5R \) - For satellite B, \( r_B = 7.5R \) First, find the ratio of the radii: \[ \frac{r_A}{r_B} = \frac{2.5R}{7.5R} = \frac{2.5}{7.5} = \frac{1}{3} \] Now, use this ratio to find the ratio of the time periods: \[ \frac{T_A}{T_B} = \left(\frac{1}{3}\right)^{3/2} = \frac{1^{3/2}}{3^{3/2}} \] Let's simplify \( 3^{3/2} \): \[ 3^{3/2} = \sqrt{3^3} = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \] So, the ratio of the time periods is: \[ \frac{T_A}{T_B} = \frac{1}{3\sqrt{3}} \] The ratio \( T_A : T_B \) is \( 1 : 3\sqrt{3} \).
Step 4: Final Answer:
The ratio of time periods of A and B is \( 1:3\sqrt{3} \).