Two satellites `A` and `B` revolve around a plant in two coplanar circular orbits in the same sense with radii `10^(4) km` and `2 xx 10^(4) km` respectively. Time period of `A` is `28` hours. What is time period of another satellite?

Explanation

To find the time period of satellite B, we can use Kepler's Third Law of planetary motion, which states that the square of the time period of a satellite is directly proportional to the cube of the semi-major axis (or radius) of its orbit.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of satellite A, \( r_A = 10^4 \) km
- Radius of satellite B, \( r_B = 2 \times 10^4 \) km
- Time period of satellite A, \( T_A = 28 \) hours

2. **Apply Kepler's Third Law:**
According to Kepler's Third Law:
\[ \frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3} \]

3. **Substitute the Known Values:**
Substitute the values of \( T_A \), \( r_A \), and \( r_B \):
\[ \frac{(28)^2}{T_B^2} = \frac{(10^4)^3}{(2 \times 10^4)^3} \]

4. **Calculate the Right Side:**
Simplify the right side:
\[ \frac{(10^4)^3}{(2 \times 10^4)^3} = \frac{10^{12}}{(2^3) \times 10^{12}} = \frac{1}{8} \]

5. **Set Up the Equation:**
Now we have:
\[ \frac{(28)^2}{T_B^2} = \frac{1}{8} \]

6. **Cross Multiply to Solve for \( T_B^2 \):**
Cross multiplying gives:
\[ 28^2 = \frac{T_B^2}{8} \]
Therefore,
\[ T_B^2 = 8 \times 28^2 \]

7. **Calculate \( T_B^2 \):**
Calculate \( 28^2 \):
\[ 28^2 = 784 \]
Thus,
\[ T_B^2 = 8 \times 784 = 6272 \]

8. **Find \( T_B \):**
Taking the square root:
\[ T_B = \sqrt{6272} \]

9. **Simplify \( \sqrt{6272} \):**
We can simplify \( \sqrt{6272} \):
\[ \sqrt{6272} = \sqrt{64 \times 98} = 8\sqrt{98} = 8 \times 7\sqrt{2} = 56\sqrt{2} \]

10. **Final Result:**
Therefore, the time period of satellite B is:
\[ T_B = 56\sqrt{2} \text{ hours} \]