The period of revolution of a satellite orbiting Earth at a height 4R above the surface of Warth is x hrs, where R is the radius of earth. The period of another satellite at a height 1.5R from the surface of the Earth is

Options

A. `(x)/(2sqrt(2))` hrs
B. `(x)/(sqrt(2))` hrs
C. `sqrt(2)x` hrs
D. `(x)/(2)` hrs

Answer

To solve the problem, we need to find the period of revolution of a satellite at a height of 1.5R above the Earth's surface, given that the period of a satellite at a height of 4R is x hours.

### Step-by-Step Solution:

1. **Identify the Radius of the Orbits:**
- The radius of the Earth (R) is given.
- For the first satellite at a height of 4R above the surface, the total radius (R1) from the center of the Earth is:
\[ R_1 = R + 4R = 5R \]
- For the second satellite at a height of 1.5R above the surface, the total radius (R2) from the center of the Earth is:
\[ R_2 = R + 1.5R = 2.5R \]

2. **Apply Kepler's Third Law:**
- According to Kepler's Third Law, the square of the period of revolution (T) of a satellite is proportional to the cube of the semi-major axis (R) of its orbit:
\[ T^2 \propto R^3 \]
- Therefore, we can write:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]

3. **Substitute the Values:**
- We know \(T_1 = x\) (the period of the first satellite).
- Substituting \(R_1\) and \(R_2\):
\[ \frac{T_2^2}{x^2} = \frac{(2.5R)^3}{(5R)^3} \]

4. **Simplify the Equation:**
- Simplifying the right side:
\[ \frac{(2.5)^3}{(5)^3} = \frac{15.625}{125} = \frac{1}{8} \]
- Thus, we have:
\[ \frac{T_2^2}{x^2} = \frac{1}{8} \]

5. **Solve for T2:**
- Taking the square root of both sides:
\[ \frac{T_2}{x} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]
- Therefore:
\[ T_2 = \frac{x}{2\sqrt{2}} \]

6. **Final Result:**
- The period of the second satellite at a height of 1.5R from the surface of the Earth is:
\[ T_2 = \frac{x}{2\sqrt{2}} \text{ hours} \]