The time period of a satellite of the earth is 10 hours. If the separations between the earth and the satellite is increased to 4 times the previous value, then what will be the new time period of the satellite ?

Options

A. 2 hours
B. 10 hours
C. 40 hours
D. 80 hours

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the time period (T) of a satellite is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this can be expressed as:

\[ T^2 \propto r^3 \]

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Let the initial time period of the satellite be \( T_1 = 10 \) hours.
- Let the initial separation (radius) between the Earth and the satellite be \( r_1 \).

2. **Determine the new separation**:
- According to the problem, the separation is increased to 4 times the previous value. Therefore, the new separation is:
\[ r_2 = 4r_1 \]

3. **Apply Kepler's Third Law**:
- According to Kepler's law, we can write the relationship between the time periods and the separations as:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

4. **Substituting the values**:
- Substitute \( r_2 = 4r_1 \) into the equation:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{(4r_1)^3} \]
- This simplifies to:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{64r_1^3} = \frac{1}{64} \]

5. **Cross-multiplying**:
- Cross-multiply to find the relationship between \( T_1 \) and \( T_2 \):
\[ T_1^2 = \frac{1}{64} T_2^2 \]

6. **Taking square roots**:
- Taking the square root of both sides gives:
\[ T_2 = 8 T_1 \]

7. **Calculating the new time period**:
- Substitute \( T_1 = 10 \) hours into the equation:
\[ T_2 = 8 \times 10 = 80 \text{ hours} \]

### Final Answer:
The new time period of the satellite \( T_2 \) is **80 hours**.