Radius of orbit of a satellite is R and T is time period. Find T' when orbit radius increase to 9R

Answer

To solve the problem of finding the new time period \( T' \) when the radius of the satellite's orbit increases from \( R \) to \( 9R \), we can use Kepler's Third Law of planetary motion, which states that the square of the period of orbit \( T \) is directly proportional to the cube of the semi-major axis (orbital radius) of the orbit.

### Step-by-Step Solution:

1. **Understand Kepler's Third Law**:
According to Kepler's Third Law, the relationship can be expressed as:
\[ T^2 \propto R^3 \]
This means:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
where \( T_1 \) and \( R_1 \) are the initial period and radius, and \( T_2 \) and \( R_2 \) are the new period and radius.

2. **Assign Values**:
Let:
- Initial radius \( R_1 = R \)
- New radius \( R_2 = 9R \)
- Initial period \( T_1 = T \)
- New period \( T_2 = T' \)

3. **Set Up the Equation**:
Substitute the known values into the equation:
\[ \frac{T^2}{T'^2} = \frac{R^3}{(9R)^3} \]

4. **Simplify the Right Side**:
Calculate \( (9R)^3 \):
\[ (9R)^3 = 729R^3 \]
Thus, the equation becomes:
\[ \frac{T^2}{T'^2} = \frac{R^3}{729R^3} = \frac{1}{729} \]

5. **Cross Multiply**:
Rearranging gives:
\[ T'^2 = 729T^2 \]

6. **Take the Square Root**:
To find \( T' \), take the square root of both sides:
\[ T' = \sqrt{729}T = 27T \]

### Final Answer:
The new time period \( T' \) when the orbit radius increases to \( 9R \) is:
\[ T' = 27T \]