The radius of the nearly circular orbit of mercury is `5.8xx10^(10)` m and its orbital period is 88 days. If a hypothetical planet has an orbital period of 55 days, what is the radius of its circular orbit ?

Explanation

To solve the problem, we will use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (R) of its orbit. Mathematically, this is expressed as:

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

This can be rewritten as:

\[ \frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \]

Where:
- \( R_1 \) and \( T_1 \) are the radius and period of Mercury.
- \( R_2 \) and \( T_2 \) are the radius and period of the hypothetical planet.

### Step-by-step Solution:

1. **Identify the given values:**
- Radius of Mercury's orbit, \( R_1 = 5.8 \times 10^{10} \) m
- Orbital period of Mercury, \( T_1 = 88 \) days
- Orbital period of the hypothetical planet, \( T_2 = 55 \) days

2. **Convert the periods from days to seconds:**
- \( T_1 = 88 \) days \( = 88 \times 24 \times 60 \times 60 \) seconds
- \( T_2 = 55 \) days \( = 55 \times 24 \times 60 \times 60 \) seconds

Calculating these:
- \( T_1 = 88 \times 86400 = 7614720 \) seconds
- \( T_2 = 55 \times 86400 = 4752000 \) seconds

3. **Use Kepler's Third Law:**
\[ \frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \]

Rearranging gives:
\[ R_2^3 = R_1^3 \cdot \left(\frac{T_2^2}{T_1^2}\right) \]

4. **Calculate \( R_1^3 \):**
\[ R_1^3 = (5.8 \times 10^{10})^3 \]

5. **Calculate \( T_2^2 \) and \( T_1^2 \):**
\[ T_2^2 = (4752000)^2 \]
\[ T_1^2 = (7614720)^2 \]

6. **Substituting the values into the equation:**
\[ R_2^3 = (5.8 \times 10^{10})^3 \cdot \left(\frac{(4752000)^2}{(7614720)^2}\right) \]

7. **Taking the cube root to find \( R_2 \):**
\[ R_2 = \sqrt[3]{R_1^3 \cdot \left(\frac{T_2^2}{T_1^2}\right)} \]

8. **Substituting the calculated values to find \( R_2 \):**
After performing the calculations, we find:
\[ R_2 \approx 4.24 \times 10^{10} \text{ m} \]

### Final Answer:
The radius of the circular orbit of the hypothetical planet is approximately \( R_2 = 4.24 \times 10^{10} \) m.