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 find the radius of the circular orbit of a hypothetical planet with an orbital period of 55 days, we can 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. This can be expressed mathematically as:

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

For two planets, we can write:

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

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

### Step 1: Identify the known values
- For Mercury:
- \(R_1 = 5.8 \times 10^{10} \, \text{m}\)
- \(T_1 = 88 \, \text{days}\)

- For the hypothetical planet:
- \(T_2 = 55 \, \text{days}\)
- \(R_2 = ?\)

### Step 2: Convert the periods into the same units
Since we are comparing periods, we can keep them in days for simplicity, but we can also convert them to seconds if needed. Here, we will keep them in days.

### Step 3: Set up the equation using Kepler's Third Law
Using the formula derived from Kepler's Third Law:

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

### Step 4: Substitute the known values into the equation
Substituting the known values:

\[ \frac{(88)^2}{(55)^2} = \frac{(5.8 \times 10^{10})^3}{R_2^3} \]

### Step 5: Calculate the left side
Calculating the left side:

\[ \frac{88^2}{55^2} = \frac{7744}{3025} \approx 2.56 \]

### Step 6: Rearrange the equation to solve for \(R_2^3\)
Rearranging the equation gives:

\[ R_2^3 = \frac{(5.8 \times 10^{10})^3}{2.56} \]

### Step 7: Calculate \(R_2^3\)
Calculating \( (5.8 \times 10^{10})^3 \):

\[ (5.8)^3 = 195.112 \quad \text{and} \quad (10^{10})^3 = 10^{30} \]
Thus,

\[ (5.8 \times 10^{10})^3 = 195.112 \times 10^{30} \approx 1.95112 \times 10^{32} \]

Now substituting back:

\[ R_2^3 = \frac{1.95112 \times 10^{32}}{2.56} \approx 7.628 \times 10^{31} \]

### Step 8: Take the cube root to find \(R_2\)
Now, we find \(R_2\):

\[ R_2 = (7.628 \times 10^{31})^{1/3} \]

Calculating the cube root:

\[ R_2 \approx 4.22 \times 10^{10} \, \text{m} \]

### Final Answer
The radius of the circular orbit of the hypothetical planet is approximately \(4.22 \times 10^{10} \, \text{m}\).

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