Assuming that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. The length of the martian year is days is

Options

A. `(1.52)^(2//3)xx365`
B. `(1.52)^(3//2)xx365`
C. `(1.52)^(2)xx365`
D. `(1.52)^(3)xx365`

Explanation

To find the length of the Martian year in days, we can use Kepler's third law of planetary motion. Here's a step-by-step solution:

### Step 1: Understand Kepler's Third Law
Kepler's third law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, this is expressed as:

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

This can be written as:

\[ \frac{T_m^2}{T_e^2} = \frac{a_m^3}{a_e^3} \]

Where:
- \( T_m \) = orbital period of Mars
- \( T_e \) = orbital period of Earth
- \( a_m \) = orbital radius of Mars
- \( a_e \) = orbital radius of Earth

### Step 2: Define the Known Values
From the problem, we know:
- The orbital radius of Mars is 1.52 times that of Earth:
\[ a_m = 1.52 \times a_e \]
- The orbital period of Earth is:
\[ T_e = 365 \text{ days} \]

### Step 3: Substitute the Known Values into Kepler's Law
Using the relationship between the orbital radii, we can substitute \( a_m \) into Kepler's law:

\[ \frac{T_m^2}{T_e^2} = \frac{(1.52 \cdot a_e)^3}{a_e^3} \]

This simplifies to:

\[ \frac{T_m^2}{T_e^2} = (1.52)^3 \]

### Step 4: Solve for \( T_m \)
Now, we can express \( T_m \) in terms of \( T_e \):

\[ T_m^2 = (1.52)^3 \cdot T_e^2 \]

Taking the square root of both sides gives us:

\[ T_m = T_e \cdot (1.52)^{3/2} \]

### Step 5: Substitute \( T_e \) to Find \( T_m \)
Now, substituting \( T_e = 365 \) days:

\[ T_m = 365 \cdot (1.52)^{3/2} \]

### Step 6: Calculate \( (1.52)^{3/2} \)
First, calculate \( (1.52)^{3/2} \):

1. Calculate \( 1.52^{3} \):
\[ 1.52^3 \approx 3.51 \]

2. Calculate \( (1.52)^{3/2} \):
\[ (1.52)^{3/2} \approx \sqrt{3.51} \approx 1.87 \]

### Step 7: Final Calculation
Now, substitute back into the equation for \( T_m \):

\[ T_m \approx 365 \cdot 1.87 \approx 683.55 \text{ days} \]

### Conclusion
Thus, the length of the Martian year is approximately **684 days**.

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