A satellite of Sun is in a circular orbit around the Sun, midway between the Suna and earth. Find the period of this satellite.

Options

A. the period of the planet is nearly 229 days 129 days
B. the period of the planet is nearly 129 days
C. the speed of the planet is greater than the escape velocity of the earth
D. the acceleration of the planet is two times the acceleration of the earth

Answer

To find the period of a satellite orbiting the Sun midway between the Sun and the Earth, 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 period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit around the Sun. Mathematically, this is expressed as:

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

### Step 2: Establish the Known Values
We know that the Earth orbits the Sun with a period of 1 year (T_E = 1 year) at a distance (r) of 1 astronomical unit (AU). Thus, we can write:

\[ T_E^2 = k \cdot r^3 \]

Where \( k \) is a constant.

### Step 3: Calculate the Constant \( k \)
For the Earth:

\[ 1^2 = k \cdot (1)^3 \Rightarrow k = 1 \]

### Step 4: Determine the Distance of the Satellite
The satellite is located midway between the Sun and the Earth. Therefore, the distance of the satellite from the Sun (r') is:

\[ r' = \frac{r}{2} = \frac{1 \text{ AU}}{2} \]

### Step 5: Apply Kepler's Law for the Satellite
Using Kepler's Third Law for the satellite, we have:

\[ T_{satellite}^2 = k \cdot (r')^3 \]

Substituting \( r' = \frac{1}{2} \):

\[ T_{satellite}^2 = k \cdot \left(\frac{1}{2}\right)^3 = k \cdot \frac{1}{8} \]

Since we found \( k = 1 \):

\[ T_{satellite}^2 = \frac{1}{8} \]

### Step 6: Solve for \( T_{satellite} \)
Taking the square root of both sides gives:

\[ T_{satellite} = \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \text{ years} \]

### Step 7: Convert to Days
To convert the period from years to days, we multiply by the number of days in a year (approximately 365 days):

\[ T_{satellite} = \frac{365}{2\sqrt{2}} \approx \frac{365}{2 \times 1.414} \approx \frac{365}{2.828} \approx 129 \text{ days} \]

### Final Answer
The period of the satellite is approximately **129 days**.

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