The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will be

Options

A. `1//2` year
B. `2sqrt(2)` years
C. `4` years
D. `8` years

Answer

To solve the problem of how the period of revolution changes when the distance between the Earth and the Sun is doubled, 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 of revolution (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit around the Sun. Mathematically, this can be expressed as:
\[ T^2 \propto r^3 \]
This means:
\[ \frac{T^2}{T'^2} = \frac{r^3}{r'^3} \]
where \( T \) is the original period, \( T' \) is the new period, \( r \) is the original distance, and \( r' \) is the new distance.

### Step 2: Define the Variables
Let:
- \( T = 1 \) year (the original period of revolution)
- \( r \) = original distance between the Earth and the Sun
- \( r' = 2r \) (the new distance, which is double the original)

### Step 3: Apply Kepler's Law
Using the relationship from Kepler's Third Law:
\[ \frac{T^2}{T'^2} = \frac{r^3}{(2r)^3} \]

### Step 4: Simplify the Equation
Substituting \( r' = 2r \):
\[ \frac{T^2}{T'^2} = \frac{r^3}{8r^3} = \frac{1}{8} \]

### Step 5: Solve for the New Period
Rearranging the equation gives:
\[ T'^2 = 8T^2 \]
Taking the square root of both sides:
\[ T' = \sqrt{8}T = 2\sqrt{2}T \]

### Step 6: Substitute the Original Period
Since \( T = 1 \) year:
\[ T' = 2\sqrt{2} \times 1 \text{ year} = 2\sqrt{2} \text{ years} \]

### Final Answer
The new period of revolution when the distance is doubled is:
\[ T' = 2\sqrt{2} \text{ years} \]

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