Question

The average mean distance of the earth from the sun is 149.6 ×106 km and the period of the earth is 1.0 year. The average mean distance of Mars from the sun is 227.9 ×106 km. Using Kepler’s third law, calculate the period of Mars.

A. 124.1 years

B. 290.5 years

C. 29.8 years

D. 1.9 years

Answer 1

Kepler's third law states that the square of the period of a planet (\( T \)) is proportional to the cube of the semi-major axis of its orbit (\( a \)) around the Sun. This can be expressed mathematically as:

\[
\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}
\]

Where:
- \( T_1 \) and \( a_1 \) are the period and semi-major axis of Earth.
- \( T_2 \) and \( a_2 \) are the period and semi-major axis of Mars.

Given:
- \( a_1 = 149.6 \times 10^6 \text{ km} \)
- \( T_1 = 1 \text{ year} \)
- \( a_2 = 227.9 \times 10^6 \text{ km} \)

We need to find \( T_2 \).

First, we can plug in the values into Kepler's third law:

\[
\frac{1^2}{(149.6 \times 10^6)^3} = \frac{T_2^2}{(227.9 \times 10^6)^3}
\]

Rearranging the equation to solve for \( T_2^2 \):

\[
T_2^2 = \left( \frac{(227.9 \times 10^6)^3}{(149.6 \times 10^6)^3} \right) \times 1^2
\]

Calculating the ratio of the distances:

\[
\frac{(227.9)^3}{(149.6)^3} = \left( \frac{227.9}{149.6} \right)^3
\]

Calculating the ratio:

\[
\frac{227.9}{149.6} \approx 1.524
\]

Now calculating \( (1.524)^3 \):

\[
(1.524)^3 \approx 3.531
\]

So now we have:

\[
T_2^2 \approx 3.531
\]

Taking the square root to find \( T_2 \):

\[
T_2 \approx \sqrt{3.531} \approx 1.88 \text{ years}
\]

Thus, the period of Mars is approximately 1.9 years.

The answer is **D. 1.9 years**.