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 Pluto from the sun is 1427 ×106 km. Using Kepler’s third law, calculate the period of Pluto.

A. 2.9 years

B. 29 years

C. 124 years

D. 248 years

Answer 1

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be expressed 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 average distance from the sun of the first planet (Earth, in this case).
- \( T_2 \) and \( a_2 \) are the period and average distance from the sun of the second planet (Pluto).

Given:
- \( a_1 = 149.6 \times 10^6 \) km (Earth's average distance from the sun)
- \( T_1 = 1.0 \) year (Earth's period)
- \( a_2 = 1427 \times 10^6 \) km (Pluto's average distance from the sun)

We need to find \( T_2 \), the period of Pluto.

First, calculate \( a_1^3 \) and \( a_2^3 \):

\[
a_1^3 = (149.6 \times 10^6)^3
\]
\[
a_2^3 = (1427 \times 10^6)^3
\]

Using the formula from Kepler's third law:

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

Rearranging for \( T_2 \):

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

Now substituting the values:

1. Compute \( \frac{a_2^3}{a_1^3} \):

\[
\frac{a_2^3}{a_1^3} = \left( \frac{a_2}{a_1} \right)^3 = \left( \frac{1427 \times 10^6}{149.6 \times 10^6} \right)^3
\]

This simplifies to:

\[
\frac{1427}{149.6} \approx 9.527
\]
Now, we can raise this to the power of 3:

\[
9.527^3 \approx 860
\]

2. Now plug this into the period equation:

\[
T_2 = 1 \cdot \sqrt{860} \approx 29.32 \text{ years}
\]

So, rounding off, the approximate period of Pluto is:

**B. 29 years**