To solve this question, we need to use Kepler's Third Law of Planetary Motion, which relates the period of orbit to the size of the orbit's semimajor axis. Kepler's Third Law is expressed mathematically as:

\(T^2 \propto a^3\)

Here, \(T\) is the orbital period (time taken for one complete orbit around the sun), and \(a\) is the semimajor axis of the planet's orbit. This law implies that the square of the period of orbit is directly proportional to the cube of the semimajor axis of the orbit.

Let's analyze the information given in the problem:

• The period of orbit for the planet is 8 times that of Earth, i.e., \(T_{\text{planet}} = 8T_{\text{earth}}\).

Using Kepler's Third Law, we can write:

\({\left(\frac{T_{\text{planet}}}{T_{\text{earth}}}\right)^2 = \left(\frac{a_{\text{planet}}}{a_{\text{earth}}}\right)^3}\)

Substitute \(T_{\text{planet}} = 8T_{\text{earth}}\) into the equation:

\({\left(\frac{8T_{\text{earth}}}{T_{\text{earth}}}\right)^2 = \left(\frac{a_{\text{planet}}}{a_{\text{earth}}}\right)^3}\)

After simplification, this becomes:

\({8^2 = \left(\frac{a_{\text{planet}}}{a_{\text{earth}}}\right)^3}\)

\({64 = \left(\frac{a_{\text{planet}}}{a_{\text{earth}}}\right)^3}\)

Taking the cube root on both sides:

\(\sqrt[3]{64} = \frac{a_{\text{planet}}}{a_{\text{earth}}}\)

\(4 = \frac{a_{\text{planet}}}{a_{\text{earth}}}\)

This implies that the semimajor axis of the planet's orbit is 4 times that of Earth's.

Therefore, the correct answer is: The semimajor axis of the planet's orbit is four times as compared to that of the Earth.