To determine the ratio of a satellite's orbital radius to that of the Earth's orbital radius using Kepler's Third Law of Planetary Motion, we can start with the law states that the square of the orbital period of a planet (or satellite) is directly proportional to the cube of the semi-major axis of its orbit.
The law can be expressed mathematically as:
\[ T^2 \propto r^3 \]
where:
- \( T \) is the orbital period,
- \( r \) is the orbital radius.
We can set up a ratio for the Earth and the artificial satellite orbiting the Sun.
Let:
- \( T_e \) be the period of the Earth (1 year),
- \( r_e \) be the orbital radius of the Earth (1 Astronomical Unit, AU),
- \( T_s \) be the period of the satellite (8 years),
- \( r_s \) be the orbital radius of the satellite.
According to Kepler's Law for both the Earth and the satellite, we have:
\[ \frac{T_s^2}{T_e^2} = \frac{r_s^3}{r_e^3} \]
Substituting the values of the periods:
- \( T_s = 8 \) years,
- \( T_e = 1 \) year.
\[ \frac{(8)^2}{(1)^2} = \frac{r_s^3}{(1)^3} \]
This simplifies to:
\[ \frac{64}{1} = \frac{r_s^3}{1} \]
Thus:
\[ r_s^3 = 64 \]
Taking the cube root of both sides gives us:
\[ r_s = 4 \]
Thus, the ratio of the satellite's orbital radius to that of the Earth's orbital radius is:
\[ \frac{r_s}{r_e} = 4 \]
Therefore, the final answer is:
\[ \boxed{4} \]
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