Kepler's Third Law for Planet Distance

This problem requires applying Kepler's Third Law of Planetary Motion, which relates a planet's orbital period ($T$) to the semi-major axis ($R$) of its orbit around the Sun. The law states that the square of the period is proportional to the cube of the semi-major axis, or $\frac{T^2}{R^3} = \text{constant}$.

Applying Kepler's Law

We can compare the Earth and planet X:

• Let $T_E$ be the Earth's period and $R_E$ be Earth's distance from the Sun.
• Let $T_X$ be planet X's period and $R_X$ be planet X's distance from the Sun.

According to Kepler's Third Law:

$ \frac{T_E^2}{R_E^3} = \frac{T_X^2}{R_X^3} $

Calculations

We are given:

• $T_X = 8$ years
• $T_E = 1$ year (standard orbital period for Earth)
• $R_E = 1.5 \times 10^{11}$ m

We need to find $R_X$. Rearranging the formula to solve for $R_X^3$: $ R_X^3 = R_E^3 \times \left(\frac{T_X}{T_E}\right)^2 $

Substitute the values:

$ R_X^3 = (1.5 \times 10^{11} \ m)^3 \times \left(\frac{8 \ years}{1 \ year}\right)^2 $

$ R_X^3 = (1.5^3 \times (10^{11})^3) \times 8^2 \ m^3 $

$ R_X^3 = (3.375 \times 10^{33}) \times 64 \ m^3 $

$ R_X^3 = 216 \times 10^{33} \ m^3 $

Now, take the cube root to find $R_X$: $ R_X = \sqrt[3]{216 \times 10^{33}} \ m $

$ R_X = \sqrt[3]{216} \times \sqrt[3]{10^{33}} \ m $

$ R_X = 6 \times 10^{11} \ m $

Therefore, the distance of planet X from the Sun is $6 \times 10^{11}$ m.