To solve this we are going to use Kepler's third law: [tex] \frac{T_{1}^2}{T_{2}^2} = \frac{R_{1} ^3}{R_{2} ^3} [/tex]
where
[tex]T_{1}[/tex] is the period of revolution of the first object
[tex]T_{2}[/tex] is the period of revolution of the second object
[tex]R_{1}[/tex] is the distance from the sun to the first object
[tex]R_{2}[/tex] is the distance from the sun to the second object
We can infer that our first object is the Earth, Since it takes one year to revolve around the sun, its period of revolution is 1 year; therefore: [tex]T_{1}=1[/tex]. We also know that the distance from the Sun to the Earth is 1 AU, so [tex]R_{1}=1[/tex].
Similarly, we can infer that our second object is the other planet. Since we know that the planet is 4 AU from the sun, [tex]R_{2}=4[/tex].
Lets replace the values in our formula to find [tex]T_{2}[/tex]:
[tex] \frac{T_{1}^2}{T_{2}^2} = \frac{R_{1} ^3}{R_{2} ^3} [/tex]
[tex] \frac{1^2}{T_{2}^2 } = \frac{1^3}{4^3} [/tex]
[tex] \frac{1}{T_{2}^2} = \frac{1}{64} [/tex]
[tex]T_{2}^2=64[/tex]
[tex]T_{2}= \sqrt{64} [/tex]
[tex]T_{2}=8[/tex]
We can conclude that the period of revolution of the planted located 4 AU form the Sun is 8 years; therefore, the correct answer is A) 8.
