We are asked to determine the perihelion distance of Halley's comet. To do that we will use the following formula:
[tex]P=a(1-e)[/tex]Where:
[tex]\begin{gathered} P=\text{ perihelion distance} \\ a=\text{ distance of semi-major ax}is \\ e=\text{ excentricity} \end{gathered}[/tex]To determine the distance of the semi-major axis we will use Kepler's third law
[tex]T^2=a^3[/tex]Where:
[tex]\begin{gathered} T=\text{ period in earth years } \\ a=\text{ distance of semi-major ax}is\text{ in Astronomical Units. } \end{gathered}[/tex]Now, we solve for "a" by taking the cubic root on both sides:
[tex]\sqrt[3]{T^2}=a[/tex]Now, we substitute the value of "T":
[tex]\sqrt[3]{(67years)^2}=a[/tex]Solving the operations:
[tex]16.5AU=a[/tex]Now we substitute this value in the formula for the perihelion:
[tex]P=(16.5AU)(1-0.97)[/tex]Solving the operations:
[tex]P=0.5AU[/tex]Therefore, the perihelion distance is 0.5 Astronomical Units.
