Answer:
Option C - 9 AU
Step-by-step explanation:
To find : What is the average distance from Planet B to the star?
Solution :
According to kepler's law,
The squares of the sidereal periods (of revolution) of the planets are directly proportional to the cubes of their mean distances from the Sun.
i.e. [tex]P^2\propto S^3[/tex]
We have given,
The average distance from the star to Planet A is [tex]S_1=4[/tex] AU.
It takes 432 Earth days for Planet A to orbit the star i.e. [tex]P_1=432[/tex]
It takes 1,460 days for Planet B to complete an orbit i.e. [tex]P_2=1460[/tex]
Substitute the values in [tex](\frac{P_1}{P_2})^2=(\frac{S_1}{S_2})^3[/tex]
[tex](\frac{432}{1460})^2=(\frac{4}{S_2})^3[/tex]
[tex]0.0875=(\frac{4}{S_2})^3[/tex]
Taking root cube both side,
[tex]\sqrt[3]{0.0875}=\frac{4}{S_2}[/tex]
[tex]0.444=\frac{4}{S_2}[/tex]
[tex]S_2=\frac{4}{0.444}[/tex]
[tex]S_2=9.00[/tex]
The average distance from Planet B to the star is 9 AU.
Therefore, Option C is correct.
Answer:
C. 9 AU
Step-by-step explanation:
The period (T) of Planet A is 432365=1.2 Earth years. The period of Planet B is 1,458365=4.0 Earth years. Kepler's Third Law states that
a3=kT2
Substituting the known values of a and T for Planet A, we have
43=k⋅1.22
64=k⋅1.4
k=45.7
Using this value for k, we can solve for the distance from Planet B to the star
a3=45.7⋅4.02
a3=731.2
a=9.0AU
An alternative solution is to notice that a3T2=k for both orbits, which means that the ratio a3T2 is the same for both orbits. So, 434322=x314582. Solving this proportion gives x3=729, so x=9.0AU.
