Answer :
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
[tex]T^2\alpha R^3\\T^2=kR^3.......................(1)[/tex]
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
[tex]\frac{T^2}{R^3}=k.......................(2)[/tex]
Let the orbital period of the earth be [tex]T_e[/tex] and its mean distance of from the sun be [tex]R_e[/tex].
Also let the orbital period of the planet be [tex]T_p[/tex] and its mean distance from the sun be [tex]R_p[/tex].
Equation (2) therefore implies the following;
[tex]\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)[/tex]
We make the period of the planet [tex]T_p[/tex] the subject of formula as follows;
[tex]T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)[/tex]
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
[tex]R_p=16R_e...............(5)[/tex]
Substituting equation (5) into (4), we obtain the following;
[tex]T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}[/tex]
[tex]R_e^3[/tex] cancels out and we are left with the following;
[tex]T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)[/tex]
Recall that the orbital period of the earth is about 365.25 days, hence;
[tex]T_p=64*365.25\\T_p=23376days[/tex]