Question

A space expedition discovers a planetary system consisting of a massive star and several spherical planets. The planets all have the same uniform mass density. The orbit of each planet is circular.

In the observed planetary system, Planet A orbits the central star at the distance of 2R and takes T hours to complete one revolution around the star. Planet B orbits the central star at the distance of R. Which of the following expressions is correct for the number of hours it takes Planet B to complete one revolution around the star?

a. 1/â8T
b. 1/2T
c. 1/3â4T
d. 2T
e. â8T

Answer 1

Answer:

T/√8

Explanation:

From Kepler's law, T² ∝ R³ where T = period of planet and R = radius of planet.

For planet A, period = T and radius = 2R.

For planet B, period = T' and radius = R.

So, T²/R³ = k

So, T²/(2R)³ = T'²/R³

T'² = T²R³/(2R)³

T'² = T²/8

T' = T/√8

So, the number of hours it takes Planet B to complete one revolution around the star is T/√8

Answer 2

The correct expression for the number of hours it takes Planet B to complete one revolution around the star is $\frac{T}{\sqrt{8} }$.

The given parameters;

• position of planet A from central star, = 2R
• time taken for Planet A = T
• number of revolutions at the given time = 1 rev
• position of planet B from central star, = R

From Kepler's law, the period of planet is related to radius as follows;

$\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \\\\T_2 ^2 = \frac{T_1^2 \times R_2^3}{R_1^3} \\\\T_2 = \sqrt{\frac{T_1^2 \times R_2^3}{R_1^3} } \\\\T_2 = \sqrt{\frac{T_1^2 \times R_2^3}{R_1^3} }\\\\T_2 = T_1 \sqrt{\frac{ R_2^3}{(2R_2)^3} }\\\\T_2 = T_1 \sqrt{\frac{R_2^3}{8R_2^3} } \\\\T_2 = T_1 \sqrt{\frac{1}{8} } \\\\T_2 = \frac{T_1}{\sqrt{8} }\\\\T_B = \frac{T_A}{\sqrt{8} } = \frac{T}{\sqrt{8} }$

Thus, the correct expression for the number of hours it takes Planet B to complete one revolution around the star is $\frac{T}{\sqrt{8} }$.

Learn more about Kepler's law here: https://brainly.com/question/24173638