Planet A and planet B are in circular orbits around a distant star. Planet A is 8.0 times farther from the star than is planet B .What is the ratio of their speeds ?
sqrt(1/8)=ans
To determine the ratio of their speeds, we can use Kepler's Third Law, which states that the square of a planet's orbital period is proportional to the cube of its average distance from the star.
Let's assume that the orbital period of planet B is T_B and the average distance from the star is R_B. The orbital period of planet A is T_A, and the average distance from the star is R_A.
Since the ratio of their distances is given as 8.0 (R_A = 8.0 * R_B), we can write the equation as:
T_A^2 / R_A^3 = T_B^2 / R_B^3
Let's assume the ratio of their speeds is V_A / V_B.
We know that speed is distance over time, so:
V_A = 2πR_A / T_A
V_B = 2πR_B / T_B
Let's substitute these equations into the previous equation:
(T_A^2 / R_A^3) = (T_B^2 / R_B^3)
((2πR_A / V_A)^2 / R_A^3) = ((2πR_B / V_B)^2 / R_B^3)
Simplifying the equation further:
(V_A^2 / R_A) = (V_B^2 / R_B)
We can substitute for the ratios given earlier:
(V_A^2 / (8.0 * R_B)) = (V_B^2 / R_B)
Simplifying again:
V_A^2 = V_B^2 / 8.0
To find the ratio of their speeds, we take the square root of both sides:
V_A / V_B = √(1/8.0)
V_A / V_B = 1 / √8.0
Converting the square root:
V_A / V_B = 1 / (2.828)
Simplifying the fraction:
V_A / V_B ≈ 0.354
Therefore, the ratio of their speeds is approximately 0.354.
To find the ratio of their speeds, we first need to understand the relationship between the distance of a planet from its star and its orbital speed.
According to Kepler's Third Law of Planetary Motion, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the star. In equation form, it can be written as:
(T1/T2)^2 = (R1/R2)^3
Where T1 and T2 are the periods of planet A and planet B respectively, and R1 and R2 are their respective distances from the star.
Since we know that planet A is 8.0 times farther from the star than planet B, we can say:
R1 = 8 * R2
Substituting this into Kepler's Third Law equation, we get:
(T1/T2)^2 = (8 * R2 / R2)^3
Simplifying the equation gives us:
(T1/T2)^2 = 8^3
(T1/T2)^2 = 512
Taking the square root of both sides:
T1/T2 = √512
Now, the orbital speed of a planet is given by the formula:
V = (2πR) / T
Where V is the orbital speed, R is the distance from the star, and T is the period of the orbit.
Since we are interested in the ratio of their speeds, we can divide the orbital speeds of the two planets:
(V1/V2) = (2πR1 / T1) / (2πR2 / T2)
Canceling out the 2π and simplifying further gives us:
(V1/V2) = (R1 / T1) / (R2/T2)
Substituting the relationship between R1 and R2, we get:
(V1/V2) = (R2 / T1) / (R2/T2)
(V1/V2) = ( T2 / T1)
Now, since it's given that the periods of planet A and planet B are the same, T1 = T2, we have:
(V1/V2) = 1
So, the ratio of their speeds is 1:1. Planet A and planet B have the same speed in their circular orbits around the distant star.