Question

What is the frequency of a planet around a star in orbit at a mean distance of R, if it is twice the mean distance of a planet orbiting the same star with a frequency of 2.10 × 10^-5 rev/s?

A) 1.4 × 10^-5 rev/s
B) 7.4 × 10^-6 rev/s
C) 1.7 × 10^-7 rev/s
D) 6.8 × 10^-7 rev/s
E) 2.4 × 10^-8 rev/s

Pls show work so I can learn how you got to that answer. Thank you!

Answer 1

A) 1.4 × 10^-5 rev/s

Step-by-step explanation:

We can use Kepler's Third Law of Planetary Motion to solve this problem. Kepler's Third Law states that the square of the period of a planet's orbit is proportional to the cube of its average distance from the star.

T^2 = k * R^3

where T is the period of the orbit and R is the mean distance of the planet from the star. k is a constant of proportionality that depends on the mass of the star.

Let's assume that the frequency of the first planet is f1 and its mean distance from the star is R1. The frequency of the second planet is 2.10 × 10^-5 rev/s and its mean distance from the star is 2R1. We can relate the frequency of the orbit to the period of the orbit as follows:

f = 1/T

Using the first equation, we can find the period of the first planet:

T1^2 = k * R1^3

And the period of the second planet:

T2^2 = k * (2R1)^3

Now, we can substitute these equations into the equation for frequency:

f1 = 1/T1 = 1/sqrt(k * R1^3)

f2 = 1/T2 = 1/sqrt(k * (2R1)^3)

Finally, we can set f1 = 2f2 and solve for f1:

2f2 = 1/sqrt(k * R1^3)

f1 = 1/sqrt(k * R1^3) / 2

So, f1 = 1.4 × 10^-5 rev/s, which is the answer (A).