Berkeley Physics Question (possible error)

[Nasem]

For book #1, secton II (Force & Motion, Gravitation)

Example 2.9b is confused the heck out of me.... I can't figure out why Answer A is the correct answer


When Johannes Kepler, 17th-centry astronomer and smart guy, predicted a relationship between the mean orbital radius and orbital period of a planet about the Sun, he found that the period squared is:

A) proportional to the cube of the orbital radius
B) proportional to the square of the orbital radius
C) proportional to the orbital radius
D) proportional to the square root of the orbital radius

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[Marjan Islam]

When I first looked at that, I was like "...man, how am I supposed to know this?!" But then 2 pages later, I learned it was Kepler's 3rd Law or something. So I guess just memorize it, that that's how it is.

 

[Nasem]

I saw that too...

But there is an algebraic way of arriving at keplers 3rd law.

You see, we start off with:
w = 2pi / T (this is the formuls for Period)

and we know that the angular speed w = v / r

so really we have

v/r = 2pi / T..... but how do you get kepler's 3rd law from here...

 

[MBHockey]

It tells you how they arrived at that answer two pages later under "Kepler's Third Law"

They just substituted v = (2*Pi*R)/T into the equation in example 2.9a for summing of forces for the planet and setting it equal to (m*v^2)/r (which is just f = ma, Newton's 2nd law)

The equation is then reduced and the relationship between T^2 and R^3 can then be ascertained. It is not an error.