What is the proof that periods cancel out in finding angular displacement?

[Bashyboy]

Homework Statement

http://answers.yahoo.com/question/index?qid=20120411050123AA0ar9P

Homework Equations

3. The Attempt at a Solution [/b
Only I have one question concerning the answer that "E" provides in the link: why do periods cancel out?

 

[gneill]

Homework Statement

http://answers.yahoo.com/question/index?qid=20120411050123AA0ar9P

Homework Equations

3. The Attempt at a Solution [/b
Only I have one question concerning the answer that "E" provides in the link: why do periods cancel out?

Are you referring to this section of the solution?:

(θ_x)(T_x) = (θ_y)(T_y) = (θ_y)(T_x / √27)

The periods cancel, so

(θ_x) = (θ_y) / √27​

The center expression, "(θ_y)(T_y)", is just an intermediate step, and "(T_y)" is then replaced by "(T_x / √27)" to yield the relationship

(θ_x)(T_x) = (θ_y)(T_x / √27)

T_x occurs on both sides of the equality and can be canceled algebraically.

 

[Bashyboy]

Actually, there is something before that, that I don't quite understand. Why are we allowed to set up this proportion (I never really understood proportions, so perhaps you could help me understand):

$\frac{r_x^3}{T_x^2} = \frac{r_y^3}{T_y^2}$?

 

[gneill]

Actually, there is something before that, that I don't quite understand. Why are we allowed to set up this proportion (I never really understood proportions, so perhaps you could help me understand):

$\frac{r_x^3}{T_x^2} = \frac{r_y^3}{T_y^2}$?

That's an application of Kepler's Third Law: for all planets, the square of the period is proportional to the cube of the orbit radius, or $T^2 \propto r^3$.

 

[Bashyboy]

Well, what confuses me is, how is the left-side of the equation equal to the right-side?

 

[gneill]

Well, what confuses me is, how is the left-side of the equation equal to the right-side?

Write the two proportionalities as equalities by introducing a constant of proportionality:

$r_x^3 = k T_x^2$

$r_y^3 = k T_y^2$

Now rearrange:

$\frac{r_x^3}{T_x^2} = k$

$\frac{r_y^3}{T_y^2} = k$

Both fractions equal k, so set them equal to each other.

 

[Bashyboy]

So, both fractions always equal the same constant, no matter which pair of planets are being considred? Why is that so?

 

[gneill]

So, both fractions always equal the same constant, no matter which pair of planets are being considred? Why is that so?

Kepler found the relationship empirically, hence his third law.

When Newton came along and developed his theory of gravity, he proved Kepler's relationship using it and his other laws.

 

[Bashyboy]

Okay, so there several proofs of this fact. Thank you for indicating that.