# Scaling The Solar System By A Factor $\alpha$

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1. Jan 13, 2017

### XanMan

I am given the solution to the first part of the problem, however not the second part - would appreciate for someone to double check my work! Cheers.

1. The problem statement, all variables and given/known data

If a scale model of the solar system is made using materials of the same respective average density as the sun and planets, but reducing all linear dimension by a factor $\alpha$, how will the velocity and period scale with $\alpha$? Assume a circular orbit in your calculations.

2. Relevant equations

$$F = \cfrac{GMm}{r^2}$$F = \cfrac{mv^2}{r}

3. The attempt at a solution

Starting off with the basic relationship required:
$$\cfrac{GMm}{r^2} = \cfrac{mv^2}{r}$$

Then: $v^2 = \cfrac{GM}{r} \propto \cfrac{G \rho R^3}{r}$, where $R$ is the radius of the Sun.

In general: $$v^2 \propto \cfrac{R^3}{r}$$

Re-scaling the solar system by $\alpha$:

$$v^2 \propto \cfrac{\alpha^3 R^3}{\alpha r}$$

i) Therefore $v \propto \alpha$

Now, $v = \cfrac{2 \pi r}{T}$. Thus:
$$\cfrac{2 \pi r}{T} \propto \alpha$$

ii) Therefore $T \propto \cfrac{1}{\alpha}$

2. Jan 13, 2017

### BvU

Well, apparently $v\propto\alpha$ is the right answer. If that is so, then I don't understand why you don't let $r\propto\alpha$ in part ii).

Oh, and a belated !

3. Jan 13, 2017

### XanMan

Thank you! I'm not quite sure what you mean exactly. With some reasoning I could have also said that since $v \propto \alpha$, then $r \propto \alpha$. It thus follows that $T \propto \cfrac{1}{\alpha}$.

4. Jan 13, 2017

### BvU

What I mean is

In part 1 you use $r' = \alpha r$ to find $v'= \alpha v$
Why don't you do that in part ii) ? to find that $T'= {2\pi r'\over v'} = ..$

5. Jan 13, 2017

### XanMan

In which case $T' = T$ - thus $T$ is independent of $\alpha$? What is wrong in my reasoning that results in me obtaining $T \propto \cfrac{1}{\alpha}$?

6. Jan 13, 2017

### BvU

You left $r$ as is. In part i) you inserted $\alpha r$:
 By the way: I had seen this part ii) result before, here

7. Jan 13, 2017

### XanMan

Ah, how silly of me! Looked at the result you forwarded as well - got it now. Thank you so much and all the best for the new year!

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