# Homework Help: Motion of Two Bodies Orbiting the Sun

1. Mar 14, 2014

### Bashyboy

Hello everyone. In my computational physics class, I am supposed to use numerical method to solve for equations of motion of the earth and moon orbiting the sun, at which I place the origin of my coordinate system. However, I have a physics related question. Before I pose the query, allow me to define some notation and some conventions:

In this problem, we are assuming circular orbits. Let $v_{es} = r_{es} \omega_{es}$ be the tangential speed of the earth relative to the sun, the distance between the sun and earth (which would be 1, as we are working in astronomical units), and the rate at which the earth rotates around the earth, respectively.

Let $v_{me} = r_{me} \omega_{me}$ be the tangential speed relative to the earth, the distance between earth and moon, and the rate at which the moon rotates around the earth, respectively.

To find the tangential velocity of the moon with respect to the sun, would I simply compute the quantity $v_{ms} = v_{es} + v_{me} \implies v_{ms} = r_{es} \omega_{es} + r_{me} \omega_{me}$?

2. Mar 14, 2014

### BvU

Not a good plan. Make a simple drawing and you will see why.

3. Mar 14, 2014

### Bashyboy

Well, I only need to use the formula to find the initial conditions, which is when all three planets are along the x-axis.

4. Mar 14, 2014

### Bashyboy

I drew a picture, but I am not quite sure I follow. I would like to understand what is wrong with this method, however.

What if I somehow used the chain rule? I want $v_{MS}$, which can be obtained from the formula $v_{MS} = r_{MS} \omega_{MS}$. Is it possible that I could could somehow compute $\frac{d \theta_{MS}}{dt}$ in terms of the other derivatives?

Wait! I think I might have figured it out. Just give me a few moments to type up my response.

Last edited: Mar 14, 2014
5. Mar 14, 2014

### Bashyboy

To find the tangential velocity of the moon relative to the sun, at time t=0, which is when all of the planets lie along the x-axis, would I use the equation $\mathbf{v}_{ms} = \mathbf{v}_{me}+ \mathbf{v}_{es}$? If so, I could take magnitude of this this, which would be the length of side
of the triangle $v_{ms}$. The three vector form some triangle. Using the law of cosines, I can find $v_{ms}$:

$v_{ms} = \sqrt{v^2_{me} + v_{es}^2 - 2v_{me}v_{es} \cos \theta}$ (I choose the positive root, as we are dealing with speeds). Since all of the planets lie along the x-axis at t=0, the time at which this calculation gives $v_{ms}$, then velocities of the moon and earth must be directed in the y-direction; therefore, the angle between the two velocities is zero.

$v_{ms} = \sqrt{(2 \pi)^2 + (6.499221914)^2 - 2(2 \pi)(6.499221914)}$.

Does this appear correct?

Last edited: Mar 14, 2014
6. Mar 14, 2014

### BvU

Ah, new info: you want the magnitude of $v_m$ under conditions
• S,e,m line up (in that order)
• Se plane coincides with em plane.

7. Mar 14, 2014

### Bashyboy

So, would the first formula I proposed work?

8. Mar 14, 2014

### BvU

gives you a starting speed at your t=0, yes.

Seems very weird to me. Would vme be $2\pi$? and ves be only a little more than that? And the minus sign is for $\cos\pi$?

I remember a previous post on this subject that isn't quite comparable, but contains some equations that might come in handy...

9. Mar 14, 2014

### Bashyboy

Yes, the reason why I have speeds like 2pi is because I am using astronomical units, in which the gravitational cosntant G = 4 pi^2.

10. Mar 14, 2014

### BvU

Not familiar with a.u., but it seems to me moon wrt earth (1 km/s) is a lot slower than earth wrt sun (30 km/s)...
Ah, I see: 1 au = r sun-earth. So you have 2 pi au /year for the sun. And here's me thinking 2 pi m/s ...

Still: moon in au/yr is 6.499221914 seems dead wrong in two respects: there's a 10% differnece between peri- and apogee, so 10 digits is nonsense. And the value is a factor 30 off... your turn for Unless you prove me wrong and then it's double - nothing

If your program does circular orbits only, isn't that somewhat contradictory with "solving the equations of motion" ?

11. Mar 15, 2014

### Bashyboy

I do not understand, how is having a circular orbit contradictory with solving the equation of motion? Isn't traveling on a circular orbit a type motion, and can't this particular type of motion be described with equations? If so, then we would have equations of motion.

Also, here is how I calculated the velocity of the moon relative to the sun:

Because the planets undergo circular motion, the rate at which they rotate through angles is constant, that is, the angular velocity is constant. \

Radius between earth and sun: $r_{es} = 1~AU$

Angular velocity of earth about the sun: $\omega_{es} = \frac{2 \pi ~rad}{1~yr}$

Radius between moon and earth: $r_{me} = 2.57 \times 10^{-6} ~AU$\

Angular velocity of the moon about the earth: $\omega_{me} = \frac{2 \pi~rad}{27 ~days} \times \frac{365 ~days}{1~yr} = 84.93935693 ~\frac{rad}{yr}$.

Therefore,

$v_{ms} = (1~AU)(\frac{2 \pi ~rad}{1~yr}) + (2.57 \times 10^{-6} ~AU)( 84.93935693 ~\frac{rad}{yr}) = 6.283403601$

12. Mar 15, 2014

### BvU

I haven't understood the 6.499221914.

From radius earth orbit 1.5e11 m and 365.25 days I get 29.78 km/s
From radius moon orbit 3.85e8 m and 27.32 days I get 1.025 km/s.

If the first in your au units is $2\pi$, the second should be around 0.2 and the sum some 6.5

Where did you find $r_{me} = 2.57 \times 10^{-6} ~AU$ ?

13. Mar 15, 2014

### BvU

Perhaps we differ in our interpretation of "to solve for equations of motion". To me it means solving $\vec F=m\,\vec a$ with a given $\vec r$ and $\vec v$ at t=0.

That the outcome is close to a circular orbit for earth around sun, I can well believe. And -- when moving with the earth -- the moon's orbit will be close to a circle as well. But that is the outcome, not the input.