Motion of Two Bodies Orbiting the Sun

[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}?

 

[BvU]

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

 

[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.

 

[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.

 

[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?

 

[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.

In that case the vector addition becomes a simple magnitude addition.

 

[Bashyboy]

So, would the first formula I proposed work?

 

[BvU]

v_{ms} = v_{es} + v_{me} \implies v_{ms} = r_{es} \omega_{es} + r_{me} \omega_{me}

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

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

Seems very weird to me. Would v_me be $2\pi$? and v_es 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...

 

[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.

 

[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" ?

 

[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

 

[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 ?

 

[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.