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Offline Aquila

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Planetary movement
« on: 06/09/2009 14:00:11 »
Let`s say a planet has a linear movement with a constant speed.
Let`s say this planet has a moon that is moving around the planet on a perfect circular path.
I wonder: how does the absolute distance the moon moved for depend on time.
The absolute speed of the moon is composed by 2 vectors that change by sinusoid functions meaning there is an acceleration of the acceleration of the moon...
Hopefully, i`d like to find this function and see how it looks like.
Thank you!


 

Offline Vern

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Planetary movement
« Reply #1 on: 06/09/2009 20:07:41 »
Well, I have just made a library of graphics functions that work with the SDL window managing library. It has a gravity function that works for simulations but the physics is not exactly correct. Here is the library.

Here is the gravity function:
Code: [Select]
void gravity(ball & sp, ball & sp0){ // g of 1920 with y = -2
  double x = sp0.x, y = sp0.y, line = 0;
  double g = 10000.00;
  double dx = sp.x - sp0.x;
  int huec;
  double dy = sp.y - sp0.y;
  double dc = sqrt(dx * dx + dy * dy);
  double nx = dx / dc;
  double ny = dy / dc;
  huec = 0x000000;
  if(sp.dtoc < dc - .5){ huec = 0xff0000; }
  if(sp.dtoc > dc + .5){ huec = 0x00ff00; }
  sp.dtoc = dc;
  sp.vx = sp.vx   - g * nx / (dc * dc);
  sp.vy = sp.vy   - g * ny / (dc * dc);
  if(sp.control > 0){
    while(line++ < dc){ putpixel(x, y, huec); x += nx; y += ny; }
  }
}
 

Offline Soul Surfer

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Planetary movement
« Reply #2 on: 07/09/2009 09:11:08 »
For the simple scenario that you have painted the moon just has a fixed velocity with a constant direction equal to the velocity of the planet that adds continuously to the  fixed velocity but variable direction of the moon in a free fall orbit around the planet.  This is essentially no different form if the planet was stationary. 

Much more interesting is if the planet is orbiting a star and both of the orbits are elliptical.  The moon is then responding to two gravitational forces, that of the planet and that of the star and both of them are variable in value and direction.  You can see them as successive corrections to the simple theory.  Firstly the moon is orbiting the planet.  Then the moon is orbiting the star with the planet in the same orbit as the planet and then there is a small correction because the moon is not on the same position as the planet and so the gravitational force from the star is slightly different in size and direction.
 

Offline Bored chemist

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Planetary movement
« Reply #3 on: 07/09/2009 21:11:31 »
I'm fairly sure that the scenario is impossible.
When the moon is to the right of the planet it will pull it to the right. When it's to the left it will pull the planet to the left. If the planet keeps getting pulled left and right it cannot move in a straight line unless it has an infinite mass. If it has, then the orbital velocity of the moon would also need to be infinite.

As Soul surfer says, the system is very similar to a sationary planet with a moon- they both orbit round their common centre of mass. You can have that centre of mass traveling in a straight line if you like.
 

Offline Soul Surfer

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Planetary movement
« Reply #4 on: 08/09/2009 23:58:31 »
BC I think that Aquila was assuming that the "moon" was a negligible mass with respect to the planet and its effect on the orbit of he planet could be neglected.  This is of course not true of the earth's moon because the centre of gravity of the earth moon system is in the space between the earth and moon.  I also was tending to assume this in my second order description in the shape of the orbits of the three bodies involved.

As always when you get down to high precision analysis a great many more effects have to be taken into account.
 

Offline syhprum

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Planetary movement
« Reply #5 on: 09/09/2009 09:06:23 »
The center of gravity of the Earth Moon system is not between the Earth and Moon but below the surface of the Earth.
 

Offline Soul Surfer

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Planetary movement
« Reply #6 on: 09/09/2009 10:14:56 »
Sorry If I got it wrong I always thought that is was pretty close to the earth but just above the surface let's check it out

Mass of earth = 6 e24 kilograms
mass of moon  = 7.36 e22 Kilograms
 
Therefore ratio =  81.5 

radius if earth  = 12700/2  = 6350 km

distance of moon  384,000 Km  divided by mass ratio 81.5  =  4710

if the mass ratio was 60 the centre of gravity would be at the surface of the earth.

I stand corrected it's about 1600 km below the surface of the earth at the point where the moon is overhead.

I must admit I didn't work it out but relied on memory and what seemed to be about right.  You tend not to realise that although the moon is only between a third and a quarter of the earth's diameter the mass difference is the cube of this and underestimate the ratio of their masses.  Three cubed is 27 and four cubed is 64 but the ratio is more than this because the moon is a lot less dense than the earth because it has no iron core.
 

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Planetary movement
« Reply #6 on: 09/09/2009 10:14:56 »

 

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