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Calculating elliptical orbits 
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#19
May1009, 04:33 PM

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There's only one problem with using Kepler's law (well, two problems). He doesn't know the period and he doesn't know the semimajor axis.



#20
May1009, 06:20 PM

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Somewhere here must lie the anwer. I have problems with the unities a=>L/T , R=>L , Mu=>L^3/(WT^2), V^2=>L^2/T^2 so V^2/Mu=>W/L, with L=lenght T=time and W=Weight with this formula correct R can be calculated from V and a. greetings Janm 


#21
May1009, 08:20 PM

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No, it cannot.
Suppose you find that, given the instantaneous velocity and acceleration at some time epoch, one possible explanation of the explanation is a mass [itex]m[/itex] located some distance [itex]d[/itex] away, in the direction of the acceleration vector. The problem is this solution is not unique. A mass [itex]4m[/itex] located some distance [itex]2d[/itex] yields exactly the same acceleration, as does any mass, distance pair of the form [itex]\kappa^2m, \kappa d[/itex]. 


#22
May1109, 02:26 AM

P: 224

Hello D H
You are right if velocity and acceleration is given at only one time but: greetings Janm 


#23
May1109, 06:09 AM

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I assumed that the 2 prefix was in the sense of 2vectors, meaning two dimensional vectors.



#24
May1109, 05:59 PM

P: 224

I have not thought otherwise. So x,y plane z=0. By the way you stated that the acceleration wil be senkrecht to the velocity. In that I cannot concur. The centrifugal force is senkrecht to the velocity, but the acceleration (F/m) is the total of attraction and centrifugal force, which are only paralel if the object is following a circle. The radial part of the acceleration gives the falling and climbing resp. to and out the gravitational centre. greetings Janm 


#25
May1109, 06:55 PM

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There is no centrifugal force in an inertial frame. Why invoke the concept? The centrifugal force only arises in a rotating reference frame. The only rotating reference frame that makes sense from an orbital sense is the frame with origin at the center of mass rotating at the mean orbital rate. If the objects are in a circular orbit, the objects are stationary in this frame: Zero velocity, zero acceleration. Not very useful. If the objects are not in a circular orbit, this rotating frame creates a real mess: Now you have coriolis forces to deal with due to the nonzero velocities. The best way to look at most orbits is from the point of view of an inertial frame. The one exception is looking at pseudo orbits about one of the libration points. We aren't doing that here. Forget about centrifugal force. 


#26
May1209, 10:27 AM

P: 224

The thread was opened with a=v^2/r, you state a_grav=gM/r^2. Please tell me that you are not serious declining centrifugal force and don't see that the first acceleration defines the centrifugal acceleration from the velocity and the radius of curvature (the best circle fitted to the curve? My problem with this problem is how can the radius of curvature be found if only velocity and acceleration is known! greetings Janm 


#27
May1209, 10:50 AM

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[/QUOTE]...and don't see that the first acceleration defines the centrifugal acceleration from the velocity and the radius of curvature[/QUOTE] Centrifugal force only exists in a rotating frame. I strongly suggest that you forget about centrifugal force until you get the basics down. 


#28
May1209, 11:36 AM

P: 22

Hi
"2 acceleration" u mean we have R'' ? (R'' = d^{2}R/dt^{2}) or we have GM/R^2 = acceleration ? + velocity 


#29
May1209, 11:46 AM

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Unless the OP is misusing notation, 2acceleration and 2vector means the acceleration's and velocity's x and y components are known (with the z component zero).
The problem is indeterminate. All you know is the direction in which the massive object lies. The distance to that object is [itex]r=\sqrt{GM/\boldsymbol a}[/itex]. You don't know the mass; it is a free parameter. Vary the mass and you can make the distance be any value from zero to infinity. That you also know the velocity doesn't help a bit. 


#30
May1209, 12:43 PM

#31
May1209, 12:50 PM

P: 22

oh.... Latex errors ... at the end of Velocity that is e[tex]\theta[/tex] not that
Click on the Latex images and read there and for : [tex]\theta[/tex] = 180  A it was a exception for this picture it depens on which quarter u want to calculate it may become like 180 + A 


#32
May1209, 01:08 PM

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Motion about an ellipse can be written in parametric equations [tex]x= a cos(\omega t)[/tex] [tex]y= b sin(\omega t)[/tex] where a and b are the semiaxes in the x and y directions, respectively. Then the velocity vector is [tex]\vec{v}= a\omega sin(\omega t)\vec{i}+ b\omega cos(\omega t)\vec{j}[/tex] and the acceleration vector is [tex]\vec{a}= a\omega^2 cos(\omega t)\vec{i} b\omega^2 sin(\omega t)\vec{j}[/tex] Of course, the crucial point is determining [itex]\omega[/itex]. You can use the fact that the force vector, and so the acceleration vector, must be directed toward one focus of the ellipse to determine that. 


#33
May1209, 01:48 PM

P: 22

and for finding "e"
i think this is the answer : (in ellipse) (e/Sin[theta] + Cotg[theta]) = Vy/Vx 


#34
May1209, 05:49 PM

P: 224

Are we differentiating again? I miss in the acceleration vector one term in the i direction a*sin(wt) and in the j direction b*cos(wt)! But overall the shape of the curve is okee but the parameter is not the regular wt. The parameter is a function of t and follows in some way out of the acceleration. greetings Janm 


#35
May1309, 03:52 AM

P: 22

Ive said how to find "w" in my post "page 2" next to Picture
also "wt" which is [tex]\theta[tex] 


#36
May1309, 10:40 AM

P: 224

Your way comes close to the answer I think and hope. With er you mean in direction of radial, so approach and recede. Your e(theta) is the angular movement. I can see that in my mechanics and properties of matter third edition page 16. It is indeed universal mechanics of a plane curve. The moment you state that the transverse part is zero isn't that the moment that you state that the sun is in 0*er + 0*e(theta)? What I mean to say is the question was velocity and acceleration given at any moment. From that the threadstarter wants to calculate the place of the sun (without using Kepler...). greetings Janm 


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