Calculating elliptical orbits

GetPhysical

You can use Kepler's law p^2=a^3.

The period squared is equal to the semi-major axis cubed.

D H

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 semi-major axis.

JANm

Hello tony873004
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

D H

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].

JANm

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

So if you take two points in time and intersect the two given accelerations the kappa you mention can be calculated!
greetings Janm

D H

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

JANm

Hello D H
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

D H

I said no such thing (and I had to look that up. senkrecht=normal) My posts in this thread:
There is no mention of the acceleration vector being normal to the velocity vector.


Dang. I thought US schools were the only ones that completely and thoroughly botched the job of teaching orbits. There is absolutely no reason to invoke the concept of centrifugal force in explaining orbits. Doing so leads to erroneous concepts.

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 non-zero 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.

JANm

Hello D H
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

D H

The thread was opened with a=v^2/r as an introductory example.

I am seriously declining centrifugal force. It is not needed. It is not real. There is no such thing as the centrifugal force in an inertial frame.
[/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.

You can't. How many times do I have to tell you this?

Mahbod|Druid

Hi

"2- acceleration" u mean we have R'' ? (R'' = d²R/dt²)

or

we have GM/R^2 = acceleration ?


+ velocity

D H

Unless the OP is misusing notation, 2-acceleration and 2-vector 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.

Mahbod|Druid

and w = V Sin X / R
:edited X:

and from Polar Direction we have :

R er
=> derivate (d f(x) / dt i`m not sure about the English word)

Velocity = R' er + Rw e(theta)
=> derivate again for acceleration

R'' er + R'w e[tex]\theta[/tex]+ R'w e[tex]\theta[/tex]+ Rw' e[tex]\theta[/tex]-Rw² er

as we know Force([tex]\theta[/tex]) = 0 , Force(Radius) = K (something u said in ur question "2- acceleration")
[tex]\theta[/tex] = 180 - A (forgot to show on picture)

all of the sentences with e[tex]\theta[/tex] equals to Zero

so we will have :

R'' er -Rw² = K

and if i havent made any mistake in my paper i have calculated R'' :
R'' = (V Sin(X) (eCos[tex]\theta[/tex]+ e^2))/R

then multiple both side of R'' - Rw^2 = K by R
luckily it didnt become a quartic function :D (how ever if it would there was one shifty sentence in function -R-.)


and for the eccentricity (spell?) (e) it is possible to calculate but i havent study ellipses yet so i don't know atm but i will think about it
(in the first page somebody said something about this but i don't know what that is)

Mahbod|Druid

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

HallsofIvy

Since an elliptical orbit is planar, it is sufficient to assume the orbit is in the xy plane.

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 semi-axes 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.

Mahbod|Druid

and for finding "e"

i think this is the answer : (in ellipse)

-(e/Sin[theta] + Cotg[theta]) = |Vy|/|Vx|

JANm

Hello HallsofIvy
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