Can You Determine the Center of Mass Distance Using Instantaneous Orbital Data?

[DiamondGeezer]

I've been doing some calculations but am obviously too dumb to work out something which should be straightforward (and no, this isn't homework)

In a circular orbit, the acceleration is related to the tangential velocity and the radius via

a=v^2/r

In an elliptical orbit, both the velocity and the acceleration are constantly changing. At two points in the orbit, the radial acceleration is zero, for example.

Q: Is it possible to infer the distance to the center of mass, knowing the 2-velocity and 2-acceleration at any instance of time?

 

[kof9595995]

what do you mean by" 2-velocity and 2-acceleration"

 

[DiamondGeezer]

OK- 3-acceleration and 3-velocity then.

I only mention 2-velocity and 2-accleration because in an elliptical orbit (like a planet around the sun) the orbit is executed on a plane.

 

[Nabeshin]

OK- 3-acceleration and 3-velocity then.

I only mention 2-velocity and 2-accleration because in an elliptical orbit (like a planet around the sun) the orbit is executed on a plane.

I think kof was more confused by your wording, as am I, than anything else.

Do you mean if you make 2 (or 3) measurements of velocity/acceleration of a satellite, you can infer the orbit? (P.S all orbits are executed in a plane, not just ellipses)

 

[DiamondGeezer]

I think kof was more confused by your wording, as am I, than anything else.

Do you mean if you make 2 (or 3) measurements of velocity/acceleration of a satellite, you can infer the orbit? (P.S all orbits are executed in a plane, not just ellipses)

NO.

3-acceleration and 3-velocity means that you can take measurements of acceleration and velocity in 3 orthogonal directions.

 

[Nabeshin]

NO.

3-acceleration and 3-velocity means that you can take measurements of acceleration and velocity in 3 orthogonal directions.

Well, if you knew the full acceleration vector, and knowing it's due completely to gravity, a simple Newton's 2nd law and universal law of gravitation should yield the distance to the COM.

 

[DiamondGeezer]

Well, if you knew the full acceleration vector, and knowing it's due completely to gravity, a simple Newton's 2nd law and universal law of gravitation should yield the distance to the COM.

Prove it.

 

[Nabeshin]

Prove it.

I don't know if latex is working...

F=m2a=Gm1m2/r^2

r=(Gm1/a)^(1/2)

That's all there is to that one. The thing is, you don't usually have an acceleration measurement.

 

[h4tt3n]

Actually, you can calculate the entire elliptical orbit from just the state vectors, ie. the position and velocity vectors. I can't remember how I did it just now, but a few years ago I made a 2d gravitational simulation program which did just that.

Cheers,
Mike

 

[DiamondGeezer]

I don't know if latex is working...

F=m2a=Gm1m2/r^2

r=(Gm1/a)^(1/2)

That's all there is to that one. The thing is, you don't usually have an acceleration measurement.

Not what I'm looking for: the G and M1 data are not known accurately enough for my purposes.

What I need is a vector equation relating velocity, acceleration and distance to center of mass in the case of an elliptical orbit.

 

[Nabeshin]

Not what I'm looking for: the G and M1 data are not known accurately enough for my purposes.

What I need is a vector equation relating velocity, acceleration and distance to center of mass in the case of an elliptical orbit.

Umm, G is known to pretty high precision so you must be looking for a damn good measurement. Like h4tt3n said, you can calculate the elliptical orbit from the state vectors, but this involves the position vector which it sounds like you don't have.

Off the top of my head I can't think of any vector equations relating the three quantities you mentioned (Measurements of these values are probably going to be a lot more imprecise than the measurement of G though...). I'll think about it more in the morning, but if you could describe what you need this for, that information might prove useful.

 

[tony873004]

You can compute your semi-major axis with
a = (2 / R - V ^ 2 / Mu) ^ -1
where mu is G*M, G is the gravitational constant, and M is the mass of the sun.
where R is sqrt(Rx^2+Ry^2+Rz^2), where R is the magnitude of your position vector
where V is sqrt(Vx^2+Vy^2+Vz^2), where V is the magnitude of your velocity vector

You can then compute your eccentricity in the following manner:

Hx = Ry * Vz - Rz * Vy
Hy = Rz * Vx - Rx * Vz
Hz = Rx * Vy - Ry * Vx
H = Sqr(Hx ^ 2 + Hy ^ 2 + Hz ^ 2)

p = H ^ 2 / Mu
q = Rx * Vx + Ry * Vy + Rz * Vz ' dot product of r*v

E = Sqr(1 - p / a) ' eccentricity

With your SMA and your eccentricity, you should be able to answer your question.

 

[DiamondGeezer]

You can compute your semi-major axis with
a = (2 / R - V ^ 2 / Mu) ^ -1
where mu is G*M, G is the gravitational constant, and M is the mass of the sun.
where R is sqrt(Rx^2+Ry^2+Rz^2), where R is the magnitude of your position vector
where V is sqrt(Vx^2+Vy^2+Vz^2), where V is the magnitude of your velocity vector

You can then compute your eccentricity in the following manner:

Hx = Ry * Vz - Rz * Vy
Hy = Rz * Vx - Rx * Vz
Hz = Rx * Vy - Ry * Vx
H = Sqr(Hx ^ 2 + Hy ^ 2 + Hz ^ 2)

p = H ^ 2 / Mu
q = Rx * Vx + Ry * Vy + Rz * Vz ' dot product of r*v

E = Sqr(1 - p / a) ' eccentricity

With your SMA and your eccentricity, you should be able to answer your question.

But I'm not trying to calculate the eccentricity and I don't know what the value of the semi-major axis is!

I'm trying to work out how far the centre of mass is given the velocity and acceleration at a point in time and knowing that the orbit is elliptical!

 

[Chronos]

Keplers law still works pretty well.

 

[D H]

Umm, G is known to pretty high precision

G is one of the least well-known physical constants: four decimal places of accuracy. We know the fine structure constant to eleven digits or so. Astronomers rarely use G; they use μ=G*M instead because the standard gravitational parameter μ can be measured independently of either G or M and can be measured to a high degree of accuracy. We know μ for the Sun to ten places of accuracy, for example.

But I'm not trying to calculate the eccentricity and I don't know what the value of the semi-major axis is!

I'm trying to work out how far the centre of mass is given the velocity and acceleration at a point in time and knowing that the orbit is elliptical!

You can't if all you know is the velocity and acceleration. Think of it this way: The gravitational acceleration toward an object of mass M and distance r is -GM/r², directed toward the mass. Double the distance and quadruple the mass and you get exactly the same acceleration, but a very different orbit.

 

[JANm]

But I'm not trying to calculate the eccentricity and I don't know what the value of the semi-major axis is!

I'm trying to work out how far the centre of mass is given the velocity and acceleration at a point in time and knowing that the orbit is elliptical!

Hello Diamondgeezer
I wanted to react on your first alinea, but now I have read the second it has become more complicated. I have to do that in short time because if I take to long I am asked to login again and after that my quote has dissapeared, a very boresome technical problem I have encounted earlier but have to live with it.

 

[JANm]

OK to your first alinea I wanted to say it is a beginvalue problem then, but that you state in your second alinea. The velocity known is OK but the acceleration has two parts Newtonian attraction and centrifugal acceleration. The remark of Chronos about Kepler is correct... So I would say you need at least the surface of the ellipse. Or could you add the actual distance to the sun to your list of beginvalues. Knowing that it is an ellipse is a limiting value and not a number. From that you only know that the total energy = potential energy + kinetic energy is below zero
greetings Janm

 

[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]

You can compute your semi-major axis with
a = (2 / R - V ^ 2 / Mu) ^ -1
where mu is G*M, G is the gravitational constant, and M is the mass of the sun.

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 m located some distance d away, in the direction of the acceleration vector.

The problem is this solution is not unique. A mass 4m located some distance 2d yields exactly the same acceleration, as does any mass, distance pair of the form \kappa^2m, \kappa d.

 

[JANm]

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

Q: Is it possible to infer the distance to the center of mass, knowing the 2-velocity and 2-acceleration at any instance of time?

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]

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

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 parallel 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]

By the way you stated that the acceleration wil be senkrecht to the velocity.

I said no such thing (and I had to look that up. senkrecht=normal) My posts in this thread:

You can't if all you know is the velocity and acceleration. Think of it this way: The gravitational acceleration toward an object of mass M and distance r is -GM/r², directed toward the mass. Double the distance and quadruple the mass and you get exactly the same acceleration, but a very different orbit.

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.

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 m located some distance d away, in the direction of the acceleration vector.

The problem is this solution is not unique. A mass 4m located some distance 2d yields exactly the same acceleration, as does any mass, distance pair of the form \kappa^2m, \kappa d.

There is no mention of the acceleration vector being normal to the velocity vector.

The centrifugal force is senkrecht to the velocity, but the acceleration (F/m) is the total of attraction and centrifugal force

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]

. There is absolutely no reason to invoke the concept of centrifugal force in explaining orbits. Doing so leads to erroneous concepts.
Forget about centrifugal force.

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]

Hello D H
The thread was opened with a=v^2/r

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

Please tell me that you are not serious declining centrifugal force ...

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.

My problem with this problem is how can the radius of curvature be found if only velocity and acceleration is known!

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 r=\sqrt{GM/|\boldsymbol a|}. 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]

http://www.picamatic.com/view/3620071_ellipse/

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\theta+ R'w e\theta+ Rw' e\theta-Rw² er

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

all of the sentences with e\theta equals to Zero

so we will have :

R'' er -Rw² = K

and if i haven't made any mistake in my paper i have calculated R'' :
R'' = (V Sin(X) (eCos\theta+ 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 haven't 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\theta not that

Click on the Latex images and read there

and for :
\theta = 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]

what do you mean by" 2-velocity and 2-acceleration"

OK- 3-acceleration and 3-velocity then.

I only mention 2-velocity and 2-accleration because in an elliptical orbit (like a planet around the sun) the orbit is executed on a plane.

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
x= a cos(\omega t)
y= b sin(\omega t)
where a and b are the semi-axes in the x and y directions, respectively.

Then the velocity vector is
\vec{v}= -a\omega sin(\omega t)\vec{i}+ b\omega cos(\omega t)\vec{j}
and the acceleration vector is
\vec{a}= -a\omega^2 cos(\omega t)\vec{i}- b\omega^2 sin(\omega t)\vec{j}
Of course, the crucial point is determining \omega. 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]

and the acceleration vector is
\vec{a}= -a\omega^2 cos(\omega t)\vec{i}- b\omega^2 sin(\omega t)\vec{j}

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

 

[Mahbod|Druid]

Ive said how to find "w" in my post "page 2" next to Picture

also "wt" which is \theta

 

[JANm]

Velocity = R' er + Rw e(theta)
=> derivate again for acceleration
R'' er + R'w e\theta+ R'w e\theta+ Rw' e\theta-Rw² er

Hello Mahbod|Druid
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

 

[Mahbod|Druid]

http://www.picamatic.com/view/3620071_ellipse/

all of the sentences with e\theta equals to Zero

so we will have :

R'' er -Rw² = K

and if i haven't made any mistake in my paper i have calculated R'' :
R'' = (V Sin(X) (eCos\theta+ 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-.)

Hi

as Gravity Force always points to Sun (from Earth to sun) and there is no other force
there will be no acceleration in e[theta]vector (e[theta] vector makes 90 degree with er)
so there will be no coriolis force or Rw' that's why i said all the sentences with e[theta] in it equals to zero


and here is the function we found after all those calculating

R'' - Rw^2 = K

and R'' = (V Sin(X) (eCos\theta+ e^2))/R

so multiple both sides of R'' - Rw^2 = K by R


and find R (distance from Sun)
we have [theta] and Distance so direction of sun is found

************************************************************

However HallsOfIvy way was much easier and ofc better
we will have "a" and "b" also wt (wt = [theta]) so here we find ellipse and one of the focuses will be direction of Sun
and we can find Distance , etc

 

[JANm]

and for finding "e"

i think this is the answer : (in ellipse)

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

Hello Mahbod|druid
The dimension of your e is dimensionless I find out from the formula! You used e as exponential: in some way giving direction and normals (senkrechtigkeit). There is speach of using e as ellipticity which indeed is a numberles dimension. So what is it?
greetings Janm

 

[belliott4488]

You guys seem to be ignoring the basic observation D H made a while ago - the problem is indeterminate if you know only velocity and acceleration, because a given acceleration can be found at an infinite number of locations if you allow the central mass to vary.

Are you assuming that this orbit is around the Sun, so that you know its mass? Otherwise, this problem can't be solved.

 

[Mahbod|Druid]

it is not indeterminate
since Velocity vectors and Acceleration vectors are known

its like saying
at the distance of R and Mass of central star = M , we will have acceleration = K
also at the distance of 2R and Mass of central star = 4M we will have same acceleration = K

but we know that a satelite is orbiting the star in a Circular orbit with velocity V
for R satelite needs V^2 / R acceleration not to fall
and for 2R satelite needs V^2/2R acceleration not to fall
so the satelite with velocity V will fall on "2R-ish" star and that's incorrect
but on distance "R" it won't fall

and it is uniqe answer to that Question because we know Satelite is not falling


to Jamn:
I found eccentricity like this :

d(R Sin[theta]) / d(R Cos[theta]) = dy/dx
and derivative of Velocity and the path of orbit are same

also R = L/1+eCos[theta]

and don't forget this : d(xy) = dx*y + xy'*dx

 

[D H]

it is not indeterminate

Yes, it is.

but we know that a satelite is orbiting the star in a Circular orbit with velocity V

You do not know that the object is in a circular orbit.

 

[Mahbod|Druid]

I just gave an example for circular because it was easier

the Velocity given by Questioner(?) and vectors given must be currect
if not it will fall on the star / run away from star
and the Questioner already mentioned that it is a Ellipse so it is uniqe

look @ page 2 or hallosivy
they will give a uniqe answer (2 but one of them is correct)

 

[D H]

I just gave an example for circular because it was easier

If you know the orbit is circular, then yes, knowing the acceleration and velocity let's you know the orbital radius. However, you do not know that the orbit is circular. All you know are the acceleration and velocity vectors. They might not even be orthogonal to one another. (If they aren't, the orbit definitely isn't circular.)

the Velocity given by Questioner(?) and vectors given must be currect
if not it will fall on the star / run away from star

You are again assuming a circular orbit (I think; I don't quite get what you mean by "it will fall on the star / run away from star").

Planets follow elliptical, not circular, orbits. That means that at times the distance between the planet and the star is decreasing and at other times, increasing. Some comets have extremely elliptical orbits. Their closest approach to the Sun (perihelion) can be inside Mercury's orbit and they can go beyond Pluto's orbit at the other end of their orbit (apohelion). Some comets, such as Comet McNaught, will never come back because they are on an escape trajectory.

and the Questioner already mentioned that it is a Ellipse so it is uniqe

Just knowing the acceleration and velocity does not yield a unique solution. In fact, if all you know are the acceleration and velocity you do not even know the orbit is an ellipse. It might well be a parabola or a hyperbola.

 

[Mahbod|Druid]

Sorry for my poor English ... :shy:

here in this pic :

Do you agree that eccentricity is uniqe for a given Velocity(+ vectors) and acceleration (+ vectors) ?

and we can find eccentricity from :
e/Sin[theta] + Cotg[theta] = -Vy/Vx


and we can find w :
V Sin(X) / R
(X was shown in page 2)
where "w" depens on the distance from Sun

and :
R'' = (V Sin(X) (eCos+ e^2))/R
it also depens on the distance

so :
R'' -Rw^2 = K

put R'' and w in the equation :

(V Sin(X) (eCos+ e^2))/R - R*(VSinX)^2/R^2 = K
=>
R =( (VSinX)(eCos + e^2) - (VSinX)^2 )/K



Because "w and R" depens on the distance from "Sun" it will give a uniqe answer
just like Circular example did

(little calculation/typing errors are possible like in finding derivatives of functions ...)

 

[D H]

Do you agree that eccentricity is uniqe for a given Velocity(+ vectors) and acceleration (+ vectors) ?

One last time, NO.

Since you guys just cannot comprehend an abstract argument, here are some specific numbers for a very simple case. Suppose an object's velocity is 29.8 kilometers/second, it's acceleration is 5.93 millimeters/second², and that the velocity vector is perpendicular to the acceleration vector. This object might be

• 1 AU from a star with a mass of 1 solar mass (circular orbit, orbital radius = 1AU).
• 2/3 AU from a star with mass = 4/9 solar mass (perihelion, semi-major axis = 4/3 AU, eccentricity = 0.5)
• 2 AU from a star with mass = 4 solar mass (apohelion, semi-major axis = 4/3 AU, eccentricity = 0.5)
• 5/9 AU from a star with mass = 25/81 solar mass (perihelion, semi-major axis = 25/9 AU, eccentricity = 0.8)
• 5 AU from a star with mass = 25 solar mass (apohelion, semi-major axis = 25/9 AU, eccentricity = 0.8)

... and so on. There are an infinite number of choices. All you know, even in this simplified example, is that the orbit's semi-latus rectum is 1 AU. The eccentricity can be anything from 0 to infinity. (Eccentricity > 1 simply means a hyperbolic orbit.)

The problem is indeterminate.

 

[Mahbod|Druid]

Aha got it

I assumed we know what :
Acceleration(y) and Acceleration(x)

thats why [theta] was known in my calculation


but if u just told that ay/ax is not known it wouldn't get this long :D

thanks 4 ur time

 

[JANm]

Hello people
In the first place I must say this is a nice thread isn't it? Secondly I am proud to introduce Differential Geometry to you all. Usually it starts with giving three but in this case two functions of t (together called k(t), which denote the place of an object in a non-rotating frame. The first derivative in time we know let us call it vx and vy. Then is spoken of a natural parametrisation. I explain this mostly: if the curve is the iron a train can ride on then natural parametrisation is the equidistant wooden things the iron is connected to something spatial which you can count. The km-poles along a highway. Why is this parametrisation introduced? Because the velocity of objects along the curve is then still free to be chosen! All right. Vx and vy are in the direction of the curve. The natural parametrisation is kdot(s)=(vx*i +vy*j)/sqrt(vx^2+vy^2), all seen as function of t. Kdot(s) is a unity vector...
Secondly is stated that differentiation of Kdot(s) gives the curvature x(s)! x(s)=|kdotdot(s)|

That gives the principal normal (head normal?) h(s).
h(s)=kdotdot(s)/x(s)

Fitting the best circle to the curve, 1/x(s)is the radius of that.

For a three dimensionalcurve there is also a binormal but in this case that is constant the direction of z, so k... Binormal constant is also called the torsion =0

With aid of the formula of lagrange the curvature can be calculated with knowledge of velocity and acceleration as a function of time...

so x(t)=sqrt((<v,v>*<a,a>-<v,a>^2)/<v,v>^3)

the inproduct of v and a is the part of acceleration along the curve the rest of the acceleration is the bending part...

greetings Janm

 

[D H]

I assumed we know what :
Acceleration(y) and Acceleration(x)

thats why [theta] was known in my calculation

We do know ax and ay. That is the 2-vector that was mentioned in post #1.

 

[D H]

Hello people
In the first place I must say this is a nice thread isn't it? Secondly I am proud to introduce Differential Geometry to you all. ...

That's very nice, JANm, except for a couple of things: In general, the radius of curvature is not equal to the distance to the central mass, and the vector from a point on an orbit toward the center of curvature does not even point toward the central mass.

 

[JANm]

That's very nice, JANm, except for a couple of things: In general, the radius of curvature is not equal to the distance to the central mass, and the vector from a point on an orbit toward the center of curvature does not even point toward the central mass.

Hello D H
Where are your problems? So used to the heliocentric model that when it is possible that the sun is not exacly in the centre but mechanically spoken is an object a little misplaced in such a way that it counterbalances all the planets. Is that a problem? Must frames be fixed to material objects?
The threadmaker tells you that velocity as a function of time is given. Integrating that you get a curve. Differentiating by the way you get the acceleration, so that is actually redundant information. The curve is flat. Integrating the velocity one will find out that the placevectors are periodic functions. The amplitude of kx and ky are not the same.
If you take t0 as the one with maximum absolute velocity
maxv=max(sqrt(vx^2+vy^2)) than you have started in the perihelium. Then the ellips is neatly x,y orientated.
Please search somewhere for ellips in a encyclopedy or on the internet and you will find how the radius of curvature of an ellipse is. Smallest near perihelium and aphelium and large in the two directions of the short axis. Indeed the radius of curvature is not pointed to matererial substance. The focus is very near to the Sun, but that is what the threadgiver wants to find out and he doesn't want to start with that.
greetings Janm