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

[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

 

[D H]

The original poster does not say that velocity as a function of time is given. You are once again trying to change the problem to something that is solvable.

The problem is that the velocity and acceleration vectors at one instant in time are known, and that is all that are known. I gave an abstract argument why this is an indeterminate problem in post #21 and some specific examples in post #45.

 

[belliott4488]

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?

I think you've missed the point: for a circular orbit the instantaneous normal will point to the central body, as it does at the apses of an elliptical orbit, but at all other points on an elliptical orbit it will not.

D H's other point is the key, however: we don't know either the velocity or the acceleration at any point other than the one given, so we cannot differentiate along this curve. Equivalently, since we cannot say how the acceleration changes as we move from this point, we cannot determine how far we are from the central body - we cannot fit a curve to a single point; we need (at least) two. This was illustrated by D H's numerical examples.

These are all abstract reasons for why the problem cannot be solved with the information given. Since your calculations lead you to a contradictory result, you must either find the mistake in your calculations yourself or show why these more general arguments are incorrect. Simply repeating the calculations is rather pointless.

 

[D H]

I think you've missed the point: for a circular orbit the instantaneous normal will point to the central body, as it does at the apses of an elliptical orbit, but at all other points on an elliptical orbit it will not.

Moreover, even though the normal does point toward the central body at the apses, the center of curvature is not the central body's location. The distance to the central body is a(1\pm e) at the two apses, while the radius of curvature is a(1-e^2) (the semi-latus rectum) at both apses.

 

[JANm]

Hello everybody
Think I have solved the problem.
let us say v=ui+wj and a=xi+yj, with x,y,u and w known numbers.
Let us name p=dr/dt, q=d^2r/dt^2, l=dteta/dt and m=dteta^2/dt^2. Somewhere in this thread was fully calculated and by freedom of choice:
(1) u=p, (2) x=q-r*l^2, (3) w=r*l and (4) y=r*m+2p*l those are four equations with 5 unknown. One can calculate p,q,l,m as functions of r. (1) p=u, (2) q=x+w^2/r (3) l=w/r and (4) m=y/r-2*u*w/r^2

(4) is a quadratic with a necesary real solution for r so the discriminant = y^2 - 8u*w*m >= 0
this gives (5) 8u*w*m <= y^2

from equation (3) follows r=w/l substitution in (4) gives:
m=y*l/w-2u*l^2/w substitute m in equation (5)
and you get a parabolic of l:
16u^2*l*2 - 8u*y*l + y^2 >=0 so the discriminant >=0
It happens to be that the discriminant of this parabolic in l =0

so l=y/(4u) in equation (3) you find r=4u*w/y the wanted radius.

to make the thing complete (4) gives m=y^2/(8u*w)
and (2) gives q=x+y*w/(4*u)

Resubstitute the 4: p, q, l,m and they are known. As we wished to prove.

Greetings Janm

 

[Mahbod|Druid]

The problem is indeterminate

m = 0
withouth any calculation

because the central star is a Dot Gravital point at it can not create coriolis/euler(i think)
and if there are other stars around it will be indeterminate for sure


Here is the only way that problem will not be indeterminate :
we must know that there is only ONE Star in the center
& we must know Ay and Ax in UN-ROTATED polar direction (means we must know [Theta] and theta is the degree between (a(1-e)) and "R" (the line that points to satelite)
also [theta] >< 0 (theta must not be Zero) )
there we can find Distance else we can't

 

[belliott4488]

JANm - I do not understand your solution. For one thing, it's rather difficult to read - can you used the superscript and subscript buttons that appear above the editing box when you post your response? That would help a lot.

Hello everybody
Think I have solved the problem.
let us say v=ui+wj and a=xi+yj, with x,y,u and w known numbers.
Let us name p=dr/dt, q=d^2r/dt^2, l=dteta/dt and m=dteta^2/dt^2. Somewhere in this thread was fully calculated and by freedom of choice:
(1) u=p, (2) x=q-r*l^2, (3) w=r*l and (4) y=r*m+2p*l those are four equations with 5 unknown. One can calculate p,q,l,m as functions of r. (1) p=u, (2) q=x+w^2/r (3) l=w/r and (4) m=y/r-2*u*w/r^2
...

Could you also show us the entire derivation in one post, rather than referring to unspecified previous posts? This thread is now more than one month old, and no one wants to search through all the posts trying to guess which one shows the derivation you claim.

In particular, I do not see how (1) - (4) could possibly be true in general. The Cartesian vector components u, w, x, and y clearly must depend on the direction of the i and j axes, but dr/dt and dtheta/dt should not, so how could it possibly be true in general that u = dr/dt, as implied by (1)? This would mean that no matter what velocity vector we are given, the i component must equal the time derivative of the radial component in polar coordinates.

Perhaps I've misunderstood what you meant to say, so please show us a clear derivation with all the steps -- and use those superscript and subscript buttons! ;-)

 

[D H]

Let us name p=dr/dt, q=d^2r/dt^2, l=dteta/dt and m=dteta^2/dt^2

One last time: you do not know the radius, or the angle theta. Your solution is not a solution. This problem does not have a unique solution.

 

[belliott4488]

Here's another way to put it: all we are given is derivatives, so even if we could solve for position as a function of time, we could not establish the values of the integration constants without some initial conditions for position variables, e.g. the r or theta D H mentioned, or Cartesian position components. Of course, if we had those, then the problem would be trivial, since that's what we're asked to compute.

Here's another way: we're in agreement (I believe) that the acceleration is not enough to determine r since a given acceleration can be found at any value of r for an appropriate choice of M, the mass of the central body (which is unknown). Also, the given velocity can apply to any object regardless of the r value (it's an initial condition, independent of r). Thus, any given combination of velocity and acceleration can apply to an object at any value of r whatsoever. In other words, velocity and acceleration alone are not sufficient to determine r.

We have now given several different arguments for why this problem is indeterminate. Anyone who disagrees must show how all of these arguments are incorrect. Until that is done, any alleged calculation to solve the problem must be considered suspect.

 

[JANm]

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.

Hello HallsofIvy
It is nessesary to create a testcase for the disbelievers. The shape of your curve was right but we didn't know omega t. Let us start with:
x=a cos (teta) and y = b sin (teta), teta being a function of t also.
Calculate from that the velocity and the acceleration. Fill in some teta I don't know and so give v_x, v_y, a_x and a_y.
greetings Janm

 

[belliott4488]

Hello HallsofIvy
It is nessesary to create a testcase for the disbelievers. The shape of your curve was right but we didn't know omega t.

That was a parametrization of an ellipse; it is not the physical description of motion in an elliptical orbit. Specifically, omega is not constant in an orbit.

Let us start with:
x=a cos (teta) and y = b sin (teta), teta being a function of t also.
Calculate from that the velocity and the acceleration. Fill in some teta I don't know and so give v_x, v_y, a_x and a_y.
greetings Janm

You can start with the orbit and then calculate the velocity and acceleration at any point - that's not in dispute. The point is that there are an infinite number of elliptical orbits that can give rise to the same velocity and acceleration at some point. Given only the velocity and acceleration, you cannot go the other direction and find one unique orbit that will correspond to them.