Calculating elliptical orbits

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
679
2
what do you mean by" 2-velocity and 2-acceleration"
 
  • #3
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.
 
  • #4
Nabeshin
Science Advisor
2,205
16
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)
 
  • #5
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.
 
  • #6
Nabeshin
Science Advisor
2,205
16
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.
 
  • #7
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.
 
  • #8
Nabeshin
Science Advisor
2,205
16
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.
 
  • #9
10
0
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
 
  • #10
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.
 
  • #11
Nabeshin
Science Advisor
2,205
16
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.
 
  • #12
tony873004
Science Advisor
Gold Member
1,751
141
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.
 
  • #13
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!
 
  • #14
Chronos
Science Advisor
Gold Member
11,408
738
Keplers law still works pretty well.
 
  • #15
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
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.
 
  • #16
223
0
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.
 
  • #17
223
0
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
 
  • #18
You can use Kepler's law p^2=a^3.

The period squared is equal to the semi-major axis cubed.
 
  • #19
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
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.
 
  • #20
223
0
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
 
  • #21
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
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
223
0
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
 
  • #23
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
I assumed that the 2- prefix was in the sense of 2-vectors, meaning two dimensional vectors.
 
  • #24
223
0
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 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
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
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 [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].
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.
 

Related Threads for: Calculating elliptical orbits

Replies
1
Views
758
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
623
  • Last Post
Replies
5
Views
900
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
15
Views
19K
  • Last Post
2
Replies
29
Views
34K
Top