# Calculating elliptical orbits

• #1
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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?

• #2
679
2
what do you mean by" 2-velocity and 2-acceleration"

• #3
126
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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
2,207
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
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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
2,207
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
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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
2,207
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
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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
126
0
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
2,207
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
Gold Member
1,752
143
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

• #13
126
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.
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

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
Gold Member
11,429
746
Keplers law still works pretty well.

• #15
D H
Staff Emeritus
15,415
687
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
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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
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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
31
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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
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687
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
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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
15,415
687
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 $\kappa^2m, \kappa d$.

• #22
223
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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
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687
I assumed that the 2- prefix was in the sense of 2-vectors, meaning two dimensional vectors.

• #24
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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
15,415
687
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.

• #26
223
0
. There is absolutely no reason to invoke the concept of centrifugal force in explaining orbits. Doing so leads to erroneous concepts.
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

• #27
D H
Staff Emeritus
15,415
687
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?

• #28
22
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Hi

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

or

we have GM/R^2 = acceleration ?

+ velocity

• #29
D H
Staff Emeritus
15,415
687
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.

• #30
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http://www.picamatic.com/view/3620071_ellipse/ [Broken]

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 havent 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 havent study ellipses yet so i don't know atm but i will think about it

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