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

[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.

 

[belliott4488]

Okay, JANm - here is perhaps the easiest way to resolve this dispute: let's just put our respective points of view to the test. I'll give you a velocity vector and an acceleration vector at a point in space, and you tell me how far the the center of force is from that point.

In return, I will give you an infinite number of orbits (elliptical, parabolic, and hyperbolic, if you like) with the same velocity and acceleration at that point.

Here is the challenge problem:

Velocity = 10.0 km/sec i + 0.0 j
Acceleration = 0.0 km/sec² i + 10.0 km/sec² j

How far away is the center of force?

 

[JANm]

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
The radius of curvature is not equal to the distance of the central mass because they are different. Don't no if you are referring to my last post: it should have been
Hello hallsofIvy
We need an example for the disbelievers
the shape of your solution was good and with help of D H we can now say x=ea+r sin(theta) and y= r cos (theta) please you are so good in differentiating! Would you like to calculate the first and the second derivative and fill in some theta. Than I can prove to D H that from the four given parameters v_x, v_y, a_x and a_y I can recalculate r.
Thank you
greetings Janm

 

[D H]

You were just given a sample problem in post #61, JanM. What is the orbit?

 

[JANm]

Velocity = 10.0 km/sec i + 0.0 j
Acceleration = 0.0 km/sec² i + 10.0 km/sec² j

How far away is the center of force?

This is moving in a circle with a radius of 10km.

 

[D H]

That is one of an infinite number of solutions. You do not know which is the case.

 

[belliott4488]

Here is the challenge problem:

Velocity = 10.0 km/sec i + 0.0 j
Acceleration = 0.0 km/sec² i + 10.0 km/sec² j

How far away is the center of force?

This is moving in a circle with a radius of 10km.

That solution is not unique. It implies that the central body is at a distance 10km, which in turn implies that the central body has a gravitational parameter (G*mass) of:

mu = a*r²
= (10 km/sec²)*(10 km)²
= 1000 km³/sec²

This quantity is unspecified, however, so there is no reason to assume it has this value. Any other value would produce a different distance to the central body, as follows:

A larger value of mu would imply a greater distance. Since the potential energy would be smaller in magnitude in this case, being proportional to 1/r, but the kinetic energy would be unchanged, the total energy would still be negative, and we would have a elliptical orbit with this point as the apoapsis.

A smaller value of mu would imply a smaller distance to the central body, and a correspondingly greater potential energy. At a distance of 5 km (mu = 250 km³/sec²), the potential and kinetic energy per unit mass are equal in magnitude, so the total energy is zero, and we have a parabolic trajectory. For any value of r less than 5 km (or mu < 250 km³), the total energy becomes positive and we have a hyperbolic trajectory. For values of r between 5 km and 10 km, we have elliptical orbits with this point as periapsis.

Your solution is one of an infinite number of possible solutions.

 

[D H]

For example, the vehicle might be

• At perifocus in a hyperbolic orbit about a point mass of 1.498×10²⁰ kg located at 1 km j relative to the vehicle,
• At perifocus in a parabolic orbit about a point mass of 3.746×10²¹ kg located at 5 km j,
• At perifocus in an elliptical orbit about a point mass of 9.589×10²¹ kg located at 8 km j,
• In a circular orbit about a point mass of 1.498×10²² kg located at 10 km j,
• At apofocus in an elliptical orbit about a point mass of 1.498×10²⁴ kg located at 100 km j.

All of the above examples yield the velocity and acceleration specified in post #61, as do an infinite number of other solutions.

 

[JANm]

Here is the challenge problem:

Velocity = 10.0 km/sec i + 0.0 j
Acceleration = 0.0 km/sec² i + 10.0 km/sec² j

How far away is the center of force?

Hello belliot4488
This is certainly NOT an example of an elliptic orbit. The acceleration stands normal to the velocity. There can be only to points in an ellips where that happens: in the perihelium or in the aphelium. I calculated both possibilities and get a=0. I am very interested in your elliptical solution for this situation.

By the way I found the radius of curvature for an ellips.

near perihelium and aphelium rc=a(1-e^2) and when the object is on the short axis rc=b+a^2/b, with b=a*sqrt(1-e^2)

greetings Janm

 

[belliott4488]

Hello belliot4488
This is certainly NOT an example of an elliptic orbit. The acceleration stands normal to the velocity. There can be only to points in an ellips where that happens: in the perihelium or in the aphelium. I calculated both possibilities and get a=0. I am very interested in your elliptical solution for this situation.

By the way I found the radius of curvature for an ellips.

near perihelium and aphelium rc=a(1-e^2) and when the object is on the short axis rc=b+a^2/b, with b=a*sqrt(1-e^2)

greetings Janm

I've posted my solutions, as has D H. For various values of the central body's mass (per D H solutions) or mu (per my previous solutions), you can can get elliptical solutions where the point in question is either periapsis (5 km < r < 10 km) or apoapsis (r > 10 km).

How did you obtain a=0 for your ellipses?? a=0 only at r=infinity for a gravitational potential.

 

[belliott4488]

For example, the vehicle might be

• At perifocus in a hyperbolic orbit about a point mass of 1.498×10²⁰ kg located at 1 km j relative to the vehicle,
• At perifocus in a parabolic orbit about a point mass of 3.746×10²¹ kg located at 5 km j,
• At perifocus in an elliptical orbit about a point mass of 9.589×10²¹ kg located at 8 km j,
• In a circular orbit about a point mass of 1.498×10²² kg located at 10 km j,
• At apofocus in an elliptical orbit about a point mass of 1.498×10²⁴ kg located at 100 km j.

All of the above examples yield the velocity and acceleration specified in post #61, as do an infinite number of other solutions.

Correct me if I've misunderstood (or forgotten my terminology), but I believe you meant to say that the vehicle is at periapsis (perigee, perihelion, periselene, etc.) or apoapsis, not perifocus or apofocus. Aren't the latter the foci of the conic section, and thus not part of the trajectory?

 

[belliott4488]

JANm:

Something occurs to me: you do not seem to have made any references to Newton's Laws in your arguments, e.g. to relate the distance to the central body by use of the fundamental law of gravity.

Are you attempting to solve this problem purely in terms of differential geometry? If so, I wonder if you are treating the acceleration as the second derivative of the position with respect to a parameter t, such as given here:

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.

That "t" is not time in the usual sense, at least it is not if \omega is constant, as it would typically be in a problem in differential geometry. "t" in this context is just the parameter of the curve, but an object moving on the curve at a rate given by \omegat would not move according to Newton's laws.

 

[D H]

The astronomy community uses periapsis as the general term for the point on the orbit that is closest to the dynamic focus of the orbit. The aerospace community uses perifocus. Both terms are correct; I prefer the latter.

http://www.websters-online-dictionary.org/el/elliptical+orbits.html
http://www.websters-online-dictionary.org/pe/perifocus.html
http://www.websters-online-dictionary.org/pe/periapsis.html

 

[belliott4488]

The astronomy community uses periapsis as the general term for the point on the orbit that is closest to the dynamic focus of the orbit. The aerospace community uses perifocus. Both terms are correct; I prefer the latter.

Huh ... I didn't know that. Ironically, I'm from the aerospace world (NASA), which is where I learned them as peri/apoapsis, but that might have been a local aberration. I used to work at the Goddard Space Flight Center, so maybe all the astronomers beat the aerospace engineers into submission there. ;-)

 

[D H]

This is certainly NOT an example of an elliptic orbit.

It most certainly can be an elliptical orbit. (It can also be parabolic, hyperbolic or circular.)

For example, suppose the vehicle is located at -8 km j relative to a point mass of 9.589×10²¹ kg and is moving at 10 km/s i relative to the point mass+vehicle center of mass. This corresponds to an elliptical orbit with eccentricity e=0.25, semi-major axis a=10.667 km, and the argument of perifocus is 270 degrees (this is an equatorial orbit, so the argument of perifocus is arbitrarily measured counterclockwise from the +x axis). When the vehicle is at perifocus it will have a velocity of 10 km/s i and an acceleration of 10 km/s² j, exactly as specified.

 

[JANm]

Something occurs to me: you do not seem to have made any references to Newton's Laws in your arguments, e.g. to relate the distance to the central body by use of the fundamental law of gravity.

Hello Belliot4488
I use the acceleration=v^2/rc - Newtonatraction/m (1)
and total energy/m=v^2/2 - actual potential/m (2)
OK I use a and e, so thanks D H for your example with M, a and e given. my mechanics book tells : total energy/m = - G*M/(2*a)
actual potential = -G*M/(a-e*a)^2
I calculate GM from (2) and put it in (1) and get acceleration=0 !
In perihelium and in aphelium the acceleration must be different from 0 otherwise you get a circle!
I will calculate with the example of D H a little further...
greetings Janm

 

[JANm]

This corresponds to an elliptical orbit with eccentricity e=0.25, semi-major axis a=10.667 km.

Hello D H
semimajor axis is 32/3 km isn't it? I always have to get used to the decimal english point... with this eccentricity b=31/3 km and rc=10 km. So if I calculate v^2/rc = 10 km/sec^2 and all the acceleration is in the centrifugal force so there is no room for the Newton atraction.
By the way D H my impulsive answer to Belliot4488 that it should be a circle is of course wrong. The acceleration to the outside would expand this "circle" drastically. I am sorry I react so edgy and not by calculating first and then remarking after...

I have the strong feeling that this enormous acceleration normal to the velocity cannot be the solution of an ellipse. It looks more like the great attractor: a uniform acceleration in one direction, but I could be wrong...
greetings Janm

 

[D H]

When you do something like getting a=0 or "no room for the Newton atraction" it suggests you are doing something wrong. So, what is the right way to look at this?

One approach is via specific energy (energy divided by mass):

E = \frac 1 2 v^2 - \frac {\mu}r

If the object is following an elliptical orbit,

E = -\frac 1 2 {\mu}a

and thus

\frac{\mu}a = 2\frac {\mu}r - v^2

While the velocity is given, neither the gravitational parameter nor the radius are known. We do however know the acceleration. By Newton's law of gravitation,

|acc| &amp;= \frac{\mu}{r^2}

and thus

\aligned<br /> \mu &amp;= |acc|r^2 \\<br /> a &amp;= \frac {|acc|r^2} {2|acc|r - v^2}<br /> = \frac {r} {2- v^2/(|acc|r)}<br /> \endaligned

Substituting the known velocity of 10 km/s and acceleration of 10 km/s² yields

\aligned<br /> E &amp;= 50\,\text{km}^2/\text{s}^2 -\,r\,10\,\text{km}/\text{s}^2 \\<br /> a &amp;= \frac {r} {2- 10\text\,{km})/r}<br /> \endaligned

Setting r equal to 8, 10, and 100 km yields semi-major axis a values of 32/3, 10, and 1000/19 km, respectively. Setting r to 5 km yields an orbital energy of zero (a parabola). Values less than 5 km yield positive orbital energies (hyperbolic orbits). The elliptical orbit model is not valid for these short distances.



Bottom line: There is no unique solution.

 

[belliott4488]

Hello Belliot4488
I use the acceleration=v^2/rc - Newtonatraction/m (1)

This is not correct. By "Newtonatraction/m" I assume you mean the gravitational acceleration, -GM/r², which is correct, but the v²/r term is not. That term is the general form of the acceleration in a circular orbit, but even if this were known to be a circular orbit (which it is not), that acceleration would be equal to the gravitational acceleration - there is only one force (gravity) and hence only one acceleration. If you wanted to, you could set them equal and solve for the value of GM that would result in the given velocity in a circular orbit, but that does not seem to be what you're doing here.

The only reason I can imagine why you would add those two terms is if you are in a rotating coordinate system, so that the v²/r is the so-called "centrifugal acceleration" that results from the fictional centrifugal force. This is completely inappropriate in this case for a number of reasons: 1) we are working in inertial coordinates, so there are no "centrifugal forces", 2) we are given no reason to assume a circular orbit in inertial space, 3) even if you wanted to work in a rotating frame of reference for some reason, you would need to specify the center of rotation and the angular rate, neither of which are suggested by the given information.

and total energy/m=v^2/2 - actual potential/m (2)

This is correct, if by "actual potential/m" you mean absolute value (i.e. the magnitude) of the gravitational potential, GM/r.

OK I use a and e, so thanks D H for your example with M, a and e given. my mechanics book tells : total energy/m = - G*M/(2*a)

Again, that's true only for a circular orbit. You can easily derive that from
a = GM/r = v²/r
for a circular orbit, combined with the expressions for potential and kinetic energy.

actual potential = -G*M/(a-e*a)^2
I calculate GM from (2) and put it in (1) and get acceleration=0 !
In perihelium and in aphelium the acceleration must be different from 0 otherwise you get a circle!
I will calculate with the example of D H a little further...
greetings Janm

I suspect that you are getting acceleration = 0 because of your inclusion of the centrifugal acceleration. That is applicable only in a rotating frame, as I said above, where the acceleration will naturally be zero. This is because the rotating frame is chosen such that a given vehicle moving in a circular orbit in inertial space will be at rest in the rotating frame. That's the point of the fictitious centrifugal force - it must be invoked in order to balance the gravitational acceleration, which must always be present, so that the total acceleration is zero.

In the inertial coordinates where our velocity and acceleration vectors are defined, however, there is only the acceleration due to gravity, which is certainly not zero - in fact, it's given as part of the supplied information, so any calculation you perform that shows that it is zero is clearly mistaken. You can't start with a non-zero acceleration, do some math (correctly), and end up with a zero acceleration.

 

[JANm]

Hello Belliot 4488 and D H
There are two statements I have to make; firstly I think you both are right that if you give an example of acceleration and velocity at the point of perihelium or aphelium that r is indeterminate. I did some calculation with e=0,25 and get a=94/15=6 4/15 km, r_min=4,7 km and r_max=8 5/6 km, a part from that GM=376 and calculated by head that is the approximate GM for the moon. This r_min < 5; I don't know why that is a limit for you...
The object has to be smaller then let's say 3 km, which is the schwartschild radius of the sun, so the object is larger than its R_s it looks like some sort of neutronmoon. Estimated that the density 10^9* density of our moon. Think that professor Hawking could be interested in such a dense object.

Secondly gentlemen I really have to object to your negation of centrifugal force. I tell it once again: if you have a curve there is a touching line, a touching surface and A TOUCHING CIRCLE.
The velocity has the direction of the toutching line, the plain is called the osculating plane, change of the osculating plane is called torsion. In our example the osculating plane remains the same so torsion: tau=0.

The best touching circle has radius of curvature rc. The acceleration is the vector sum of the centrifugal force mv^2/rc and the gravitational force G*M*m*vec(r)/r^3. There are two points in which the centrifugal force has the same direction as the centrifugal force, those are aphelium and perihelium. In the perihelium the object goes too fast for a circular object so a remaining acceleration drives the object away from the gravitator. In the aphelium the object goes to slow for a circular orbit so it falls toward the gravitator.

Hope you really understand the last alinea!

greetings Janm

 

[belliott4488]

Hello Belliot 4488 and D H
There are two statements I have to make; firstly I think you both are right that if you give an example of acceleration and velocity at the point of perihelium or aphelium that r is indeterminate. I did some calculation with e=0,25 and get a=94/15=6 4/15 km, r_min=4,7 km and r_max=8 5/6 km, a part from that GM=376 and calculated by head that is the approximate GM for the moon. This r_min < 5; I don't know why that is a limit for you...
The object has to be smaller then let's say 3 km, which is the schwartschild radius of the sun, so the object is larger than its R_s it looks like some sort of neutronmoon. Estimated that the density 10^9* density of our moon. Think that professor Hawking could be interested in such a dense object.

This was a completely fictional example for the purposes of examining the question. There is no physical significance to the numbers I picked, so there is no need to question how physically realistic any specific solution might be.

The reason for the 5 km limit is that for a central body at any closer distance the total energy is greater than zero, so the the orbit will be unbounded, that is, a hyperbola.

Secondly gentlemen I really have to object to your negation of centrifugal force. I tell it once again: if you have a curve there is a touching line, a touching surface and A TOUCHING CIRCLE.
The velocity has the direction of the toutching line, the plain is called the osculating plane, change of the osculating plane is called torsion. In our example the osculating plane remains the same so torsion: tau=0.

I've studied differential geometry and the concept of the osculating circle has been familiar to me for some twenty years. I suspect DH is familiar with it as well. Unfortunately, it has no bearing on this problem.

You have a single velocity vector, which you could use to approximate the next relative position after a time delta-t. You could also use the acceleration to approximate the new velocity vector after delta-t. Then you are finished. You have no way of knowing the acceleration at the next point (because you do not know the location of the central body), so you can no longer integrate the derivatives in order generate the curve. You simply don't have enough information to calculate the radius of the osculating circle.

You seem to be using the acceleration in an attempt to do this, but you are assuming a circular orbit when you set a = v^2/r.

The best touching circle has radius of curvature rc. The acceleration is the vector sum of the centrifugal force mv^2/rc and the gravitational force G*M*m*vec(r)/r^3.

No, it is not. This is very much incorrect. In a rotating reference frame you can invoke a centrifugal force, but it does not exist in an inertial frame - ever! This is a basic fact from elementary mechanics, and if you do not understand it, then it is not surprising that you make so many other incorrect claims.

Non-inertial reference frames are used when it is helpful to go to a frame in which a moving object is at rest. If that object is actually accelerating, then the frame must also accelerate, so it is non-inertial, and you will observe the fictional "inertial forces", the centrifugal and coriolis forces. In our case, this would be very difficult since the object is not known to be in a circular orbit, so the reference frame in which is is stationary would be very difficult to define. If you could do that, however, then you would find that these two inertial forces would indeed be equal and opposite to the gravitational force, so that the object would remain stationary in this reference frame. In order to do this, however, you would need to know the object's acceleration everywhere on its trajectory, which you do not.

There are two points in which the centrifugal force has the same direction as the centrifugal force,

? I assume this is a typographical error.

those are aphelium and perihelium. In the perihelium the object goes too fast for a circular object so a remaining acceleration drives the object away from the gravitator. In the aphelium the object goes to slow for a circular orbit so it falls toward the gravitator.

This is the way you might describe the dynamics in a non-inertial reference frame. Not only is there no reasonable way to define such a frame for this problem, there is also no need to do so.
In fact, it is merely the object's momentum that carries it in an orbit other than a circular one for the cases you mention. All you need are the object's position and momentum and the true forces acting on it, and its path is completely determined. In this case, there is only one force, gravity, and that is all that appears in the equations of motion.

Hope you really understand the last alinea!

greetings Janm

JANm - I strongly suggest that you go and review some elementary mechanics, specifically orbital or two-body mechanics. There are plenty of on-line resources that can explain these concepts to you.

 

[D H]

The easy point first:

This r_min < 5; I don't know why that is a limit for you...
The object has to be smaller then let's say 3 km, which is the schwartschild radius of the sun, so the object is larger than its R_s it looks like some sort of neutronmoon.

You are making too much out of Belliot's sample problem. He picked some rather non-realistic numbers. Ignore that. Just pretend the universe is Newtonian for this problem.

Regarding the 5 kilometers: It's not a limit. Look at my examples in [post=2215767]post #67[/post]. I intentionally gave an example with a distance of 1 kilometer. Once again, forget the fact that the numbers are not realistic. Look at the numbers from the perspective of Newtonian mechanics.

That 5 km separation is very significant from the perspective of the kind of orbit the spacecraft is in. If the point mass is located exactly five kilometers away, the given velocity and acceleration make the total mechanical energy zero. At distances less than five kilometers the mechanical energy is positive; at distances greater than five kilometers the mechanical energy is negative. When the energy is negative the orbit is an ellipse (a circle being a special case of an ellipse). When the energy is positive the orbit is a hyperbola; in a sense it isn't really an orbit. Zero mechanical energy and non-zero angular momentum means the orbit is a parabola. That is the significance of the 4 km separation.

Secondly gentlemen I really have to object to your negation of centrifugal force. I tell it once again: if you have a curve there is a touching line, a touching surface and A TOUCHING CIRCLE.

There is a good reason that Frenet–Serret formulae are not used much in classical orbital mechanics. They make an absolute mess of the physical equations of motion.

The best touching circle has radius of curvature rc. The acceleration is the vector sum of the centrifugal force mv^2/rc and the gravitational force G*M*m*vec(r)/r^3.

What you have done here is to create an incredibly nasty reference frame.

• The center of curvature is moving, and is doing so non-uniformly (it orbits about the central mass). This means you need to incorporate into the equations of motion an inertia force equal to the additive inverse of the product of the mass and the second time derivative of the location of the center of curvature.
• The frame is rotating, and doing so at a weird rate (it is the rate at which the unit tangent vector is rotating). This means you need to add a centrifugal force to your equations of motion.
• The frame is rotating non-uniformly. This means you need to add a fictitious force equal to the additive inverse of the product of the mass and derivative of the unit tangent vector's rotation rate (i.e., the unit tangent vector's angular acceleration).
• The radius of curvature isn't constant. The spacecraft is moving back and forth along the x axis in this horrendous frame. This means you need to add a coriolis force to your equations of motion.

In short, the acceleration is not "the acceleration is the vector sum of the centrifugal force mv^2/rc and the gravitational force G*M*m*vec(r)/r^3."


The acceleration in the inertial frame is G*M/r^2, directed toward the central mass. There is no centrifugal force, nor an inertial force, a coriolis force, or an angular acceleration force. My advice: Until you fully understand orbits from the perspective of an inertial reference from, it would best suit you to forget about centrifugal force. You will make mistakes until you get the basics right.

 

[JANm]

[*]The frame is rotating, and doing so at a weird rate (it is the rate at which the unit tangent vector is rotating). This means you need to add a centrifugal force to your equations of motion.
[*]The frame is rotating non-uniformly. This means you need to add a fictitious force equal to the additive inverse of the product of the mass and derivative of the unit tangent vector's rotation rate (i.e., the unit tangent vector's angular acceleration).
.

Hello D H
Please don't start with me with the principle of Mach. Herbert Dingle was also very found of it. (Non) uniformly rotating frames have side-effects. In Meteorology it has to be used and when it is done properly the weather-forecasts are fine. Somewhere was already mentioned the coriolis force which by the way comes from a uniformly rotating frame. The Earth rotates 366,2425 times with respect to the stars in a very regular way explained by the law of conservation of angular momentum. The fact that the sun only rises 365,2425 times a year is because the Earth makes one turn around the sun in a yaer.

The Earth moves in a approximate circle. If the radius would be 150 km then the difference with a circle is no more then a metre. The very special effect is that the sun is in the focus of this curve. If the sun is closest that happens at appr. 5 januari you have perihelium. That is why australia has such very hot summers and we have mild winters.
Perihelium means that the Earth sweeps away from the sun:
because rotational force exceeds the gravitational attraction.

30 years ago I wrote a scription about iceages and studied the efect that the halfyear -pi/2 < teta < pi/2 can be different from the other "half". Milankovitsch says that that is the most important fact for very large scale climatology. To stay with the Earth please calculate the days between 21 september and 21 march and you will see that the halfyaers are not of the same lenght.

The problem with this scription was and is that it uses the three laws of Kepler which seems to be out of date, or something so I was not "allowed" to use this knowledge in htis thread.
greetings Janm

 

[D H]

That's very nice, JanM.

Where did I mention Mach's principle? (I didn't. This is a purely Newtonian problem, and how momentum arises is outside the scope of this problem.)

How does this write-up address the Frenet-Serret frame, shifted to the center of curvature, which you are implicitly using to calculate centrifugal force? This frame has a rotation rate equal to ||{\boldsymbol v}||/\kappa, the ratio of the velocity vector magnitude to the radius of curvature. This is not any of the standard orbital rates.

What in the world do Milankovitch cycles have to do with the topic at hand?

The only things wrong with Kepler's laws are (1) they are a consequence of Newtonian gravity, and (2) they implicitly assume the orbiting body has negligible mass compared to the central body. This is a classical mechanics problem, so issue #1 is not a problem here. There are simple corrections to Kepler's laws to get around the problem of an orbiting body with non-negligible mass, so that isn't a problem either.

 

[JANm]

OK then
Kepler 1: Planets move in ellipse orbits around the sun, with the sun in a focus.
e=sqrt(a^2-b^2)/a and r=a*(1-e^2)/(1+cos(nu))
(r,nu) are the heliocentric polar coordinates.
Kepler 2: r covers equal surfaces in equal times:
(1/2)r^2*d(nu)/dt=pi*a*b/P
Kepler 3: P^2/a^3=4*pi^2/GM
a value which is equal for all planets.
Those are the three laws.
We have noticed that with given semi-major axis: a the energy remains constant. But K3 gives that the period p is then also constant.
So disturbances change e, b and pi_p and don't change a or p.

K3 gives us:
p=2pi*a*sqrt(a/GM)=2*(22/7)*(94/15)*sqrt(94/(15*376)=
(22/7)*(94/15)*sqrt(1/15)=4,5 seconds.

Nice to know the period, but now we need a velocity and a acceleration in a point which is not perihelium or aphelium.
Janm

 

[belliott4488]

OK then
Kepler 1: Planets move in ellipse orbits around the sun, with the sun in a focus.
e=sqrt(a^2-b^2)/a and r=a*(1-e^2)/(1+cos(nu))
(r,nu) are the heliocentric polar coordinates.
Kepler 2: r covers equal surfaces in equal times:
(1/2)r^2*d(nu)/dt=pi*a*b/P
Kepler 3: P^2/a^3=4*pi^2/GM
a value which is equal for all planets.
Those are the three laws.
We have noticed that with given semi-major axis: a the energy remains constant. But K3 gives that the period p is then also constant.
So disturbances change e, b and pi_p and don't change a or p.

What's pi_p? Otherwise, this is all very nice.

Of course, Newton showed that Kepler's laws applied to any pair of bodies bound by gravitational attraction, so that means that they apply to our problem, just with a different (unknown) value of M, the mass of the central body, which we are not told is the Sun.

K3 gives us:
p=2pi*a*sqrt(a/GM)=2*(22/7)*(94/15)*sqrt(94/(15*376)=
(22/7)*(94/15)*sqrt(1/15)=4,5 seconds.

You're losing me ... it looks like you're setting a = 94/15 and GM = 376 ... where do these values come from?

Nice to know the period, but now we need a velocity and a acceleration in a point which is not perihelium or aphelium.
Janm

Well, we would know the period if we knew both the semimajor axis a and the gravitational parameter GM, but we don't know either one.

I don't understand how this helps to solve the problem we've posed to you.

Why have you not answered any of the questions that D_H and I have asked? Instead of responding to our arguments, you just continue to make new assertions, many of which are true, but not clearly relevant, others of which are false, for reasons we have given. Please respond directly to the objections we have raised.

 

[JANm]

What in the world do Milankovitch cycles have to do with the topic at hand?

Hello D H
Milankovitch used the changements of the socalled Secular parameters to calculate the change of incoming radiation of the sun
using bands on the Earth of 10 degrees. As I already explained change in perihelium point in the year gives change in northern and southern summer and winter halfyears.
The obliquity is the axis of Earth's rotation to the plane of rotation. Changes in obliquity, Changes in eccentricity, Changes in perihelium are the secular parameters. They cannot be fortold with the Kepler equations which are just exact ways to make the ellipse complete. Perhaps the thread-giver wants to calculate these by NOT USING Kepler I don't know. By the way Milankovitch did all his scientific work in prison during the war.
greetings Janm

 

[D H]

For crying out loud, JANm!

This is a freshman physics problem, at best. As such, we are not concerned with general relativity, nor third body perturbations, nor non-spherical mass distributions. Baby steps. You need to learn to crawl before you can learn to walk, and you need to learn to walk before you can learn to run.

You yourself implicitly assumed a Keplerian world by using ellipses. Real orbits are not really ellipses. They're fairly close to ellipses, but they are not ellipses.

 

[D H]

Not really ellipses. Fixed, and thanks.

 

[JANm]

What's pi_p? Otherwise, this is all very nice.

You're losing me ... it looks like you're setting a = 94/15 and GM = 376 ... where do these values come from?

pi_p is the point of perihelium in the language of spring-point? Is that the right name?

If you take e =0,25 and your example v=10 km/s and acc=10 km/s^2
a is normal to v, then follows a =94/15, r_min=4,7 and GM=376.
This is perihelium and there

-GM/(2a)=v^2/2 - GM/(a(1-e) and

acc= v^2/rc - GM/(a(1-e)^2

So there is the example of a solution of an ellipse. I have admitted there needs to be extra information in the case that the point taken is perihelium or aphelium. So that is an exeption to the solution I still mean to have if another point on the ellipse is taken.

Greetings Janm

 

[D H]

Try again, Jan. You cannot deduce the semi-major axis from the acceleration and velocity. Let's see what can be deduced. This development assumes Keplerian orbits about a central body of mass M.

Nomenclature:

• α Semi-latus rectum. The semi-latus rectum and semi-major axis are related via \alpha = a(1-e^2), where e is the orbit's eccentricity.
  
  
• θ The true anomaly, the angular displacement between the ray emanating from the central body to the perifocal point of the orbit and the ray emanating from the central body to the orbiting body's current location, measured positive in the direction of the orbiting object's motion.
  
  
• μ The central body's standard gravitational parameter, \mu\equiv GM.
  
  
• a The semi-major axis.
  
  
• a The orbiting object's acceleration vector. To avoid confusion between the acceleration and semi-major axis, I will use ||a|| to denote the magnitude of the acceleration.
  
  
• r The distance between the orbiting object and central object.
  
  
• v and v The orbiting object's velocity vector and its magnitude, measured with respect to a non-rotating frame with origin at the central object's center of mass.

I'll use the vis-viva, orbit, and gravitational attracton equations as the basis for this analysis, respectively

\aligned<br /> v^2 &amp;= \mu\left(\frac 2 r - \frac 1 a\right) \\<br /> \frac{\alpha} r &amp;= 1 + e\cos\theta \\<br /> ||{\boldsymbol a}|| &amp;= \frac \mu{r^2}<br /> \endaligned

Note that the a in the vis-viva equation is the semi-major axis, not the acceleration. Combining the above with the relation between the semi-latus rectum and semi-major axis yields

\aligned<br /> v^2 &amp;= \frac{\mu}{\alpha}(1+e\cos\theta)^2<br /> \left(1+\left(\frac{e\sin\theta}{1+e\cos\theta}\right)^2\right) \\<br /> ||{\boldsymbol a}|| &amp;= \frac \mu{\alpha^2}(1 + e\cos\theta)^2<br />

From which

<br /> \frac{v^2}{||{\boldsymbol a}||} =<br /> \alpha\left(1+\left(\frac{e\sin\theta}{1+e\cos\theta}\right)^2\right)<br />

Note that for an object an perifocus or apofocus, the above simplifies to

\alpha = \frac{v^2}{||{\boldsymbol a}||}

Thus for the example specified in [post=2214819]post #61[/post], we do know that the semi-latus rectum is 10 km. We do not know the semi-major axis. It can take on any value from 10 km and above, depending on the eccentricity.

I have admitted there needs to be extra information in the case that the point taken is perihelium or aphelium. So that is an exeption to the solution I still mean to have if another point on the ellipse is taken.

Wrong. There is no general solution. All knowing the acceleration gives you is the direction toward the central body and the ratio \mu/r^2. The gravitational acceleration depends on position and mass only; velocity is not part of the equation.


Look at it this way: Suppose, in a completely different context (i.e., forget this problem in terms of answering the question that follows), all that you know about two parameters x and y is that the ratio between them is a known quantity c: x/y=c. Given that information alone, can you deduce the value of y? The answer is no. You don't have enough information. The same situation applies to this particular problem. You know the value of a ratio, but you don't know the value of either of the parameters involved in the ratio.

 

[JANm]

Try again, Jan. You cannot deduce the semi-major axis from the acceleration and velocity.

Hello D H
You did a lot of homework. I will adjust my next writings to this profound way of defining. Indeed there was trouble with acceleration and semimajor axis and perhaps others. I will read this thread more profoundly some other time...

If you say try again and then say that what I have to try is impossible that does not sound inviting. I say now for the third time.

In the case that velocity and acceleration are given on the umbilical points of the curve a more indefinite stage appears than in the most common stage that the velocity and acceleration are not normal to each other.

I am very bothered by the fact that the threadgiver does not give a general example at this moment of discussion. I am discussioned to defeat with an example measurement would never give...

If you are in an elliptical orbit around an object and you measure velocity and acceleration than it is impossible to measure exact at the moments of the umbilical points of the curve.

Umbilical points in differential geometry are difficult. In differential geometry it is nicer to calculate on a geoid than to calculate on a bolar object. Information of curvature in different directions being different on a geoid and exact the same on a bolar object LOOSES INFORMATION.

I HAVE ADMITTED THAT THERE ARE MANY SOLUTIONS POSSIBLE if you take umbilical points. Moving in a circle is the most probable solution if given values of velocity are normal to acceleration, because moving in a circle means that velocity is ALWAYS normal to acceleration.

B4488 and DH you are teasing the differential geometrist and for the very last time I need an example where acceleration and velocity are NOT normal to each other. Then I can test the calculations (which took me very much time I must say; doesn't matter, but after all my work I want a sensible testcase...
greetings Janm

 

[belliott4488]

What's pi_p? Otherwise, this is all very nice.

You're losing me ... it looks like you're setting a = 94/15 and GM = 376 ... where do these values come from?

pi_p is the point of perihelium in the language of spring-point? Is that the right name?

If I understand you, you're referring to the argument of perigee, which is the angle in the orbital plane between the ascending node and the perigee (perifocus or periapsis in general), measured at the central body. The ascending and descending nodes are the intersections of the orbit with the x-y plane of the inertial coordinate system. (By the way, the English spellings are "apehelion" and "perihelion" for orbits about the Sun.)

If you take e =0,25 and your example v=10 km/s and acc=10 km/s^2
a is normal to v, then follows a =94/15, r_min=4,7 and GM=376.
This is perihelium and there

-GM/(2a)=v^2/2 - GM/(a(1-e) and

acc= v^2/rc - GM/(a(1-e)^2

Okay, so you have arbitrarily chosen to set e = 0.25 (as we write on my side of the Atlantic Ocean - 0,25 in Haarlem). This selects one of the infinite number of possible solutions. (e = 0 gives the circular orbit solution you selected earlier.)

So there is the example of a solution of an ellipse. I have admitted there needs to be extra information in the case that the point taken is perihelium or aphelium. So that is an exeption to the solution I still mean to have if another point on the ellipse is taken.

Greetings Janm

Okay - maybe we're now in closer agreement. You have shown that the solution is known if we assume a circular orbit, but if the orbit is general (ellipse, parabola, or hyperbola), then we would need additional information, such as a second point of the orbit, in order to find a unique solution.

 

[JANm]

(By the way, the English spellings are "apehelion" and "perihelion" for orbits about the Sun.)

Hello Belliott4488
You mean that if you have an ellipse for instance of the vulcano active moon of Jupiter Io to Jupiter YOU WOULD NOT CALL THE CLOSEST POINT OF IO TO JUPITER THE PERIHELIUM, and after it he's turned 180 degrees in whatever subsytem rotation planet curvature system you and DH work: YOU STILL WOULD NOT CALL THAT APHELIUM
Is that what I understand with discussion with you both?
I do not want to speak with you for a whole week. I am sich and tired of your pityfull only circular excample witch can only point to a circle of 10 km.

AND I AM VERY ANGRY WITH YOU THAT YOU DIDN"T ADMIT THAT
e=0, r=a=b=10km, GM=1000 is actually really a solution to the problem. Statistically the only sensible one.

A physisician is NOTHING if he can't admit that another physician is RIGHT AT ONE TINY LITTLE BIT of moment

Ih

 

[belliott4488]

JanM - I thought from your previous post that you had agreed that this problem is indeterminate, but now it seems that you have returned to your earlier position, despite all the simple reasons we have given in opposition to it (which you continue to ignore).

...
I am very bothered by the fact that the threadgiver does not give a general example at this moment of discussion. I am discussioned to defeat with an example measurement would never give...

...

B4488 and DH you are teasing the differential geometrist and for the very last time I need an example where acceleration and velocity are NOT normal to each other. Then I can test the calculations (which took me very much time I must say; doesn't matter, but after all my work I want a sensible testcase...
greetings Janm

I am surprised! You are a mathematician, no? I would not have expected that you would require a physically meaningful set of conditions for what is essentially a mathematics problem. But no matter - I believe the following should be more realistic, and the velocity and acceleration vectors are not normal to each other, as you requested. (I'm still setting the components in the z-direction to zero so that we can stay in the x-y plane; I trust you will not object). Here is the new problem:

velocity v = 0 km/sec i + 10 km/sec j
acceleration a = -0.6 km/sec² i - 0.1 km/sec² j

I'm pretty sure that's a more physically realistic example. It is close to values I might use in my daily work.

So, given the new velocity vector and acceleration vector, do you still claim to be able to tell us how far this vehicle is from the central body?

 

[belliott4488]

Hello Belliott4488
You mean that if you have an ellipse for instance of the vulcano active moon of Jupiter Io to Jupiter YOU WOULD NOT CALL THE CLOSEST POINT OF IO TO JUPITER THE PERIHELIUM, and after it he's turned 180 degrees in whatever subsytem rotation planet curvature system you and DH work: YOU STILL WOULD NOT CALL THAT APHELIUM

No, I would not - the correct words in English are "aphelion" and "perihelion", which refer only to orbits around the Sun (Greek \alpha\pi o (apo) for "away" and \eta\epsilon\lambda\iota o\sigma (helios) for "Sun". For Earth orbit you have apogee and perigee (Geos = Earth), for the Moon, aposelene and periselene, etc. I don't know the words for Jupiter-centered orbits off-hand, but the general terms are apoapsis and periapsis (or apofocus and perifocus, as DH has explained).

Is that what I understand with discussion with you both?
I do not want to speak with you for a whole week. I am sich and tired of your pityfull only circular excample witch can only point to a circle of 10 km.

Actually, the 10 km circle was your solution. As DH and I both told you, there are an infinite number of elliptical orbits with the given velocity and acceleration, including ones that will have semimajor axes of thousand of kilometers, if that's what you want. Just let GM increase without bound, and you get ever-larger distances to the central body for the same acceleration. Of course, if you place anybody at such a point and give it the velocity I stated, it must enter some kind of orbit - it must go somewhere, mustn't it?

AND I AM VERY ANGRY WITH YOU THAT YOU DIDN"T ADMIT THAT
e=0, r=a=b=10km, GM=1000 is actually really a solution to the problem. Statistically the only sensible one.

Of course we admitted that your solution is valid - we simply denied that it was the only solution.

Now you are speaking of statistics? Is this how you solve problems in Physics? You consider the infinite set possible solutions of an indeterminate problem and then apply probability to pick one solution? Wow. I would simply call the problem "indeterminate" and leave it at that.

A physisician is NOTHING if he can't admit that another physician is RIGHT AT ONE TINY LITTLE BIT of moment

Ih

Well, in English a "physician" is a medical doctor; the scientists are called "physicists", but I assume you are trying to insult me and not your medical care-giver.

But I am relieved to hear that I am more than "nothing" since I have noted many times when you were correct, for example, your circular solution is a correct solution to the problem given -- it is, however, only one of an infinite number of perfectly physically correct solutions.

You, on the other hand, seem never to have considered the possibility that you might be wrong in your reasoning. Instead of responding to a number clear objections that both DH and I have raised, you simply ignore those objections and present another incorrect argument - or you diverge onto an unrelated subject in which you can refer to a lot of impressive mathematics that has no bearing on the current problem.

You don't seem to be at all interested in learning how to improve your understanding of the subjects we've brought up, so if you want to end this exchange here, that's fine with me. Until you are interested in seeing where you have gone wrong (and you are definitely wrong - you have made mistakes for which I would have given my first-year Physics students very poor marks), then this conversation is really a waste of all of our time.

 

[D H]

Hello Belliott4488
You mean that if you have an ellipse for instance of the vulcano active moon of Jupiter Io to Jupiter YOU WOULD NOT CALL THE CLOSEST POINT OF IO TO JUPITER THE PERIHELIUM,

I have not criticized you regarding your abuse of terminology because you don't know the terminology. But since you asked -- I absolutely would not call the closest point of Io to Jupiter the perihelium point. Perihelion (apparently perihelium in German) means closest to the Sun, and the Sun only. If I were talking about the closest point on Io's orbit about Jupiter to Jupiter I would either use the correct specific term (perijove) or one of the correct generic terms (perifocus or periapsis).

AND I AM VERY ANGRY WITH YOU THAT YOU DIDN"T ADMIT THAT
e=0, r=a=b=10km, GM=1000 is actually really a solution to the problem.

Neither of us said r=10 km is not a solution to the problem. In fact, just the opposite:

That is one of an infinite number of solutions. You do not know which is the case.

That solution is not unique.