Is tangential velocity in elliptical orbits stable?

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Andreas C
Hi, my name is Andreas, and I am new to this forum. I had a couple of questions regarding orbital dynamics, since I am not very familiar with the field:

Is tangential velocity in elliptical orbits stable like in circular orbits? If not, how exactly does it change? How do you calculate orbital velocity from tangential velocity? What's the mechanism that keeps elliptical orbits from dissipating?

drvrm
Is tangential velocity in elliptical orbits stable like in circular orbits? If not, how exactly does it change? How do you calculate orbital velocity from tangential velocity? What's the mechanism that keeps elliptical orbits from dissipating?

well, i will like to know what you mean by stable velocity.
my guess is that by stable you mean the magnitude being same at each point on its path.
the evidence is that you call it stable in circular orbits.
if you look at the motion of a body on elliptical path its angular velocity is changing ,so the speed will vary -keeping areal velocity constant as per Keplar's law.
the change in velocity does not mean that the planets are dissipating its total energy-the kinetic energy may be changing but the potential energy is also changing and its a conservative system.
the area swept by the the radius vector drawn from center of the force to the body sweeps equal area in equal time , that means the Areal velocity is constant
mathematically it means 1/2. r^2 . d(theta)/dt = constant. so its tangential velocity will be changing - it will be slower when r is large and faster when r is smaller.

Andreas C
well, i will like to know what you mean by stable velocity.
my guess is that by stable you mean the magnitude being same at each point on its path.
the evidence is that you call it stable in circular orbits.

That's 100% correct. I kinda struggle with the terminology, because English is not my first language.

if you look at the motion of a body on elliptical path its angular velocity is changing ,so the speed will vary -keeping areal velocity constant as per Keplar's law.

Oooh, thanks now I get it. I think I kinda confused the two. By the way, is there a simple derivation of this from Newton's laws?

drvrm
By the way, is there a simple derivation of this from Newton's laws?

if you calculate the area swept by the radius vector say in infinitesimal time dt -then the radius vector must have turned by an angle d(theta) and have swept ds a length of an arc- one can use the relation d(theta)= ds /r
the small triangle formed with base ds has almost same r as two sides ;
so the area can be written as dA = 1/2 . base . altitude = 1/2. ds. r = 1/2. r.d(theta).r = 1/2. r^2,d(theta)
the motion has happened in dt time
Therefore dA/dt = 1/2 . r^2. d(theta)/dt
but the angular momentum of the body is m. r^2. d(theta)/dt and is a constant of motion as no external torque is acting on the body and mass m of the body being constant - r^2. d(theta)/dt must be constant , therefore areal velocity must be constant.

Andreas C
Andreas C
Thanks, that's a good answer. I am slightly confused though on the matter of orbital and tangential velocity. How do I calculate the orbital velocity if I know the tangential velocity in an ellipse?

drvrm
orbital velocity
what is your definition of orbital velocity?
if it is orbital speed an average speed of completing one rotation then its a scalar and can be calculated ;
if you denote it by (a path length/time taken ) then the tangential velocity magnitude can give you a picture as its a curved path.

Andreas C
Andreas C
what is your definition of orbital velocity?
if it is orbital speed an average speed of completing one rotation then its a scalar and can be calculated ;
if you denote it by (a path length/time taken ) then the tangential velocity magnitude can give you a picture as its a curved path.

Yeah, it's orbital speed. I'm sorry, as I have already said, I struggle a bit with the terminology. I want to know how to calculate it if we know the tangent velocity at a given point of the ellipse. Also how to find the "instantaneous" orbital speed. Is it just equal to the magnitude of the tangent velocity or am I making a mistake here?

ogg
A circle is to an ellipse like a square is to a rectangle. Ellipses are much, much more complicated than circles or rectangles however. There is some similarity between a pendulum and a elliptical orbit. In both cases the total energy =potential+kinetic is conserved and constant (but the ratio PE:KE is changing periodically). In our Solar System, most of the planets and moons are in near circular orbits. So, circles are good approximations. Of course, this isn't true for most comets! (which are on highly elliptical or even parabolic 'orbits'). Tangental velocity is instantaneous velocity. The path length of an ellipse does NOT result in a elementary formula like the formulas for circle or rectangles. (see elliptic integrals, non-elementary functions.) These are almost never discussed in introductory physics or math courses. While velocity is constant in a circular orbit, it is NOT in an elliptical orbit. Ellipses are non-elementary.

Andreas C
Andreas C
Yeah, but the question is, how do you find the orbital speed if you know either the periapsis or the apoapsis and the tangential velocity (and the mass of each body, of course)? I guess this probably requires first finding the shape of the ellipse and the path length, and then the time it takes, right? Am I missing something here?

I don't understand the difference you make between orbital and tangential velocity. For me, these are synonymous.

Gold Member
I don't understand the difference you make between orbital and tangential velocity. For me, these are synonymous.

If you switch the word Orbital to Angular (and this may be what the OP means) the difference is that the velocity along the tangent is not the same as the radius times the angular velocity. That's just Geometry. It could be as simple as that.

Andreas C
Andreas C
If you switch the word Orbital to Angular (and this may be what the OP means) the difference is that the velocity along the tangent is not the same as the radius times the angular velocity. That's just Geometry. It could be as simple as that.

Yeah, that's probably it. I know, I've probably confused all of you very much with my poor terminology, so sorry about that.

Gold Member
What's the mechanism that keeps elliptical orbits from dissipating?
I think this could be the nub of the question, the rest being to do with definitions. There is no loss mechanism in an orbit that's well out of the drag of an atmosphere so the total Energy must remain constant. At any Point, the PE plus the KE will be the same as at any other point. In an elliptical orbit, the satellite moves from a position of minimal GPE when it is going fastest (Max KE). It then climbs away, gaining GPE and losing KE, until it is at its perigee, when its GPE is highest (i.e. furthest out of the potential well of the planet). At apogee and perigee, the tangential velocity is the same as its angular velocity times radius. Keppler's Laws were found by straight observation (and mind bending) and with no idea of the Inverse Square Law that Newton later showed to be operating and producing an elliptical orbit.

Andreas C
Didn't Newton base his inverse square law on Kepler's laws? Am I missing something here? I probably am...

Gold Member
Didn't Newton base his inverse square law on Kepler's laws? Am I missing something here? I probably am...
Yes, of course. But the ISL was how Newton explained a 'physical reason how Kepler's Laws are followed. Kepler was just describing the observed pattern of things and not the mechanics of it. At the time, it was thought that there had to be a 'motor' driving it all around (just as you always need something to keep an earthbound mechanism going). Isaac out them all straight!! Respect.

Andreas C
Papa Doyle
What a lovely discussion. Usually there is someone asking about something. But, because the person wants information, as opposed to already having it, the question asker often uses an improper term or makes a cognitive error about something considered obvious to those who know the subject a bit better. What usually follows is a smarmy diatribe from someone who clearly knows something about the answer but is more interested in pointing out what's wrong with the question (and the "what's wrong" part usually reveals almost exactly what the person wants to know anyway.)

At that point most questioners just back down -- and this repeated tone is what keeps me away from these forums. I just hate seeing it.

And so ***my hearty congratulations*** Drvrm for the intelligent, caring, and human way you have addressed this question. Let it be a standard!

Andreas C and sophiecentaur
Papa Doyle
I registered just to make that point -- never felt compelled to register for these forums in the several years I've been visiting. Thank you drvrm whomever you are. I think I'd like you even if you smelled bad.

Andreas C
Thank you everyone for the great replies! I have one more question now: how do I determine 1) the shape of an ellipse and 2) the time one revolution takes if I know the tangential velocity at a given point in the orbit and the gravitational acceleration?

Gold Member
I think you should find what you need in this link - or the one it links to, for the orbital equation.

Andreas C
I think you should find what you need in this link - or the one it links to, for the orbital equation.

Oh OK, thank you!

Jenab2
I'll assume, as I usually do, that this question refers to an object in orbit around the sun.

G = the gravitational constant = G = 6.67384e-11 m³ kg⁻¹ sec⁻²
M = the combined mass of the sun and of the orbiting body, kilograms.*
a = the semimajor axis of the elliptical orbit, meters.
e = the eccentricity of the elliptical orbit

*The mass of the sun is 1.98855e30 kilograms.

Define a constant, k, which is a speed in meters per second.
k = √{ GM / [ a (1 − e²) ] }

The instantaneous velocity vector of the orbiting body, MKS units, referred to the canonical coordinate system (sun at origin, orbit in the XY plane, the orbiting body's perihelion on +x axis) is [Vx''', Vy''', Vz'''].

Vx''' = −k sin θ
Vy''' = k (e + cos θ)
Vz''' = 0

And the speed in the orbit is simply

v = √[(Vx''')² + (Vy''')² + (Vz''')²]

Which can be checked with the Vis Viva equation:

r = a (1−e²) / (1 + e cos θ)
v = √[GM(2/r−1/a)]

For the same instant of time, the position of the object in rectangular canonical coordinates is found from

x = r cos θ
y = r sin θ
z = 0

Where the angle θ is the true anomaly of the orbiting body at that moment. The true anomaly is the angle, subtended at the sun, from a ray extending toward the perihelion of the orbit and to another ray extending toward the current position of the body in that orbit. It is measured in the plane of the orbit in the angular direction of motion.

As time passes, θ increases. In a circular orbit (e=0), θ would increase at a constant rate, whereas the separation between the sun and the orbiting body would not change. However, in an elliptical orbit (0<e<1), θ increases at a variable rate, faster when the orbiting body is near perihelion and slower when it is near aphelion. Furthermore, the separation between the primary and the orbiting body is likewise variable, though inversely to the speed in the orbit.

The constant thing is the angular momentum per unit mass. That's what causes a line from the sun to the orbiting body to sweep out equal areas in equal times.

If you want to transform the canonical velocity to velocity in heliocentric ecliptic coordinates, do this:

Rotate the triple-prime velocity vector by the argument of the perihelion, ω.

Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz''' = 0

Rotate the double-prime velocity vector by the inclination, i.

Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i

Rotate the single-prime velocity vector by the longitude of the ascending node, Ω.

Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'

The unprimed velocity vector [Vx,Vy,Vz] is the sun-relative velocity in ecliptic coordinates.

The average speed in the orbit is found by dividing the circumference of the orbit by the orbital period. The period, P, of an orbit is found from

P = 2π √[a³/(GM)]

The circumference, C, of an elliptical orbit is found by solving, numerically, this elliptical integral of the 2nd kind:

C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ

Making the average speed in orbit

vₐ = (2/π) √(GM/a) ∫(0,π/2) √(1−e²sin²θ) dθ

Note that this is NOT the same as the speed in the orbit when r=a, except when the orbit is circular (e=0). A lot of astronomy professors don't seem to know this!

v(r=a) = √(GM/a)
vₐ = √(GM/a) (2/π) ∫(0,π/2) √(1−e²sin²θ) dθ

vₐ ≠ v(r=a), in general, though it's usually close.

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Jenab2
You might want to discover the true anomaly as a function of time, so that you can input time, output true anomaly, and then use the math in my previous comment to find the heliocentric ecliptic position and the sun-relative velocity at a particular time.

The period of the orbit, P.

P = 365.256898326 a^(1.5)

The mean anomaly, m.

m''' = (t−T)/P
m'' = m''' − integer(m''')
If m'' is less than 0, then m'=m''+1 else m'=m''
m = 2πm'

where T is the time of perihelion passage and t is the time of interest (or observation).

The eccentric anomaly, u.

u₀ = m + (e − e³/8 + e⁵/192) sin(m) + (e²/2 − e⁴/6) sin(2m) + (3e³/8 − 27e⁵/128) sin(3m) + (e⁴/3) sin(4m)

i = 0

Repeat
i = i+1
E = uᵢ − e sin uᵢ − m
F = 1 − e cos uᵢ
G = e sin uᵢ
H = e cos uᵢ
A = −E/F
B = −E/(F + ½ AG)
C = −E/(F + ½ AG + ⅙ B²H)
uᵢ₊₁ = uᵢ + C
Until |uᵢ₊₁−uᵢ| < 1e-12

u = uᵢ₊₁

The canonical (triple-prime) heliocentric position vector.

x''' = a (cos u − e)
y''' = a sin u √(1− e²)
z''' = 0

(I forgot to put the triple prime on the canonical position vector in my previous post, and the software won't let me edit that post now. It should read...

For the same instant of time, the position of the object in rectangular canonical coordinates is found from

x''' = r cos θ
y''' = r sin θ
z''' = 0

Instead of lacking the ''' marks.)

The true anomaly, θ.

θ = arctan( y''' , x''' )

Notice that you can get the canonical position vector [x''', y''', z'''] directly from the eccentric anomaly, u, and then use y''' and x''' to get the true anomaly, θ. When you know what the true anomaly is, you can use it to find the sun-relative velocity at time t.

The arctan( y''', x''' ) is the two-dimensional arctangent function, related to the 1-dimension version of the function as follows:

arctan( y''' , x''' ) = atan(y'''/x''') + Q
if x'''>0 and y'''>0 then Q=0
if x'''<0 then Q=π radians
if x'''>0 and y'''<0 then Q=2π radians

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Jim60
For ease of use, wouldn’t these calculations be better place in an Excel spreadsheet, or some other math program?

Jenab2
For ease of use, wouldn’t these calculations be better place in an Excel spreadsheet, or some other math program?
I've written programs based on the math in that post. I'm not familiar with Excel, though. I can program in several versions of BASIC for the PC, in Pascal, in FORTRAN, in Python 3.x, in the calculator coding languages for the TI-89 and the Casio fx-9860gii. I haven't, as best as I can remember, ever integrated a calculating program into a web page. I've seen them, e.g., units conversion pages, a page to evaluate the Bessel function of real order N and argument x, and so on. But I haven't created any such pages myself.

Gold Member
...I haven't, as best as I can remember, ever integrated a calculating program into a web page. I've seen them, e.g., units conversion pages, a page to evaluate the Bessel function of real order N and argument x, and so on. But I haven't created any such pages myself.
I created a web page to help others create such web pages. It's Javascript with my own extended library.

Here's your math from post #21: http://orbitsimulator.com/code/tdunn/code5.html?Jenab2_01.txt
I haven't double-checked the math against known numbers, and I may have made mistakes, but at first glance, it looks good.

In addition to straight-forward Javascript, I've made the following additions (click "Help" on the webpage to see all of them)
G = 0.00000000006672
Ms = the mass of the Sun (1.98891691172468E+30)
AU = 149597870691
pi = 3.14159265358979
trigMode = 'degrees'; no need to convert your trig to radians after this line.
pythag3(x, y, z) compute the magnitude of a 3-component vector
cls(); clears the output window
print ("text");
print (variable);
print("text" + variable);

Press the "Run" button to execute the code.
If you make modifications, you can "Save As" under a new name.
To make a link to your saved page do this:
Assume you Saved As with a filename of "abcd" (no quotes).
Your URL will be http://orbitsimulator.com/code/tdunn/code5.html?abcd.txt
As a precaution, also save your code as a text file on your own computer, in case I decide to pull the plug on this project.

Jenab2
θ
trigMode = 'degrees'; no need to convert your trig to radians after this line.
While I can't be certain of it, lacking familiarity with Java, it may be that it will still be necessary to change angles from degrees into radians. The reason is that some of the equations are transcendental, having the variable to be solved for as both linear terms and as arguments of trigonometric functions. Kepler's equation:

m = u − e sin u

...is perhaps the best-known example, when given m and e, solving for u.

// user inputs
M = Ms; //mass equals mass of Sun
a = 1 * AU; // semi-major axis in AU
e = 0; // eccentricity
theta = 90; // mean anomoly?
omega = 100; // argument of periapsis
Omega = 120; // longitude of ascending node;
i = 10; // inclination

// do the math
trigMode = 'degrees';
k = sqrt(G*M / (a * (1 - e*e)));
Vxppp = -k * sin(theta);
Vyppp = k * (e + cos(theta));
Vzppp = 0;
v = pythag3(Vxppp, Vyppp, Vzppp);
From the quote above:
theta = 90; // mean anomoly?

Theta (θ) in my equations is the true anomaly.

Gold Member
The program uses Javascript, not Java. They're not related.

trigMode = 'degrees'
This isn't part of Javascript. I added this.
I also added the trig and inverse functions. Javascript uses Math.sin(arg).
They work like this:

Code:
function sin(arg) {
if (trigMode == 'degrees') {arg = arg * pi / 180}
return Math.sin(arg);
}
same for cos(arg), tan(arg)
Code:
function asin(arg) {
var a = Math.asin(arg);
if (trigMode == 'degrees') {a = a * 180 / pi}
return a;
}
same for acos(arg), atan(arg), atan2(arg,arg)

Thanks, I fixed the comment
theta = 90; // true anomaly

Jenab2
Check then.

Given:
mean anomaly, m = 1.0 radian
eccentricity, e = 0.5

Find the eccentric anomaly, u.

m = u − e sin u

u₀ = 1.501832843888963 ← Danby's Fourier series gives a pretty good first approximation.
u₁ = 1.498701135566615
u₂ = 1.498701133517848
u₃ = 1.498701133517848

*converged to 16 significant figures

1 radian = 57.295779513082320876798154814105°

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