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Roger's Equations

This blog features weekly an equation, formula, or constant that occurs frequently in Engineering or Science. I will try to present the subject matter in a nonformal, conversational style that can be easily followed. Criticism and corrections are encouraged, as are suggestions for future discussions.

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4 comments

Lagrange's Equations and the Central Potential - Part II

Posted November 14, 2006 5:51 PM by Roger Pink
User-tagged by 1 user

Last time we derived the equation for an orbit about a Kepler like central potential (k/r) in terms of radius (r) and angle (θ) starting from the Lagragian for the system.

L=T - U

where T is the total kinetic energy of the system and U is the total potential energy for the system. This produced a Lagrangian that looked like;

which we then solved (in many steps) to get;

where u=1/r, r is the distance from the center of the potential, e is the eccentricity of the orbit, θ-θ0 is an angular distance travelled in that orbit, L is the angular momentum, m is the mass, and k is the gravitational constant.

This entry I'd like to see if we can draw some conclusions regarding orbits from this equation and others that came up during our derivation in Part I.

Taking another look at the Lagragian, if we were to add U (potential energy) to T (kinetic energy), rather than subtract it as we do in the Lagragian, we would have an expression for the total energy of the system that looked like this;

we say the value is constant because we've accounted for the energy of the system and we are ignoring outside perturbations. Since Energy is conserved, T + U is as well.

The equation for energy above becomes;

E= 1/2m(dr/dt)2 + (L2/2mr2) - k/r

where L comes from our derivation in Part I in which we show;

L = (mr2(/dt))

For convenience, lets say

Ueff= (L2/2mr2) - k/r)

giving us

.

where the r-dot in the equation above is (dr/dt). If we take the equation;

dθ=(/dt)(dt/dr)dr

which is just a complicated way of saying dθ=dθ,

and since dt/dr = 1/(dr/dt) and we know from Part I that /dt = L/mr2, we can substitute in our dθ=dθ equation above and get the complicated expression;

θ-θ0 = ± ∫ (2m(E+(k/r)-(L2/2mr2)))-1/2 (1/r2) dr

the left hand side of the equation dθ has been integrated producing θ-θ0 . The equation above is a complicated integral for θ in terms of r. Skipping straight to the solution of the integral above, and reorganizing in a particular way that I will explain in a second, we get;

α/r = 1 + (1 + (2EL2/mk2))1/2 cos(θ-θ0)

where α = L2/mk. This is the solution of the integral above, reorganized (with α/r on the right) so that we can see that it is in the form of an equation for a conic section with a focus at the origin;

α/r = 1 + ε Cos(θ-θ0)

where ε is the eccentricity of the conic section. So we now have an equation that relates energy and eccentricity of an orbit;

ε = (1+ 2EL2/mk2)1/2

There are four types of conic sections;

so there will be four types of orbits, and because the conic sections are described by eccentricity, and we've related energy to eccentricity, we can predict the orbit shape based on energy, see below;

If E>0 .................................then ε>1.................Hyperbolic Orbit
If E=0 .................................then ε=1................ Parabolic Orbit
Umin<E<0 ............................then 0<ε<1..............Elliptical Orbit
E = Umin ......................................then ε = 0 ...............Circular Orbit

If your wondering what Umin is, its the minimum value of Ueff which I introduced earlier in the equation;

Ueff= (L2/2mr2) - k/r)

See the diagram below with the orbit types;

It's worth noting that circular orbits are just a special kind of elliptical orbit where the minor and major axis are equal. With that in mind, notice in the diagram above that only elliptical orbits are "closed" which is to say they repeat themselves. Hyperbolic and Parabolic orbits are what we think of as deflections or scattering. So the four situations listed above are a complete set of descriptions of the possible orbits, which was derived using only energy considerations. Taking a closer look at the closed orbits (Circular and Elliptical), we can use the geometry of an ellipse to give us insight as to how Energy effects the shape of a closed orbit.

An ellipse has a major and minor axis. The major axis by definition is;

2a = rmin + rmax

which, using the equation above for radius is equal to;

2a=α/(1+ε) + α/(1 - ε)

Since the maximum value for cos θ is +1, which corresponds to rmin and the smallest value for cos θ is -1 which corresponds to rmax. This is equal to;

2α/(1+ε)2 = k/E

which means that the length of the major axis is the strength of the gravitational field divided by the energy of the orbiting object. The higher the energy of an orbit, the more elliptical it becomes. A circular orbit is the lowest energy orbit you can have for a given rmin (distance from the source of the potential). In the diagram below, the elliptical orbit is the more energetic one. In fact, you could create the elliptical orbit from the circular one by applying thrust (kinetic energy). The location of where the thrust was fired would be the rmin of the new higher energy elliptical orbit. The eccentricity of the new elliptical orbit will simply depend on how much energy has been added by the thrust, see the diagram below;

In fact, this is how space travel has been done for all our space prob missions like Voyager 1 and 2 etc. See the diagram below;

Hopefully you can see in the diagram above that as voyager 1 and 2 each moved from planet to planet, the orbits were elliptical. When the probes would get to a new planet, gravity assist, which adds (or removes) energy from the probe, was used to change the direction and eccentricity of the old orbit in order to point it at a new planet. The Cassini probe that went to Saturn illustrates this idea as well;

Anyway, I think that's enough, though there is so much more that can be derived from the Lagrangian. You can show the area swept per time in an elliptical orbit is constant and the relation for the periods of planets. Orbital velocity, escape velocity, orbital energy, escape energy, lagrange points, etc. Maybe in the future, the far future, I think I'm set with orbits for a while ;)

Till next time.

Special thanks to the following websites;

http://scienceworld.wolfram.com/physics/CentralOrbit.html
http://farside.ph.utexas.edu/teaching/336k/lectures/node80.html
http://www.worldforge.org/project/newsletters/July2002/LagrangianP3
http://electron9.phys.utk.edu/phys513/Modules/module6/motion_in_a_central_potential.htm
http://en.wikipedia.org/wiki/Wiki


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#1

Re: Lagrange's Equations and the Central Potential - Part II

11/15/2006 3:34 AM

Hi Roger, brilliant post again! Just a minor oversight - you wrote: "When the probes would get to a new planet, thrust was fired to change the direction and eccentricity of the old orbit in order to point it at a new planet."

When reaching intermediate planets, NASA uses gravity assist to change the orbital energy relative to the Sun and not thrust. Cassini in fact used two Venus swingbys, one Earth and one Jupiter swingby to gather enough energy to reach Saturn. There they used a combination of thrust and "reverse gravity assist" to put the craft into orbit.

Thrust is normally only used to make minor course corrections to achieve precise swingby distances.

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#2
In reply to #1

Re: Lagrange's Equations and the Central Potential - Part II

11/15/2006 9:43 AM

Jorrie,

You're right, the energy of the spacecraft is increased in flybys by stealing some of the planet's angular momentum which increases the speed and thus the energy of the probe. No thrust is necessary during the flyby, only along the way to make course corrections (a lot of small things out there tugging on the spacecraft). I'll fix it in the entry above.

I found a picture of the orbits of Cassini when it got to Saturn, pretty cool.

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#3
In reply to #1

Re: Lagrange's Equations and the Central Potential - Part II

11/16/2006 6:50 AM

Hi Jorrie,

There is just one thing I fail to see about gravity assist when people say e.g. "Cassini will increase its speed to 60,000 kph after getting some gravity assist from Jupiter, and follow its way to, say Neptune."

There can be a gravity assist only when approaching. But after you pass, this is gravity loading, isn't it? Gravity is a conservative field. You just lose some potential energy and gain some kinetic, but not for long as you go deep in the universe.

So, I think gravity assist is misleadingly used as adding net energy, but it is actually used just for just changing direction, right? Or, are there some tricks?

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#4
In reply to #3

Re: Lagrange's Equations and the Central Potential - Part II

11/16/2006 7:52 AM

Hi Cameo, you wrote: "So, I think gravity assist is misleadingly used as adding net energy, but it is actually used just for just changing direction, right? Or, are there some tricks?"

The trick lies in the fact that the planet is in orbit around the Sun and not moving freely in empty space. It is possible to make the spacecraft free-fall towards the planet for a longer time than what it takes to reach the same distance again going away from the planet.

I think you will agree that if that is accomplished, the craft has 'stolen' some kinetic energy from the planet! This is the essence behind gravity assist.

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