How Can Kepler's 3rd Law and Related Equations Be Applied to Orbital Mechanics?

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SUMMARY

This discussion focuses on the application of Kepler's 3rd Law and related equations in orbital mechanics. The key equation, P^2 = (4π^2a^3)/(GM), illustrates the relationship between the orbital period (P) and the semi-major axis (a) for a one-body problem, where M is the mass of the sun. The two-body problem is represented by P^2 = (4π^2a^3)/(G(M+m)), where m is the mass of the orbiting body, which can be neglected if m << M. Additionally, the discussion covers velocity equations for elliptical orbits, including those at perihelion and aphelion, and the general escape velocity equation, v = √(2GM/r).

PREREQUISITES
  • Understanding of Kepler's Laws of Planetary Motion
  • Familiarity with gravitational equations and constants (G, M)
  • Knowledge of elliptical orbits and their parameters (semi-major axis, eccentricity)
  • Basic principles of classical mechanics, particularly energy conservation
NEXT STEPS
  • Study the derivation of Kepler's 3rd Law in detail
  • Learn how to calculate orbital velocities at different points in an elliptical orbit
  • Explore the implications of the two-body problem in celestial mechanics
  • Investigate the concept of escape velocity and its applications in orbital dynamics
USEFUL FOR

Astronomy students, astrophysicists, and anyone interested in understanding the mechanics of celestial bodies and their orbits will benefit from this discussion.

forestmine
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Homework Statement



I don't have any particular problems I need help with. I'm just trying to wrap my head around some of the equations and their meanings. Any clarification would be great!

We have from Kepler's 3rd Law the following:

The period squared is proportional to the semi-major axis cubed.
Or,

P^2 = ((a^3)4*pi^2)/GM

and this equation in particular is the one body problem, where the mass of the body rotating around the sun is negligible in comparison to the sun, so M is the mass of the sun? If so, what's negligible mass?

And the two body problem would be the same thing, but,

((a^3)4*pi^2)/G(M+m)

Would the two-body problem apply for a planet's elliptical orbit around the sun?

From there, there is the equation for velocity along an elliptical orbit:

v^2 = GM (2/r - 1/a)

M in this case is the mass of the particular body whose velocity we are calculating? Say, the mass of the Earth?

And then there are these equations, which I'm a bit confused about.

Velocity at perihelion: v^2 = (GM/a)(1+e/1-e)

Velocity at aphelion: v^2 = (GM/a)(1-e/1+e)

So, if I'm given the eccentricity of the ellipse, I ought to use one of the last two equations, depending on whether it's at perihelion or aphelion? If a specific eccentricity is not given, I'm assuming I can use the equation v^2 = GM (2/r - 1/a), where r is a specified distance of either perihelion or aphelion?

And one last question. If the general equation for escape velocity is given by:

v=(2MG/r)^1/2

can I apply it in this form for elliptical orbits? If so, is M in this case again the mass of the body from which a particle or object is attempting to escape, and r is the distance out to that object?

Hopefully this is somewhat clear!

Thanks in advance!
 
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Can anyone help clear this up?
 
This is a result that comes straight out of the virial theorem or mechanical similarity, I can never remember the names of these things.
It can be used for any 1/r potential, or 1/r^2 force law system!

It states that for a potential that goes as r^k the time between any two points separated by a length d on the trajectory are like
t \approx d^{1-\frac{k}{2}}
 
forestmine said:

Homework Statement



I don't have any particular problems I need help with. I'm just trying to wrap my head around some of the equations and their meanings. Any clarification would be great!

We have from Kepler's 3rd Law the following:

The period squared is proportional to the semi-major axis cubed.
Or,

P^2 = ((a^3)4*pi^2)/GM

and this equation in particular is the one body problem, where the mass of the body rotating around the sun is negligible in comparison to the sun, so M is the mass of the sun? If so, what's negligible mass?

And the two body problem would be the same thing, but,

((a^3)4*pi^2)/G(M+m)

Would the two-body problem apply for a planet's elliptical orbit around the sun?

If m << M (which is what's meant by the mass of the planet being negligible compared to the mass of the star) then you can replace m+M with just M (i.e. the total mass of the system is dominated by the mass of the star). Since this is totally the case in our solar system, the first equation would work just fine. That's useful, because it means that if you can measure the period and semi-major axis of some satellite object that is much smaller than the thing that it is orbiting, then you can figure out, to a good approximation, the mass of the thing that it is orbiting.


forestmine said:
From there, there is the equation for velocity along an elliptical orbit:

v^2 = GM (2/r - 1/a)

M in this case is the mass of the particular body whose velocity we are calculating? Say, the mass of the Earth?

And then there are these equations, which I'm a bit confused about.

Velocity at perihelion: v^2 = (GM/a)(1+e/1-e)

Velocity at aphelion: v^2 = (GM/a)(1-e/1+e)

So, if I'm given the eccentricity of the ellipse, I ought to use one of the last two equations, depending on whether it's at perihelion or aphelion? If a specific eccentricity is not given, I'm assuming I can use the equation v^2 = GM (2/r - 1/a), where r is a specified distance of either perihelion or aphelion?

I'm sure that if you plugged in the expression for the distance of aphelion or perihelion (in terms of eccentricity) into "r" in the first equation, then you would be able to derive the last two equations. Try it.

forestmine said:
And one last question. If the general equation for escape velocity is given by:

v=(2MG/r)^1/2

can I apply it in this form for elliptical orbits? If so, is M in this case again the mass of the body from which a particle or object is attempting to escape, and r is the distance out to that object?

Hopefully this is somewhat clear!

Thanks in advance!

As you yourself have stated, this is the general equation for escape velocity. This means that it applies generally. Yes, of course M is the mass of the body being escaped from, and r is the starting distance of the object that is trying to escape. You can see that easily if you just derive the expression for the escape velocity.
 
Thanks for the replies.

I apologize for perhaps missing some of the more obvious things. I jumped into this higher level astronomy class before having taken any introductory courses, and so derivations are often skipped over, assuming we should already know those. I'll have to work through them myself.

I follow everything except:

"I'm sure that if you plugged in the expression for the distance of aphelion or perihelion (in terms of eccentricity) into "r" in the first equation, then you would be able to derive the last two equations. Try it."

Am I correct in that the expression for the distance of perihelion is r_p=a(1-e) and for aphelion, r_a=a(1+e).

Plugging one of these expressions into the expression for velocity along an elliptical orbit, I'm having trouble with the simple derivation.

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

v^2 = GM [(2a - 1+e)/a(1-e)]

And from there I'm not sure how to arrive at the equation for velocity at perihelion?
(I realize I should have posted this in the general astronomy section...sorry!)
 
forestmine said:
I follow everything except:

"I'm sure that if you plugged in the expression for the distance of aphelion or perihelion (in terms of eccentricity) into "r" in the first equation, then you would be able to derive the last two equations. Try it."

Am I correct in that the expression for the distance of perihelion is r_p=a(1-e) and for aphelion, r_a=a(1+e).

Plugging one of these expressions into the expression for velocity along an elliptical orbit, I'm having trouble with the simple derivation.

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

v^2 = GM [(2a - 1+e)/a(1-e)]

And from there I'm not sure how to arrive at the equation for velocity at perihelion?
(I realize I should have posted this in the general astronomy section...sorry!)

I think that you just made a slight error when taking a(1-e) as the common denominator for the two fractions you're adding. The numerator of the first fraction remains unchanged, since its denominator is *already* a(1-e). Therefore, your expression on top should just be (2 - 1 + e), NOT (2a - 1 + e).

To derive escape velocity, set the kinetic energy of the object equal to its gravitational potential energy. Then solve for the speed, v. The reason for this is that the object is trapped in a "well" of depth -GM/r, (which tops out at 0 energy at infinity). In order to "escape" to infinity, it must have at least this amount of kinetic energy to start with (enough to make it to the top of the well). Otherwise, it will reach zero speed at a finite distance away (i.e. it will still be partially within the well -- its kinetic energy will have been entirely converted to potential energy at a level that is still below zero). Then it will begin to fall back in (albeit very very slowly if it makes it quite far out).
 
Ok, it's all much clearer to me now.

Thank you for the help! I really appreciate it.
 

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