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

• forestmine
In summary: 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 liket \approx d^{1-\frac{k}{2}}
forestmine

## 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!

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!

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'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'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.

## 1. What is Kepler's 3rd Law?

Kepler's 3rd Law, also known as the Law of Harmonies, is a mathematical equation that describes the relationship between the orbital period and the average distance of a planet from the sun.

## 2. How is Kepler's 3rd Law calculated?

To calculate Kepler's 3rd Law, you need to know the orbital period (P) of a planet and its average distance from the sun (a). The equation is P^2 = a^3, where P is measured in years and a is measured in astronomical units (AU).

## 3. Why is Kepler's 3rd Law important?

Kepler's 3rd Law is important because it allows us to accurately predict the orbital period of a planet based on its distance from the sun. This law also helped to confirm the heliocentric model of the solar system proposed by Copernicus.

## 4. How did Kepler discover this law?

Johannes Kepler discovered this law by analyzing the data collected by his mentor, Tycho Brahe, who meticulously recorded the positions of the planets in the sky. Kepler noticed a pattern in the orbital periods and distances of the planets, leading to the development of his 3rd Law.

## 5. Does Kepler's 3rd Law apply to all planets in the solar system?

Yes, Kepler's 3rd Law applies to all planets in the solar system, as well as other celestial objects that orbit the sun, such as comets and asteroids. However, it may not be applicable to objects orbiting other stars or in other galaxies.

Replies
7
Views
1K
Replies
3
Views
2K
Replies
3
Views
1K
Replies
7
Views
1K
Replies
3
Views
1K
Replies
16
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
7
Views
456
Replies
3
Views
2K