Planetary orbits - the 2-body problem

[dyn]

Hi

I am confused about certain aspects of deriving the planetary orbit equation by considering it as 2-body problem. I will ask my first question now before i get to my other questions. In the David Tong notes on "Dynamics & Relativity" he states that a particle in central force potential obeys
ma = -∇ V(r)
He then states that this can be interpreted as 2 particles with separation r interacting through the inter-particle potential V. The origin r = 0 is the centre of mass of the 2 particles. Also m is the reduced mass of the 2 particles.
My question is why is r=0 the centre of mass position ? The vector r starts at one mass and ends at the other one so surely r = 0 corresponds to the position of one of the masses ? Also when calculating gravitational potentials the origin r=0 always seems to be the position of the particle ( or centre of a spherical mass)

Thanks

 

[dextercioby]

Well, D. Tong is assuming that some important details are known. You have two particles of masses $m_1, m_2$ and coordinate vectors describing their instantaneous position in the lab frame $\bf{r_1}(t), \bf{r_2}(t)$. The 2-body electrostatic/gravitostatic potential is $V(\vert \bf{r}_1 - \bf{r}_2\vert)$.
Then, in order to solve the coupled system of 6 ODEs from Newton's second law applied for each particle, you need to make the so-called "separation of motion" into the motion of the CoM and the motion of a virtual particle of reduced mass around the CoM. They, for simplicity, because the CoM is an IRF, you shift the description from the lab system to the CoM system and the originally coupled system of 6 ODEs transforms to uncoupled ODEs.

More to read here: https://en.wikipedia.org/wiki/Two-body_problem#Reduction_to_two_independent,_one-body_problems (off the top of my head I cannot pinpoint a textbook of classical mechanics providing all the possible details and calculations).

Later Edit: down below one finds a full treatment by the user vanhees71, so no need to look it up in a textbook of classical mechanics.

 

[dyn]

I have looked at those wikipedia notes and they are similar to mine. I don't understand how r = 0 is the position of the COM.
The equation is μ$\ddot{r}$ = F(r) where F is gravitational force between the 2 masses. For this force r=0 is at the position of one of the masses or centre of it if it is a sphere ; r=0 is not the position of the centre of mass. The variable r is the separation of the 2 masses ; this is not related to the COM

 

[PeroK]

When $r=0$, the separation of the masses is $0$, which means the masses are at the same place, namely the common COM.

In an orbital scenario $r=0$ never happens. It only happens in a direct collision.

 

[Ibix]

It's a mathematical trick, basically. It turns out that the path of mass $m_1$ in a two body system is the same as a mass $m_1m_2/(m_1+m_2)$ in orbit around a mass $m_1+m_2$ pinned (don't ask what it's pinned to) at the center of mass. That's a much easier system to handle mathematically.

I haven't looked at Tong's notes, but the Wiki article linked above is proving that result.

 

[dyn]

I think I'm getting confused between 2 real masses and the "imaginary" mass with reduced mass. When i arrive at the orbit equation

r = r₀ / ( 1 + εcosθ )

where ε is the eccentricity of the orbit ; i am assuming r is the distance from the COM located at the focus of the ellipse to one of the masses ? This means that in the general case where the COM is not located at one of the masses ; the r in the orbit equation is not the distance between the 2 masses . Is that correct ?

 

[pbuk]

the r in the orbit equation is not the distance between the 2 masses . Is that correct ?

That is correct: the r in the orbit equation for m1 is the distance between m1 and the CoM.

Edit: this is not correct for the reduced mass method as noted by the OP below

 

[dyn]

I just found in the David Morin Classical Mechanics book that the r in the orbit equation using reduced mass is the distance between the 2 masses. To find the distance between each mass and the COM use (m₂/M)r for mass m₁ and (m₁/M)r for mass m₂ where M=m₁+m₂

 

[hutchphd]

So does that clear it up for you?

 

[dyn]

I have one last question. Using the reduced mass equation, someone on the Earth could consider that the Sun is orbiting around the Earth while someone "on" the Sun could consider that the Earth is orbiting around the Sun ; is that argument valid ? If so , why is it considered that the Earth orbits the Sun ? Is it because the COM of the Earth-Sun system is inside the Sun ?

 

[hutchphd]

Well if you have to choose one or the other, that would certainly make sense. The sun outmasses the Earth by about 333,000 so it will be 279 miles from the sun center (is that right?).

 

[PeroK]

I have one last question. Using the reduced mass equation, someone on the Earth could consider that the Sun is orbiting around the Earth while someone "on" the Sun could consider that the Earth is orbiting around the Sun ; is that argument valid ? If so , why is it considered that the Earth orbits the Sun ? Is it because the COM of the Earth-Sun system is inside the Sun ?

In the simplest model of the solar system the Sun is fixed at the centre. At the next level of complexity, the Sun moves about by approximately its diameter, from the influence of the other planets, most significantly Jupiter.

In any case, in reality the Sun-Earth is not an isolated two_body problem.

Moreover, the alternative model is that the Sun orbits the Earth daily, not annually, due to the rotation of the Earth.

 

[vanhees71]

Let's solve the equations. That's easier than a lot of text!

The equations of motion for the Sun (position $\vec{x}_1$) and the Planet (position $\vec{x}_2$) are
$$
\begin{split}
m_1 \ddot{\vec{x}}_1 &=-\frac{G m_1 m_2}{|\vec{x}_1-\vec{x}_2|^3}(\vec{x}_1-\vec{x}_2),\\
m_2 \ddot{\vec{x}}_2 &=-\frac{G m_1 m_2}{|\vec{x}_1-\vec{x}_2|^3}(\vec{x}_2-\vec{x}_1)
\end{split}$$
Adding both equations leads to
$$M \ddot{\vec{R}}=0$$
with
$$\vec{R}=\frac{1}{M} (m_1 \vec{x}_1+m_2 \vec{x}_2), \quad M=m_1+m_2.$$
Multiplying the 2nd EoM. with $m_1/M$ and the 1st with $m_2/M$ and subtracting both equations leads to
$$\mu \ddot{\vec{r}}=-\frac{G m_1 m_2}{r^3} \vec{r},$$
where
$$\mu=\frac{m_1 m_2}{M}, \quad \vec{r}=\vec{r}_2-\vec{r}_1.$$
So we have reduced the equation of motion for the two-body problem to the equation of motion for its center of mass, which moves with constant velocity, and the equation of motion for a quasiparticle with mass $\mu$ ("reduced mass") with a force given by the gravitational interaction between Sun and planet. We can choose our inertial reference frame such that it stays at rest and thus $\vec{R}=\vec{0}=\text{const}$ is the origin of this new inertial reference frame.

For the relative motion you can use the other conservation laws. Since the force is a central force, angular momentum is conserved, i.e.,
$$\vec{L}=\mu \vec{r} \times \dot{\vec{r}}=\text{const}$$
and we can choose the reference frame such that $\vec{L}=L\vec{e}_3$. Then the motion is entirely in the $x_1$-$x_2$ plane. The conservation of angular momentum is also Kepler's 2nd Law according to which the $\vec{r}$ swipes out equal areas in equal times.

Further since the force has a time-independent potential
$$V(r)=-\frac{G m_1 m_2}{r}$$
the energy is conserved.
$$E=\frac{\mu}{2} \dot{\vec{r}}^2-\frac{G m_1 m_2}{r}=\text{const}.$$
Now we introduce polar coordinates
$$\vec{r}=r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix}$$
leading to
$$\dot{\vec{r}}^2=\dot{r}^2 + r^2 \dot{\varphi}^2$$
and
$$L=\mu (x_1 \dot{x}_2-x_2 \dot{x}_1)=\mu r^2 \dot{\varphi}=\text{const}$$
Then the energy-conservation equation reads
$$E=\frac{\mu}{2} \dot{r}^2 + \frac{L^2}{2 \mu r^2} - \frac{G m_1 m_2}{r}.$$
That's the equation of motion for a one-dimensional problem of a particle with mass $\mu$ moving in an effective potential with
$$V_{\text{eff}}=\frac{L^2}{2 \mu r^2} - \frac{G m_1 m_2}{r}.$$
To find the shape of the orbit we note that
$$\dot{r}=r'(\varphi) \dot{\varphi}=\frac{L}{\mu r^2} r'(\varphi)=-\frac{L}{\mu} \left (\frac{1}{r} \right)',$$
where a prime now means a derivative wrt. $\varphi$. Introducing $s=1/r$ the energy-conservation equation reads
$$E=\frac{L^2}{2 \mu}(s^{\prime 2}+s^2)-G m_1 m_2 s.$$
To solve this differential equation, it's easier to first take another derivative wrt. $\varphi$, giving
$$s' [\frac{L^2}{\mu} (s''+s)-G m_1 m_2]=0.$$
Now either $s'=0$, which means $s=1/r=\text{const}$, in which case the orbit is a circle, or
$$s''+s=\frac{G \mu m_1 m_2}{L^2}.$$
The general solution of this equation obviously is
$$s=\frac{G \mu m_1 m_2}{L^2} + C \cos(\varphi+\varphi_0),$$
where $C$ and $\varphi$ are integration constants. We can choose our coordinate system such that $s$ becomes maximal ($r$ minimal) for $\varphi=0$, i.e., we can set $\varphi_0=0$, which corresponds to the usual convention of the astronomers to count the angle relative to the perihelion (closest distance between planet and Sun).

Further $s'=-C \sin \varphi$ and thus using once more the energy-conservation equation above:
$$C=\sqrt{1+\frac{2 \mu E}{L^2}} \frac{G \mu m_1 m_2}{L^2}.$$
Finally the orbit reads
$$r=\frac{1}{s} = \frac{L^2}{G \mu m_1 m_2} \, \frac{1}{1+ \epsilon \cos \varphi}$$
with
$$\epsilon=\sqrt{1+\frac{2 \mu E}{L^2}}.$$
Obviously the orbit is a conic section. The bound orbits are obviously for $E<0$, i.e., $0 \leq \epsilon<1$, ellipses. This is Kepler's 1st Law: Both the Sun and the planet are on elliptic orbits with the center of mass as one of its foci.

 

[dyn]

Thanks for all your replies

 

[Richard R Richard]

I think I'm getting confused between 2 real masses and the "imaginary" mass with reduced mass. When i arrive at the orbit equation

r = r₀ / ( 1 + εcosθ )

where ε is the eccentricity of the orbit ; i am assuming r is the distance from the COM located at the focus of the ellipse to one of the masses ? This means that in the general case where the COM is not located at one of the masses ; the r in the orbit equation is not the distance between the 2 masses . Is that correct ?

The equation of the elliptical orbit from the closest focus to the central body

$$r (\theta) = \dfrac {a (1- \varepsilon ^ 2)} {1+ \varepsilon \cos \theta}$$

And from the other focus

$$r (\theta) = \dfrac {a (1- \varepsilon ^ 2)} {1- \varepsilon \cos \theta}$$

Where a is the semi-major axis of the orbit

If $r_{0} = a (1- \varepsilon ^ 2)$

Then $r_0$ is called a "straight half width", it takes relevance when in parabolic trajectories the semi-major axis $a$ is infinite, then $r_0$ is taken, to avoid the inconvenience.

Each mass orbits the CoM with an elliptical trajectory synchronous with the other mass, both have the periapsis at the same moment and the apoapsid at the same moment, as both trajectories have the same eccentricity and as the line that joins the masses contains the focus, the $r_0$ of each curve can be added obtaining the distance between masses. Thus $r$ represents in the joint orbit represents the distance between the two masses.

source: https://es.wikipedia.org/wiki/Excentricidad_orbital

This means that in the general case where the COM is not located at one of the masses ; the r in the orbit equation is not the distance between the 2 masses . Is that correct ?

It is precisely always the distance between the two masses.

But as it is preferable than doing as many on a reference system that is not rotating on a CoM, (it is preferred didactically static), when in $M = m_1 + m_2$ we have that $m_1 >> m_2$ then the CoM is very close to the center from $m_1$, then $r_1$ is approximated to $0$ without causing major errors, hence the mass $m_2$ has an elliptical orbit, whose focus is on the CoM which in this case will be the center of $m_1$ and the distance $r$ will be the radial distance based on that system centered on $m_1$

 

[hutchphd]

It is precisely always the distance between the two masses.

What? Wrong. Already answered.

 

[Richard R Richard]

What? Wrong. Already answered.

Do you think my answer is different from yours?

 

[hutchphd]

If it id the same why reiterate? Finis.

 

[Richard R Richard]

If it id the same why reiterate? Finis.

I repeat it, because, I think that the objective of a forum
is to seek different points of view on the same subject.
Do you agree or disagree with what I have written?

 

[vanhees71]

What? Wrong. Already answered.

Why? Of course $\vec{r}=\vec{r}_1-\vec{r}_2$ and thus $r=|\vec{r}|$ is the distance of the Sun to the planet.

 

[hutchphd]

Do you agree or disagree with what I have written?

Remind me not to (try to) do Physics past my bedtime. Apologies.
Very nice graphics!

 

[dyn]

The diagram is #15 is very helpful but why is the eccentricity of both masses the same ?

 

[hutchphd]

$$\vec{R}=\frac{1}{M} (m_1 \vec{x}_1+m_2 \vec{x}_2), \quad M=m_1+m_2.$$
Multiplying the 2nd EoM. with $m_1/M$ and the 1st with $m_2/M$ and subtracting both equations leads to
$$\mu \ddot{\vec{r}}=-\frac{G m_1 m_2}{r^3} \vec{r},$$
where
$$\mu=\frac{m_1 m_2}{M}, \quad \vec{r}=\vec{r}_2-\vec{r}_1.$$
So we have reduced the equation of motion for the two-body problem to the equation of motion for its center of mass, which moves with constant velocity, and the equation of motion for a quasiparticle with mass $\mu$ ("reduced mass") with a force given by the gravitational interaction between Sun and planet. We can choose our inertial reference frame such that it stays at rest and thus $\vec{R}=\vec{0}=\text{const}$ is the origin of this new inertial reference frame.

So $$\vec{x_1}=-\frac {m_2} {m_1} \vec{x_2}$$ in the $\vec R=0$ coordinate system

 

[dyn]

The above equation says that the the position vector of mass 1 relative to the COM is proportional to the vector of mass 2 relative to the COM but with a negative sign. Is that statement enough to prove that the 2 eccentricities are the same ?

 

[Richard R Richard]

The diagram is #15 is very helpful but why is the eccentricity of both masses the same ?

I understand that yes, they necessarily have to have the same eccentricity.

See what happens to the apoapsid and the periasid

Starting from the orbit of mass 1 we know that

$$ e = \dfrac {r_ {a_1} -r_ {p_1}} {r_ {a_1} + r_ {p_1}} $$ Ec1

If it is true the equation that @hutchphd gives you $\vec {x_1} = - \frac {m_2} {m_1} \vec {x_2}$

$$ r_ {a_1} = - \dfrac {m_2} {m_1} r_ {a_2} $$

And

$$ r_ {p_1} = - \dfrac {m_2} {m_1} r_ {p_2} $$

replacing in Ec1

$$ e_1 = \dfrac {- \dfrac {m_2} {m_1} r_ {a_2} - (- \dfrac {m_2} {m_1} r_ {a_2})} {- \dfrac {m_2} {m_1} r_ {a_2 } + (- \dfrac {m_2} {m_1} r_ {a_2})} $$

$$ e_1 = \dfrac {\cancel {- \dfrac {m_2} {m_1}}} {\cancel {- \dfrac {m_2} {m_1}}} \dfrac {r_ {a_2} -r_ {p_2}} { r_ {a_2} + r_ {p_2}} $$

$$ e_1 = \dfrac {r_ {a_2} -r_ {p_2}} {r_ {a_2} + r_ {p_2}} = e_2 $$

$$ e_1 = e_2 = e $$

 

[dyn]

Thank you

 

[dyn]

Can i just check that i understand the following points correctly ; i am hoping for 3 corrects !

1 - when using the reduced mass i obtain the orbit equation which is an ellipse. So the Earth orbits the Sun in an ellipse taking the Sun as fixed. Can i also take the Earth as fixed and say that the Sun orbits the Earth in an ellipse which is exactly the same size and shape as the previous ellipse. Is that correct ?

2 - Angular momentum is conserved because the force is central ; but is it correct to state that the angular momentum of the Sun about the COM is not equal to the angular momentum of the Earth about the COM ?

3 - Mechanical energy is conserved because the force is conservative ; but is it correct to state that the energy of the Sun is not equal to the energy of the Earth

 

[vanhees71]

ad 1) No, when working with the reduced mass you describe the Sun and the Earth as two bodies with finite mass, and describe the motion in terms of the center of mass position, which moves with constant velocity, such that you can choose an inertial reference frame where this center of mass position stays at rest in the origin of the coordinate system, and an equation of motion for the relative position $\vec{r}=\vec{r}_{\text{E}}-\vec{r}_{\text{S}}$. The bound solutions are ellipses (or as a special case circles).

What does this mean for the motion of the Earth and Sun? From $(\vec{R}=m_E \vec{r}_{E} + m_S \vec{r}_S)/(m_E+m_S)=0$ and $\vec{r}=\vec{r}_E-\vec{r}$ you get
$$\vec{r}_E=\frac{m_S}{m_S+m_E} \vec{r}, \quad \vec{r}_S=-\frac{m_E}{m_S+m_E} \vec{r},$$
i.e., the Earth and the Sun move on ellipses with one of its foci being the center of mass (i.e., the origin of our inertial reference frame).

Since $m_{E} \ll m_S$ you can approximate this as
$$\vec{r}_S \simeq 0, \quad \vec{r}_E \simeq \vec{r},$$
which is Kepler's 1st law in its original form: The Earth's orbit is an ellipse with the Sun in one of its foci.

ad 2) The total angular momentum is constant, because the gravitational interaction force is a central force, i.e.,
$$\vec{L}=m_E \vec{r}_E \times \dot{\vec{r}}_E+m_E \vec{r}_S \times \dot{\vec{r}}_S=\text{const}.$$
In our center-of-mass frame this reads
$$\vec{L}=\frac{m_E m_S}{m_E+m_S} \vec{r} \times \dot{\vec{r}}=\mu \vec{r} \times \dot{\vec{r}}.$$
The angular momenta of Earth and Sun are of course not equal
$$\vec{L}_E=m_E \vec{r}_E \times \dot{\vec{r}}_E = \frac{m_S}{m_E+m_S} \vec{L}, \quad \vec{L}_S = m_S \vec{r}_S \times \dot{\vec{r}}_S=\frac{m_E}{m_E+m_S} \vec{L}.$$

ad 3) I guess you mean the kinetic energy of Earth and Sun. They are
$$T_E=\frac{m_E}{2} \dot{\vec{r}}_E^2=\frac{\mu}{2} \frac{m_S}{m_E+m_S} \dot{\vec{r}}^2, \quad T_S=\frac{m_S}{2} \dot{\vec{r}}_S^2=\frac{\mu}{2} \frac{m_E}{m_E+m_S} \dot{\vec{r}}^2.$$

 

[dyn]

Can i just check that i understand the following points correctly ; i am hoping for 3 corrects !

1 - when using the reduced mass i obtain the orbit equation which is an ellipse. So the Earth orbits the Sun in an ellipse taking the Sun as fixed. Can i also take the Earth as fixed and say that the Sun orbits the Earth in an ellipse which is exactly the same size and shape as the previous ellipse. Is that correct ?

2 - Angular momentum is conserved because the force is central ; but is it correct to state that the angular momentum of the Sun about the COM is not equal to the angular momentum of the Earth about the COM ?

3 - Mechanical energy is conserved because the force is conservative ; but is it correct to state that the energy of the Sun is not equal to the energy of the Earth

Thank you.

As regards my 1st question , the Gregory book on CM states "in the 2-body problem the motion of P₁ relative to P₂ is the same as if P₂ were held fixed and P₁had the reduced mass μ instead of its actual mass" Now 1 and 2 can refer to the Sun and Earth or vice versa which is why my question referred to taking the Sun as fixed and then the Earth as fixed ; not rotating around the COM

As for Q3 i was referring to the total mechanical energy . ie. KE+PE

 

[dyn]

As regards Q2 is the angular momenta of the Sun and the Earth individually constant or is it only their sum that is constant ?

 

[vanhees71]

Thank you.

As regards my 1st question , the Gregory book on CM states "in the 2-body problem the motion of P₁ relative to P₂ is the same as if P₂ were held fixed and P₁had the reduced mass μ instead of its actual mass" Now 1 and 2 can refer to the Sun and Earth or vice versa which is why my question referred to taking the Sun as fixed and then the Earth as fixed ; not rotating around the COM

As for Q3 i was referring to the total mechanical energy . ie. KE+PE

It's a bit strangely formulated, but obviously your book refers to the relative motion, i.e.,
$$\mu \dot{\vec{r}}=-G m_S m_E \frac{\vec{r}}{r^3}, \quad \vec{r}=\vec{r}_E-\vec{r}_S.$$
The details of the kinematics I have given above.

The total mechanical energy is the energy of both Earth and Sun together. It doesn't make sense to attribute "total energy" to the one or the other body, because the potential is an interaction potential, i.e., a two-body quantity.

 

[dyn]

It's a bit strangely formulated, but obviously your book refers to the relative motion, i.e.,
$$\mu \dot{\vec{r}}=-G m_S m_E \frac{\vec{r}}{r^3}, \quad \vec{r}=\vec{r}_E-\vec{r}_S.$$
The details of the kinematics I have given above.

The total mechanical energy is the energy of both Earth and Sun together. It doesn't make sense to attribute "total energy" to the one or the other body, because the potential is an interaction potential, i.e., a two-body quantity.

To calculate the escape velocity of a mass from the Earth , we require that the total energy of the mass , ie KE+PE be equal to zero. Is that not a similar situation where we attribute total energy to one body in an interaction potential ?

 

[vanhees71]

Here you approximate the situation by treating the Earth simply by an external gravitational field, neglecting the motion of the Earth due to its interaction with the object, which is of course justified, because the mass of the Earth is very much larger than that of the object. The same holds for the cestial mechanics Kepler problem in our solar system, because all planets' masses are very much smaller than that of the Sun.

 

[etotheipi]

If you are doing a problem then it's useful to remember that you may convert one-body equations to two-body equations by the transformation $G \mapsto G' = \left(1 + \frac{m_1}{m_2} \right)G$

 

[pbuk]

If you are doing a problem then it's useful to remember that you may convert one-body equations to two-body equations by the transformation $G \mapsto G' = \left(1 + \frac{m_1}{m_2} \right)G$

Stop procrastinating!