# Planetary orbits - the 2-body problem

• I
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

• Delta2

dextercioby
Homework Helper
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.

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• Dale and vanhees71
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
Homework Helper
Gold Member
2020 Award
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
2020 Award
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.

• sophiecentaur and pbuk
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
Gold Member
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 Last edited:
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
hutchphd
Homework Helper
So does that clear it up for you?

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
Homework Helper
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
Homework Helper
Gold Member
2020 Award
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.

• dyn, hutchphd and sophiecentaur
vanhees71
Gold Member
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
$$\begin{split} m_1 \ddot{\vec{x}}_1 &=-\frac{G m_1 m_2}{|\vec{x}_1-\vec{x}_2|^3}(\vec{x}_1-\vec{x}_2),\\ m_2 \ddot{\vec{x}}_2 &=-\frac{G m_1 m_2}{|\vec{x}_1-\vec{x}_2|^3}(\vec{x}_2-\vec{x}_1) \end{split}$$
$$M \ddot{\vec{R}}=0$$
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
$$\vec{r}=r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix}$$
$$\dot{\vec{r}}^2=\dot{r}^2 + r^2 \dot{\varphi}^2$$
and
$$L=\mu (x_1 \dot{x}_2-x_2 \dot{x}_1)=\mu r^2 \dot{\varphi}=\text{const}$$
$$E=\frac{\mu}{2} \dot{r}^2 + \frac{L^2}{2 \mu r^2} - \frac{G m_1 m_2}{r}.$$
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:
$$C=\sqrt{1+\frac{2 \mu E}{L^2}} \frac{G \mu m_1 m_2}{L^2}.$$
$$r=\frac{1}{s} = \frac{L^2}{G \mu m_1 m_2} \, \frac{1}{1+ \epsilon \cos \varphi}$$
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.

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• Leo Liu, JD_PM, alan123hk and 6 others

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. 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##

• PeroK, vanhees71, Delta2 and 1 other person
hutchphd
Homework Helper
It is precisely always the distance between the two masses.

Do you think my answer is different from yours?

hutchphd
Homework Helper
If it id the same why reiterate? Finis.

• weirdoguy
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?

• weirdoguy
vanhees71
Gold Member
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.

• Richard R Richard
hutchphd
Homework Helper
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, Richard R Richard and vanhees71
The diagram is #15 is very helpful but why is the eccentricity of both masses the same ?

hutchphd
Homework Helper
$$\vec{R}=\frac{1}{M} (m_1 \vec{x}_1+m_2 \vec{x}_2), \quad M=m_1+m_2.$$
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

• vanhees71 and 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 ?

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

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

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

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

$$e_1 = e_2 = e$$

• vanhees71 and dyn