# What is the analytical closed-form solution of the two-body problem to verify its numerical integration results?

The purpose is to study the motion of two celestial bodies analytically. What is the closed-form of the two-body problem if I was to solve it analytically without using a numerical approximation technique.

An example where this would be useful is this question from the book Analytical Mechanics of Space Systems by Hanspeter Schaub.

Write a numerical simulation that integrates the differential equations of motion in Eq. (9.45) using a fourth-order Runge Kutta integration scheme. Using the subroutine of task (b), compare the answer of the numerical integration to the analytical two-body solution.

$$\mathbf{\ddot{r}}=-\frac{\mu}{r^3}\mathbf{r} = -\frac{\mu}{r^2}\mathbf{\hat{r}} \tag{9.45}$$

• Not going to type up an answer right now, but this exact question is answered in -- well -- pretty much every single orbital mechanics textbook out there. – Tristan Apr 2 at 17:19
• What do you miss in this Wikipedia article? See also the links to other articles there. – Uwe Apr 2 at 17:24
• @Uwe What I miss is how is the transformation of the set of elements to find the motion of a planet in arbitrary reference frame. – John Apr 2 at 19:28

This is a supplemental answer for now because while we know that a two body orbit can be reduced to a one body orbit around a central potential, doing that here will be a little distracting and I think the result for the one body in central potential looks cleaner. See also answers to Can the radial oscillations of an elliptical orbit be solved using a fictitious centrifugal potential?

Per this comment I know I've had a discussion somewhere in this site (or in Astronomy SE) where it was first explained to me that Kepler orbits do have analytical solutions you can write down for time as a function of position, even though we still do need to use numerical techniques (e.g. Newton's method) to solve position as a function of time. (see also How did Newton and Kepler (actually) do it?)

Equation 27 in Wikipedia's Kepler orbit; Properties of trajectory equation is

$$t = a \sqrt{\frac{a}{\mu}}\left(E - e \sin E \right)$$

where $$a$$ is the semimajor axis, $$\mu$$ is the standard gravitational parameter also known as the product $$GM$$, $$e$$ is the eccentricity and $$E$$ is the Eccentric anomaly.

The relationship between $$E$$ and the true anomaly $$\theta = \arctan2(y, x)$$ is

$$\tan \frac{\theta}{2} = \sqrt{ \frac{1+e}{1-e} } \tan \frac{E}{2}$$

and solving for $$E$$:

$$E(\theta) = 2 \arctan \sqrt{ \frac{1-e}{1+e} } \tan \frac{\theta}{2}.$$

plugging back in to the first equation (but not writing it all out):

$$t(\theta) = a \sqrt{\frac{a}{\mu}}\left(E(\theta) - e \sin E(\theta) \right)$$

Let's try a numerical check of this amazing result. Note that with $$a=1$$ and $$\mu=1$$ the period is $$2 \pi$$.

The last plot at bottom left shows that the analytical $$t(\theta)$$ based on $$\theta$$ from a numerically integrated orbit matches the time used in the numerical calculation for an $$e=0.8$$ elliptical orbit. There will be numerical glitches or singularities at the endpoints and for $$e=1$$ but it seems to check out nicely!  Python script:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

def deriv(X, t):
x, v = X.reshape(2, -1)
acc = -x * ((x**2).sum())**-1.5
return np.hstack((v, acc))

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

e = 0.8
a = 1.0
mu = 1.0
r_peri, r_apo = a*(1.-e), a*(1.+e)
v_peri, v_apo = [np.sqrt(2./r - 1./a) for r in (r_peri, r_apo)]
T = twopi * np.sqrt(a**3/mu)

X0 = np.array([r_peri, 0, 0, v_peri])
X0 = np.array([-r_apo, 0, 0, -v_apo])
times = np.linspace(-T/2., T/2., 1001)

answer, info = ODEint(deriv, X0, times, full_output=True)

theta = np.arctan2(y, x)
E = 2. * np.arctan(np.sqrt((1.-e)/(1.+e)) * np.tan(theta/2))
t = a * np.sqrt(a/mu) * (E - e * np.sin(E))

if True:
plt.figure()
plt.subplot(2, 1, 1)
plt.plot(x, y)
plt.plot(, , 'ok')
plt.gca().set_aspect('equal')
plt.title('y vs. x numerical')
plt.subplot(2, 1, 2)
plt.plot(times[1:-1], x)
plt.plot(times[1:-1], y)
plt.xlim(-pi, pi)
plt.title('x(t) and y(t) numerical')
plt.show()

plt.subplot(2, 2, 1)
plt.title('theta(t_numerical)')
plt.plot(times[1:-1], theta)
plt.xlim(-pi, pi)
plt.ylim(-pi, pi)
plt.gca().set_aspect('equal')
plt.subplot(2, 2, 2)
plt.title('E_analytic(theta_numerical)')
plt.plot(E, theta)
plt.xlim(-pi, pi)
plt.ylim(-pi, pi)
plt.gca().set_aspect('equal')
plt.subplot(2, 2, 3)
plt.title('theta(t_analytic)')
plt.plot(t, theta)
plt.xlim(-pi, pi)
plt.ylim(-pi, pi)
plt.gca().set_aspect('equal')
plt.subplot(2, 2, 4)
plt.title('t_analytic(t_numerical)')
plt.plot(t, times[1:-1])
plt.xlim(-pi, pi)
plt.ylim(-pi, pi)
plt.gca().set_aspect('equal')
plt.show()

• Thank you for this very informative answer! I just have one small question that might be silly; I see that you used the ODE45 to integrate the orbit of the two-body problem. How do you verify that the orbit is correct using the analytical solution of the problem? – John Apr 4 at 2:51
• I have updated this post with an image clarifying what I mean by my previous comment. – John Apr 4 at 2:58
• @John First a note: for the numerical integration I used SciPy's odeint which chooses between "1: adams (nonstiff), 2: bdf (stiff)" for the integration method" I don't know how it compares to MatLab's ODE45 but for such an easy nonstiff problem like this I think any integrator will give good results. There is also a method called RK45 en.wikipedia.org/wiki/… – uhoh Apr 4 at 4:19
• Thank you for your answer. The reason I did not post the rest of the question because it is irrelevant. Part a) and b) are related to transform the classical orbital elements to Cartesian orbital position and vice-versa. However, no analytical solution is provided or either asked to be derived (I think the author assumes we already know it). I have updated the question by adding Eq. (9.45) which he refers too. I hope its clearer now and apologies for any misunderstanding. – John Apr 4 at 14:12
• Got it! Thank you for your answer! – John Apr 5 at 16:28

The distance from the focus of attraction of an orbit can be expressed as a function of the true anomaly (angle) given by $$r(\theta)=a\frac{1-e^2}{1+ecos(\theta)}$$, where $$a$$ is the semi-major axis and $$e$$ is the eccentricity.

• How is the second mass affecting the first mass based on gravitational attraction in this equation? – John Apr 3 at 0:36
• Answers to Can the radial oscillations of an elliptical orbit be solved using a fictitious centrifugal potential? also contain some helpful information. – uhoh Apr 3 at 1:22
• Is it possible to show $t(\theta)$ as well? – uhoh Apr 3 at 23:12
• @John: The reference frame is fixed to one of the two masses (usually the larger one). In this reference frame, one mass observes the other mass rotating around it as if the first mass were motionless. Both objects exert forces on each other, but the reference frame makes it appear as if only one mass were moving with respect to the other. – Paul Apr 3 at 23:33
• @uhoh: Time as a function of true anomaly? Perhaps you mean true anomaly as a function of time? Or perhaps time between two anomalies? – Paul Apr 3 at 23:53