Inverse-square attraction and conic orbits

1. Why inverse-square attraction is special

For a general central force, the reciprocal-radius equation can remain nonlinear. Inverse-square attraction cancels its explicit $u^2$ denominator, leaving a constant driving term. The orbital geometry then follows from the elementary harmonic differential equation.

Use $\mathbf F=-k\hat{\mathbf r}/r^2$ with $k>0$, $V=-k/r$ and $V(\infty)=0$. The mass is $\mu$, angular momentum magnitude is $\ell>0$, and $E$ is relative mechanical energy. For gravity between two bodies, $k=Gm_1m_2$ and $k/\mu=G(m_1+m_2)$.

2. Solution of the orbit equation

Binet's equation becomes

A constant particular solution is $u_0=\mu k/\ell^2$. The homogeneous equation has solutions $\cos\theta$ and $\sin\theta$, as verified by twice differentiating them. Thus

Write $C=u_0e\cos\theta_0$ and $D=u_0e\sin\theta_0$, with $e=\sqrt{C^2+D^2}/u_0\geq0$. The addition formula gives $u=u_0[1+e\cos(\theta-\theta_0)]$. Choose the polar axis through the closest approach so that $\theta_0=0$ and define the semi-latus rectum

Here italic $p$ is an orbital length, while bold $\mathbf p$ later denotes linear momentum and $p_r$ denotes radial canonical momentum. The orbit is

For $e>0$, $\theta=0$ maximizes the denominator and gives periapsis $r_p=p/(1+e)$. At $\theta=\pi/2$, $r=p$, explaining the geometrical name semi-latus rectum. For $e=0$ the circle has no distinguished periapsis direction.

3. Why the equation describes a conic

With $x=r\cos\theta$, $y=r\sin\theta$, multiply the orbit equation by its denominator to get $r+ex=p$. Thus $r=p-ex$, which must remain nonnegative on the physical branch. Squaring gives

or

For $0\leq e<1$, complete the square:

This is an ellipse with semimajor axis $a=p/(1-e^2)$, semiminor axis $b=p/\sqrt{1-e^2}=a\sqrt{1-e^2}$, and center at $x=-ae$. One focus is the force center at the origin, a distance $ae$ from the ellipse center.

For $e=1$, the equation becomes $y^2=p^2-2px=-2p(x-p/2)$: a parabola with vertex at periapsis $x=p/2$. For $e>1$, put $d=e^2-1>0$ and complete the square to get

This is a hyperbola; the physical attractive-scattering orbit uses the branch satisfying the original unsquared condition $r=p-ex\geq0$. Squaring alone introduces the other geometric branch without making it part of the same physical trajectory.

4. Energy and eccentricity

From the orbital energy relation,

Since $u=(1+e\cos\theta)/p$ and $u'=-e\sin\theta/p$,

Use $\ell^2/(\mu p)=k$. Then

Therefore

The angular terms cancel because $E$ is conserved. At fixed $\ell$, the minimum allowed energy is $E_c=-\mu k^2/(2\ell^2)$, where $e=0$. Energies below this have no real orbit for that angular momentum.

This table assumes $k>0$, $\ell>0$, and zero potential at infinity. Radial collision trajectories with $\ell=0$ are degenerate and are not classified by simply setting $e=1$ in the regular conic formula.

For an ellipse, $p=a(1-e^2)$ gives

At fixed $a$, more eccentric ellipses have less angular momentum. At fixed $\ell$, changing energy changes eccentricity and size together.

5. Speed and escape

From $E=\mu v^2/2-k/r$ and $E=-k/(2a)$,

for an ellipse. This is the vis-viva relation. For a circular orbit $a=r$, it gives $v_c=\sqrt{k/(\mu r)}$. Setting escape energy to zero gives $v_{\rm esc}=\sqrt{2k/(\mu r)}=\sqrt2v_c$. This is a speed threshold in the ideal isolated inverse-square model; launch direction determines angular momentum and whether a trajectory encounters a physical surface first.

For a hyperbola, the positive geometric semimajor length $a_h$ obeys $E=k/(2a_h)$, giving $v^2=(k/\mu)(2/r+1/a_h)$. Some books use a negative signed $a$ to retain the same vis-viva formula. We use positive lengths and state the sign explicitly.

For a hyperbolic orbit the allowed angular range is $|\theta|<\theta_\infty$, where $\cos\theta_\infty=-1/e$. At its limits the denominator vanishes and $r\to\infty$. Continuing beyond that range gives an unphysical negative polar radius.

6. Applications

Example 1 · Geometry from the turning radii

An ellipse has $r_p=2\,\mathrm{length\ units}$ and $r_a=8\,\mathrm{length\ units}$. Adding and subtracting $a(1-e)$ and $a(1+e)$ gives $a=5$ and $e=(8-2)/(8+2)=0.6$. Then $p=a(1-e^2)=3.2$ and $b=a\sqrt{1-e^2}=4$. The force center is $ae=3$ units from the ellipse center.

Example 2 · Launch faster than circular speed

At radius $r_0$, a particle is launched tangentially at $v=\sqrt{3/2}\,v_c$, where $v_c^2=k/(\mu r_0)$. Its energy is $E=\mu v^2/2-k/r_0=-k/(4r_0)$ and its angular momentum satisfies $\ell^2=\mu^2r_0^2v^2=(3/2)\mu kr_0$.

Thus $a=2r_0$ and $e^2=1-3/4=1/4$, so $e=1/2$. Since the radial velocity is zero and the speed exceeds circular speed, the launch point is periapsis: $r_p=a(1-e)=r_0$. Apoapsis is $3r_0$. The radius increases initially because $\ddot r=v^2/r_0-k/(\mu r_0^2)>0$.

Example 3 · Escape in dimensionless units

Let $\mu=k=1$, launch radius $r_0=2$, and tangential speed $v_0=1$. Then $E=v_0^2/2-1/r_0=0$, $\ell=r_0v_0=2$, and $p=\ell^2=4$. The parabola is $r=4/(1+\cos\theta)$, with periapsis $r_p=2$. The particle approaches zero speed only as $r\to\infty$, not at a finite apoapsis.

Example 4 · A hyperbolic encounter

Take $\mu=k=1$, asymptotic speed $v_\infty=1$, and impact parameter $b_{\rm imp}=2$. Far away, $E=1/2$ and $\ell=\mu b_{\rm imp}v_\infty=2$. Thus $e=\sqrt{1+2(1/2)(4)}=\sqrt5$, $p=4$, and $r_p=4/(1+\sqrt5)=\sqrt5-1$.

The positive energy means the particle returns to infinity after the encounter. The periapsis is smaller than the impact parameter because attraction bends the incoming trajectory inward. The Runge–Lenz chapter will recover the deflection angle from the same constants.

7. Geometry and dynamics together

The conic equation supplies shape and orientation; $E$ and $\ell$ determine size and eccentricity; the area law supplies the time dependence. A complete orbital description needs all three. An ellipse drawn with its center at the gravitating body would misrepresent the inverse-square force except in the circular case.

Exercises

Hints and solutions are collected separately.

1. 20.1 Solve the inverse-square Binet equation and define its two geometric constants.

2. 20.2 Derive the energy–eccentricity relation.

3. 20.3 An ellipse has $r_p=3$ and $r_a=9$ in length units. Find $a,e,p$.

4. 20.4 For $\mu=k=1$ and $\ell=2$, what is the lowest allowed energy?

5. 20.5 For $\mu=k=\ell=1$, classify the orbit at $E=-1/4$.

6. 20.6 Derive the escape speed at radius $r$ for inverse-square attraction.

7. 20.7 At fixed semimajor axis, how does angular momentum depend on eccentricity?

8. 20.8 A particle is launched tangentially at half the circular speed at $r_0$. Find $a,e$ and decide which apsis is the launch point.

9. 20.9 For a hyperbola with $e=2$, find its limiting true anomaly.

10. 20.10 Why does $E=0$ not mean a parabolic particle is at rest?

11. 20.11 Use Cartesian coordinates to identify the ellipse center for $r=p/(1+e\cos\theta)$ with $e<1$.

12. 20.12 Why must $\ell=0$ be treated separately in the energy–eccentricity formula?