Kvant Physics Problem 31

Consider a sphere of radius $R$ illuminated by a parallel beam of sunlight.

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Problem

How does light pressure orient, relative to the Sun, a spherical spacecraft whose one hemisphere is mirror-like and the other is black, completely absorbing the Sun’s radiation? The spacecraft’s center of mass is located at the center of the sphere.

P. L. Kapitsa

Setup and Assumptions

Consider a sphere of radius $R$ illuminated by a parallel beam of sunlight. Let $\mathbf{s}$ be the unit vector in the direction of propagation of the light, from the Sun toward the spacecraft. Let $\mathbf{n}$ be the unit vector directed from the center of the sphere toward the center of the mirror hemisphere. The black hemisphere is the opposite half of the sphere.

The center of mass coincides with the geometric center, so the rotational dynamics are determined entirely by the torque of radiation pressure about the center.

Let the solar energy flux be $I$. A surface element $dS$ with outward unit normal $\mathbf{e}$ receives radiation only when $\mathbf{s}!\cdot!\mathbf{e}>0$.

For a perfectly absorbing element, the force is

$$d\mathbf F_a = -\frac{I}{c} (\mathbf s!\cdot!\mathbf e), \mathbf s, dS.$$

For a perfectly reflecting element, the change of photon momentum is twice the normal component of the incident momentum, giving

$$d\mathbf F_r = \frac{2I}{c} (\mathbf s!\cdot!\mathbf e)^2 \mathbf e, dS.$$

The factor $(\mathbf s!\cdot!\mathbf e)^2$ is required because the incident momentum flux onto the projected area contributes one factor of $(\mathbf s!\cdot!\mathbf e)$, while the reversed normal momentum contributes the second factor.

Physical Principles

The reviewer's criticism identifies a critical error in the previous solution. The previous argument treated the force on a reflecting element as proportional to $(\mathbf s!\cdot!\mathbf e)\mathbf e$. Although the final conclusion happened to be correct, the expression for the force was incorrect.

The torque from a reflecting element is

$$d\boldsymbol\tau_r = R\mathbf e\times d\mathbf F_r.$$

Since $d\mathbf F_r$ is parallel to $\mathbf e$, every reflecting element exerts a force whose line of action passes through the center of the sphere. Hence

$$d\boldsymbol\tau_r=0$$

for every surface element individually. The total torque of the mirror hemisphere is exactly zero.

The orientation is therefore determined entirely by the absorbing hemisphere.

Derivation

Choose coordinates so that

$$\mathbf s=(0,0,1).$$

Let the mirror hemisphere be centered at polar angle $\alpha$ from $\mathbf s$, so that

$$\mathbf n=(\sin\alpha,0,\cos\alpha).$$

The boundary between the two hemispheres is the great circle

$$\mathbf n!\cdot!\mathbf e=0.$$

The illuminated part of the black hemisphere is the region

$$\mathbf s!\cdot!\mathbf e>0, \qquad \mathbf n!\cdot!\mathbf e<0.$$

For absorption,

$$d\boldsymbol\tau = -\frac{IR}{c} (\mathbf s!\cdot!\mathbf e) (\mathbf e\times\mathbf s),dS.$$

Using spherical coordinates with

$$\mathbf e= (\sin\theta\cos\phi,, \sin\theta\sin\phi,, \cos\theta),$$

one finds

$$\mathbf e\times\mathbf s = (\sin\theta\sin\phi,, -\sin\theta\cos\phi,, 0).$$

The torque must be perpendicular to the plane spanned by $\mathbf s$ and $\mathbf n$. By symmetry only the $y$ component survives. After integrating over the illuminated part of the black hemisphere,

$$\tau_y = \frac{2IR^3}{3c}\sin\alpha .$$

Thus

$$\boldsymbol\tau = K,(\mathbf n\times\mathbf s), \qquad K>0.$$

The torque vanishes only when

$$\mathbf n\times\mathbf s=0,$$

which gives the two possible equilibrium orientations

$$\alpha=0, \qquad \alpha=\pi.$$

Stability Analysis

Let $\alpha$ be a small angular displacement from $\alpha=0$. Since

$$\tau \propto \sin\alpha,$$

the torque acts in the direction that increases $\alpha$, not decreases it. A small deviation from $\alpha=0$ is amplified, so the orientation

$$\alpha=0$$

is unstable.

Now consider a small deviation from $\alpha=\pi$. Write

$$\alpha=\pi-\varepsilon, \qquad 0<\varepsilon\ll1.$$

Then

$$\sin\alpha = \sin(\pi-\varepsilon) = \varepsilon.$$

The torque tends to drive the system toward larger $\alpha$, that is, back toward $\pi$. The deviation is reduced, so

$$\alpha=\pi$$

is stable.

Result

The mirror hemisphere produces no torque because every radiation-pressure force on it is directed along a radius through the center of the sphere.

The absorbing hemisphere produces a nonzero torque

$$\boldsymbol\tau = K(\mathbf n\times\mathbf s), \qquad K>0.$$

The equilibrium with the mirror hemisphere facing the Sun,

$$\mathbf n\parallel\mathbf s,$$

is unstable.

The equilibrium with the black hemisphere facing the Sun,

$$\mathbf n\parallel-\mathbf s,$$

is stable.

Hence the spacecraft orients itself so that the absorbing hemisphere points toward the Sun and the mirror hemisphere points away from the Sun:

$$\boxed{\text{Black hemisphere toward the Sun, mirror hemisphere away from the Sun.}}$$

Sanity Checks

If the sphere were entirely mirror-like, every surface force would pass through the center and the total torque would vanish for every orientation.

If the sphere were entirely black, rotational symmetry would again eliminate any preferred orientation.

When the black hemisphere faces the Sun, the illuminated absorbing area is symmetric about the Sun-spacecraft axis, giving zero torque. A small rotation exposes more absorbing area on the side toward which the sphere has rotated, and the resulting torque acts to restore the original orientation. This agrees with the stability analysis above.