Kvant Physics Problem 31
A rigid spherical spacecraft of radius $R$ and mass $m$ is placed in a uniform поток of solar radiation.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m14s
Source on kvant.digital
Problem
How does radiation pressure orient a spherical spacecraft relative to the Sun, one half of which is mirror-like and the other black, completely absorbing solar radiation? The center of mass of the spacecraft is located at the center of the sphere.
P. L. Kapitsa
Setup and Assumptions
A rigid spherical spacecraft of radius $R$ and mass $m$ is placed in a uniform поток of solar radiation. The Sun is assumed to be at infinite distance so that the incident rays form a parallel beam with intensity $I$ (W·m$^{-2}$). The radiation pressure scale is $p_0 = I/c$, where $c$ is the speed of light.
The sphere is divided into two hemispheres by a plane passing through its center of mass. One hemisphere is perfectly absorbing (black), the other is perfectly specularly reflecting (mirror-like). The center of mass coincides with the geometric center of the sphere.
The unknown is the equilibrium orientation of the sphere relative to the Sun direction, determined by the condition of vanishing net torque caused by radiation pressure.
Gravitational forces and other environmental effects are neglected. The sphere is assumed to be in vacuum, and self-shadowing or multiple scattering is ignored. The problem is treated in a stationary regime where the orientation is constant in time.
Physical Principles
The momentum flux carried by electromagnetic radiation is $I/c$. For a surface element with outward normal $\mathbf{n}$ and incidence angle $\theta$ relative to the incoming light direction $\mathbf{s}$ (a unit vector pointing from the Sun toward the spacecraft), the absorbed power per unit area is $I\cos\theta$, and the corresponding force on a perfectly absorbing surface element is directed along $\mathbf{s}$ with magnitude $\frac{I}{c}\cos\theta,dA$.
For a perfectly reflecting surface, the normal component of momentum is reversed. The force on a surface element is directed along the outward normal and has magnitude $\frac{2I}{c}\cos^2\theta,dA$, giving the vector form $d\mathbf{F} = \frac{2I}{c}\cos^2\theta,\mathbf{n},dA$.
The torque about the center of the sphere is given by
$$\boldsymbol{\tau} = \int \mathbf{r} \times d\mathbf{F},$$
where $\mathbf{r} = R\mathbf{n}$ for a spherical surface.
Equilibrium orientations satisfy the condition $\boldsymbol{\tau} = \mathbf{0}$.
Derivation
Let the unit vector $\mathbf{a}$ denote the symmetry axis of the sphere, perpendicular to the dividing plane between hemispheres. Let $\mathbf{s}$ be the direction of sunlight. The orientation is fully characterized by the angle $\alpha$ between these vectors, defined by $\cos\alpha = \mathbf{a}\cdot\mathbf{s}$.
Each surface element is specified by its outward normal $\mathbf{n}$. The hemisphere condition is determined by the sign of $\mathbf{n}\cdot\mathbf{a}$. The local incidence angle satisfies $\cos\theta = \mathbf{n}\cdot\mathbf{s}$ for all illuminated points with $\mathbf{n}\cdot\mathbf{s} > 0$.
The torque density on the absorbing hemisphere is
$$d\boldsymbol{\tau}_{\text{abs}} = R\mathbf{n} \times \left(\frac{I}{c}(\mathbf{n}\cdot\mathbf{s})\mathbf{s},dA\right).$$
On the reflecting hemisphere it is
$$d\boldsymbol{\tau}_{\text{mir}} = R\mathbf{n} \times \left(\frac{2I}{c}(\mathbf{n}\cdot\mathbf{s})^2\mathbf{n},dA\right),$$
and since $\mathbf{n} \times \mathbf{n} = \mathbf{0}$, the reflecting part produces no direct torque contribution in this ideal specular model about the center.
The net torque therefore arises entirely from the absorbing hemisphere and from the fact that only half of the illuminated sphere contributes absorbing pressure.
The absorbing contribution simplifies to
$$d\boldsymbol{\tau}_{\text{abs}} = \frac{IR}{c}(\mathbf{n}\cdot\mathbf{s}),(\mathbf{n}\times\mathbf{s}),dA.$$
By rotational symmetry, the total torque must lie along $\mathbf{a}\times\mathbf{s}$. The magnitude can therefore be written as
$$\tau(\alpha) = \frac{IR^3}{c},K,\sin\alpha,$$
where $K$ is a positive dimensionless constant obtained from integrating over the illuminated absorbing hemisphere.
The key structural consequence is that $\tau(\alpha)$ is proportional to $\sin\alpha$, since reversing $\mathbf{a}$ relative to $\mathbf{s}$ flips the sign of the asymmetry, and torque must vanish when $\mathbf{a}\parallel\mathbf{s}$ or $\mathbf{a}\parallel -\mathbf{s}$.
Equilibrium requires
$$\tau(\alpha)=0 \quad \Rightarrow \quad \sin\alpha = 0,$$
so
$$\alpha = 0 \quad \text{or} \quad \alpha = \pi.$$
To determine which orientation is physically realized, the sign of the restoring tendency is examined. If $\mathbf{a}$ is slightly rotated from $\alpha = 0$, the absorbing illuminated region increases on one side of the axis in a way that produces a torque increasing $\alpha$, making $\alpha = 0$ unstable. Around $\alpha = \pi$, a small deviation reduces exposure of the absorbing hemisphere toward the Sun, producing a restoring torque that decreases the deviation, making $\alpha = \pi$ stable.
The stable equilibrium therefore corresponds to $\mathbf{a} = -\mathbf{s}$, meaning the normal to the dividing plane points away from the Sun and the absorbing hemisphere is oriented toward the Sun.
Result
The equilibrium orientations satisfy
$$\mathbf{a} \parallel \mathbf{s} \quad \text{or} \quad \mathbf{a} \parallel -\mathbf{s}.$$
The stable configuration is
$$\mathbf{a} = -\mathbf{s},$$
so that the black (absorbing) hemisphere faces the Sun and the mirror hemisphere faces away.
Substituting numerical quantities is not required because the result is purely geometric and independent of $I$, $R$, $m$, and $c$.
$$\boxed{\text{Stable equilibrium: absorbing hemisphere points toward the Sun}}$$
Sanity Checks
Dimensional consistency follows since the torque density scale is $\frac{IR^3}{c}$, which has units $(\text{W·m}^{-2})(\text{m}^3)/(\text{m·s}^{-1}) = \text{N·m}$.
In the limiting case where both hemispheres are identical, the asymmetry vanishes and $K \to 0$, eliminating any preferred orientation, consistent with zero torque.
If the entire sphere were absorbing, symmetry under rotations about $\mathbf{s}$ would eliminate any transverse torque component, again consistent with no preferred orientation.
A sign error would most easily arise in assigning the direction of the absorbed momentum transfer. Reversing the direction of the force in the absorbing hemisphere would invert the stability conclusion, swapping the roles of $\alpha = 0$ and $\alpha = \pi$.