1. The Physical Model (Photon Gas)

The cube cools by radiating heat $Q$ from one face. We model the radiation leaving the surface as an isotropic photon gas. The force exerted by such radiation is equivalent to the Radiation Pressure acting on the area of the face.

The recoil force $F$ arises from the momentum carried away by the photons. For a surface emitting isotropically into a hemisphere, the effective force is related to the power $P$ by a geometric factor arising from the integration of momentum flux.

2. Deriving the Geometric Factor (1/3)

Let’s calculate the average momentum transfer. Consider a small area $A$ and photons emitted at an angle $\theta$ to the normal.

• Momentum component: The recoil momentum is opposite to the normal component of the photon’s momentum: $p_z = \frac{E}{c}\cos\theta$.
• Flux Projection: The rate at which photons cross the surface plane is proportional to their velocity component perpendicular to the surface: $v_z = c\cos\theta$.

The total force is proportional to the product of these two factors, averaged over the hemisphere:

$$ \text{Factor} = \frac{\int_{0}^{\pi/2} (\cos\theta)_{\text{mom}} \cdot (\cos\theta)_{\text{flux}} \cdot \sin\theta d\theta}{\int_{0}^{\pi/2} \sin\theta d\theta} $$

$$ \text{Factor} = \int_{0}^{\pi/2} \cos^2\theta \sin\theta d\theta = \left[ -\frac{\cos^3\theta}{3} \right]_0^{\pi/2} = \frac{1}{3} $$

Thus, the relationship between the Force (thrust) and the Power radiated is:

$$ F = \frac{1}{3} \frac{P}{c} $$

3. Impulse and Terminal Speed

The cube starts from rest. The total impulse delivered to the cube results in its final momentum.

$$ \Delta p_{\text{cube}} = \int F dt $$

Substituting $F = \frac{P(t)}{3c}$:

$$ m v = \int \frac{P(t)}{3c} dt = \frac{1}{3c} \int P(t) dt $$

The integral of Power over time is the total heat energy radiated, $Q$.

$$ m v = \frac{Q}{3c} \implies v = \frac{Q}{3mc} $$

4. Numerical Calculation

Given values:

• Total Heat $Q = 900 \text{ J}$
• Mass $m = 4.0 \text{ g} = 4.0 \times 10^{-3} \text{ kg}$
• Speed of light $c = 3 \times 10^8 \text{ m/s}$

Substituting these into the derived formula:

$$ v = \frac{900}{3 \times (4.0 \times 10^{-3}) \times (3 \times 10^8)} $$

$$ v = \frac{900}{36 \times 10^5} $$

$$ v = 25 \times 10^{-5} \text{ m/s} $$

$$ v = 0.25 \times 10^{-3} \text{ m/s} = 0.25 \text{ mm/s} $$