## FANDOM

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The Yarkovsky effect is a mechanism that can change the angular momentum of an object (asteroid) orbiting a luminous gravitating mass (star). The object absorbs radiation in the radial direction (i.e. along the line connecting it to the luminous mass), but might emit it in a different angle $\alpha$ (measured relative to the radial direction). This effect is strongest if the radiation is emitted perpendicular to the radial direction $\alpha = \frac{\pi}{2}$. This case sets an upper limit for the torque this effect can exert. Suppose the luminosity of the star is $\dot{E}_s$. The maximum energy the asteroid can absorb is

$\dot{E}_a \approx \dot{E}_s \left( \frac{R}{d} \right)^2$

if $R$ is the size of the asteroid and $d$ is its distance from the star. The force is given by $F \approx \frac{\dot{E}_d}{c}$, and the torque is therefore

$\dot{L}_{max} \approx F \cdot d \approx \frac{\dot{E}_s}{c} \left( \frac{R}{d} \right)^2 d$

This effect will only be relevant for objects that can significantly change their angular momentum within the lifetime of the system. The time scale for a significant change of the angular momentum is

$\tau_{L,min} \approx \frac{L}{\dot{L}_{max}} \approx \frac{M_a v_a d}{\frac{\dot{E}}{c} \left( \frac{R}{d} \right)^2 d} \approx \frac{\rho R d^2 c \sqrt{G M_s/d}}{\dot{E}_s}$

where $\rho$ is the mass of the asteroid, $M_s$ is the mass of the star. For an object of size $R = 1 km$, density $\rho = 1 g / cc$, at a distance of $d = 1 AU$ emitting at solar luminosity this time scale is about 10 million years.

We now turn to calculate the angle between the absorbed and emitted radiation. For that purpose we assume that the star is a hollow, infinite cylinder. The temperature on the surface of the cylinder evolves as

$\frac{\partial T}{\partial t} = \frac{\kappa}{R^2} \frac{\partial^2 T}{\partial \theta^2} - \omega \frac{\partial T}{\partial \theta} - \frac{T}{\tau} + I \left( \theta \right)$

where $\theta$ is the angle relative to the line connecting the asteroid to the star. The first term on the right hand side is heat diffusion due to conduction. The second term is due to the rotation of the cylinder (asteroid) around its axis at an angular velocity $\omega$. The third term represents cooling over a fixed time scale $\tau$, and the fourth heating. In steady state the left hand side is zero. For simplicity, we assume that the heating term is $I \left( \theta \right) = I_0 \left(1 + \cos \theta \right) = I_0 \left(1 + \exp\left(i \theta \right) \right)$. The solution would be of the form $T \left( \theta \right) = T_0 + T_1 \exp \left(i \theta \right)$. Substitution yields

$T_0 = \tau I_0$

$T_1 = \frac{\tau I_0}{1 + \frac{\kappa \omega}{R^2} + i \omega \tau} = \frac{I_0 \tau}{\sqrt{\left(1 + \tau \kappa / R^2 \right)^2 + \left( \tau \omega\right)^2}} \exp \left( i tan^{-1} \left( \frac{\omega \tau}{1 + \kappa \tau/R^2} \right)\right)$

The peak temperature is at an angle $\alpha = \tan^{-1} \left( \frac{\omega \tau}{1 + \kappa \omega/R^2} \right)$ relative to the line connecting the asteroid and the planet. This add a correction term to the torque

$\dot{L} = \dot{L}_{max} \sin \alpha$

and the time for changing the angular momentum becomes

$\tau_L \approx \frac{L}{\dot{L}} \approx \frac{\rho R d^2 c \sqrt{G M_s/d}}{\dot{E}} \frac{\sqrt{\left(\omega \tau \right)^2 + \left( 1 + \kappa \tau / R^2\right)^2}}{\omega \tau}$

$\tau_L \approx \frac{\rho d^2 c \sqrt{G M_s/d} \kappa}{\dot{E}_s R \omega}$
So this effect is inefficient if the radius is small enough. Assuming thermal conductivity of iron $\kappa = 2.3 \cdot 10^{-5} \frac{m^2}{s}$, a radius of $R = 1 mm$ and a rotation period of an hour yields a time scale of about a million years. If the rotation of the planet around itself is prograde (same as the orbital) then this effects increases the orbital angular momentum. If it is retrograde (opposite to the orbital) then it decreases the orbital angular momentum.