Gravity vs. Radiation

Suppose the distance from the ball to the center of the Sun is $r$ , the gravitational force will be:

$F_1=\frac{GM_sm}{r^2}=\frac{4GM_s \rho \pi R^3}{3r^2}.$

The number of photons absorbed by the ball in time $\Delta t$ is:

$N=\frac{P\Delta t}{E_{\gamma}} \frac{\Delta S}{S}=\frac{P \lambda \Delta t}{hc} \frac{\pi R^2}{4\pi r^2}=\frac{P \lambda \Delta t R^2}{4hcr^2}.$

Since the black ball absorbs all the incident light, each photon will transfer a momentum of $\Delta p=\frac{h}{\lambda}.$ The radiational force is:

$F_2=N\frac{\Delta p}{\Delta t}=\frac{P R^2}{4cr^2}.$

If $F_1=F_2$ , then $\frac{4GM_s \rho \pi R^3}{3r^2}=\frac{P R^2}{4cr^2}.$

$R=\frac{3P}{16GM_s \rho \pi c}=0.6\mu\text{m}$

We'll work in metres and kilograms to avoid any issues with units. Let the radius of the ball in metres be $r$ and let the distance from the ball to the sun in metres be $d$ . The volume of the ball will be $\frac{4\pi r^3}{3}$ , so its mass will be $\frac{4000\pi r^3}{3}$ . Therefore (using the formula $F=\frac{Gm_1m_2}{r^2}$ ): $\text{Weight}, W=\frac{G\times \frac{4000\pi r^3}{3} \times 2 \times 10^{30}}{d^2}.$

The $4\times 10^{26}~W$ of power produced by the sun will be evenly distributed over a sphere of surface area $4\pi d^2$ around the sun. Assuming that the ball is fairly small, the area of the sphere it obscures will be roughly flat, so the area of this sphere obscured by the ball can be approximated by the area of the great circle (the intersection of the sphere and a plane which passes through the center point of the sphere) of the ball. This is equal to $\pi r^2$ . So: $\text{Power actually absorbed by the ball}, P_a=4\times 10^{26}\times \frac{\pi r^2}{4\pi d^2}$ $\implies P_a=\frac{10^{26} r^2}{d^2},$ $\implies \text{Energy absorbed by the ball in one second}, E_1=\frac{10^{26} r^2}{d^2}.$

Now, using the fact that $E=\frac{hc}{\lambda}$ and $\lambda=\frac{h}{p}$ , we deduce that $p=\frac{E}{c}$ (alternatively we can use the relativistic momentum equation $E^2 = (pc)^2 + (m_0c^2)^2$ and let $m_0=0$ because we're dealing with photons). Hence $\text{rate of change of momentum}=\frac{E_1}{c}$ , so from earlier we know that: $\text{Rate of change of momentum}=\frac{10^{26} r^2}{d^2 c},$ but rate of change of momentum is just force, so: $\text{Force due to light}, F_L=\frac{10^{26} r^2}{d^2 c}.$

There is a counterbalance, so $F_L=W$ , hence: $\frac{10^{26} r^2}{d^2 c}=\frac{G\times \frac{4000\pi r^3}{3} \times 2 \times 10^{30}}{d^2},$ cancelling out $\frac{r^2}{d^2}$ from both sides and rearranging gives: $r=\frac{3\times 10^{26}}{c\times G\times 4000\pi\times 2\times 10^{30}}.$ We can then just substitute in our values for $c$ and $G$ to give $r=5.97\times 10^{-7}~\text{metres}$ , so $R=0.597~\text{micrometres}$ .

Last week we solved for the force due to photon radiation on the sail of a steampunk solar system cruiser. The sphere is very similar.

In short, the Sun emits photons uniformly in all directions. The fraction of the photons incident upon an object at a given distance is given by the fraction of the sky (as seen from the surface of the Sun (not recommended)) that is taken up by the object.

Looking at the ball, the man on the Sun sees an effective surface area of $\displaystyle \pi r_b^2$ at the distance $R_b$ . The surface of the ball absorbs all of the incident photons and so, absorbs all of their initial momentum. The radius of the ball, obviously, is $r_b$ .

The total number of photons coming out from the Sun is $\displaystyle \frac{P}{E_\gamma} = \frac{P\lambda}{hc}$ .

The number of photons absorbed by the ball is then $\displaystyle N_\gamma = \frac{P}{E_\gamma} = \frac{P\lambda}{hc}\frac{\pi r_b^2}{4\pi R_b^2}$ .

The total momentum delivered per unit time is then the total number of photons incident on the ball per unit time multiplied by the momentum of a single photon $h/\lambda$ (there is no photon recoil):

$\displaystyle F_\gamma = N_\gamma p_\gamma = \frac{P}{c}\frac{r_b^2}{4R_b^2}$

The gravitational force on the ball is given by $\displaystyle F_g = G\frac{M_S\rho\frac43 \pi r_b^2}{R_b^2}$

Equating the two forces, we find

$\boxed{\displaystyle r_b = \frac{3P}{16cGM_S\pi\rho}}$

With the numbers given (changing units so that $\rho = 1000 \mbox{ kg}/\mbox{m}^3$ ), we find $r_b \approx 0.597 \mu\mbox{m}$ .

These are some small balls.

We saw similar problem previous week. This is just an extension of that one.

Let us say the distance between the ball and sun is d. Note that due to the black colour the photons would be absorbed and the final momentum would be $0$ .

Initial momentum = $p$ = $\frac{h}{\lambda}$ .

Consider a small ring element subtending angle $2\theta$ on the center having area $dA = 2 \pi Rsin\theta Rd\theta = 2\pi R^2 sin\theta d\theta$ .

From the previous week's problem, No. of photons striking this element/per second( $dn$ ) = $\frac{P}{4\pi d^2}{dA}\frac{\lambda}{hc}= \frac{PR^2sin\theta \lambda d\theta}{2d^2hc}$ .

The forces along axes of rings will be added and forces perpendicular to axes would be cancelled.

$dF_{net} = dFcos\theta = pdncos\theta$

Hence, $F_{net} = \int pdncos\theta = \int_{0}^{\frac{\pi}{2}} \frac{h}{\lambda} \frac{PR^2sin\theta \lambda cos\theta}{2d^2hc} d\theta = \frac{PR^2}{4d^2c}$ .

Now, this force will balance gravitational force.

Hence, $\frac{PR^2}{4d^2c} = \frac{GM_{s} \rho \frac{4 \pi R^3}{3}}{d^2}$

$\Rightarrow R = \frac{3P}{16 \pi \rho G M_{s} c} = \boxed{0.6 \mu m}$ .

Suppose the distance from the ball to the center of the Sun is , the gravitational force will be: . The number of photons absorbed by the ball in time is: . Since the black ball absorb all the incident light, each photon will transfer a momentum of . The radiational force is: . If , then .