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I'm trying to get an estimative for how much force is applied on a solar sail by solar radiation.

So the first question is elastic or inelastic collision? Elastic colision provides a change in momentum of 2p, while an inelastic colision would have a change in momentum of p.

Photons have no mass, so $p = \frac{\text{E}}{c}$.

Then 2p should be the impulse one particle gives to the sail in I don't know how much time. I made a wild guess of 0.001s and by searching the energy of a photon and that about 2e24 photons would hit the sail per second. Doing impulse/time I've calculated that the force would be around 6 to 7 N. However, I have no idea if that is a good approximation. In wikipedia they say that a 800 m square sail would catch around 5 N of force.

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  • $\begingroup$ First, I think it's more an "absorption of energy" than collision. Second, shouldn't it be $p=\frac{E}{c^2}$? $\endgroup$ – CoilKid May 25 '15 at 3:22
  • $\begingroup$ J = kg (m/s)^2. If I divide by v squared I'm left with kg. $\endgroup$ – 0 kelvin May 25 '15 at 5:17
  • $\begingroup$ Keep in mind that the solar wind is not made up of photons. $\endgroup$ – WhatRoughBeast May 25 '15 at 10:15
  • $\begingroup$ @WhatRoughBeast The force exerted by solar radiation is much stronger than that exerted by the solar wind. $\endgroup$ – Rob Jeffries May 25 '15 at 14:38
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You are making life difficult for yourself by thinking in terms of photons. Yes, in principle you can say that the momentum carried by a photon is $E/c$, where $E$ is the photon energy, and you can then say that for a perfect reflection, there is an exchange of momentum with the sail of $2E/c$. Then multiply by how many photons strike the sail per second and that's your answer. The problem is that the Sun emits photons with a spectrum of energies. I'll come back to this approach at the end, but here is an easier one.

The momentum per second carried by radiation can also be written as $$ F = \frac{1}{c} \int \vec{S} \cdot d\vec{A}, $$ where $\vec{S}$ is the Poynting vector (the power per unit area) carried by the radiation and it is integrated over the area of the sail. For a perfect reflection, you multiply this expression by 2 to give the net force exerted on the sail.

The Poynting vector from the Sun is just $$ \vec{S} = \frac{L}{4 \pi r^2} \hat{r},$$ where $L$ is the total luminosity of the Sun, $r$ is the distance from the Sun and it points radially away from the Sun. Thus $$F_{sail} = \frac{2}{c} \frac{L}{4\pi r^2} A \cos \theta,$$ where $A \cos \theta$ is the projected area of the sail ($\theta$ is the angle between the normal to the sail area and the Sun).

Let's assume $r=1.5 \times 10^{11}$ m, $L= 3.83\times10^{26}$ W, $A=6.4\times 10^5$ m$^2$ (the wikipedia page says the sail is 800 m on a side)and $\theta=0$. In this case, I get $F_{sail} = 5.8$ Newtons.

Back to photons. The solar spectrum is dominated by visible light. If we make the assumption that all the solar photons are visible, with a wavelength of 500 nm, then each photon carries $4 \times 10^{-19}$ J. The Sun must therefore emit approximately $9.6 \times 10^{44}$ of these per second. At a distance $r$ from the Sun, the sail intercepts a fraction $A/4\pi r^2$ of these photons and receives a momentum $2E/c$ from each one. Using $A=6.4\times 10^{5}$ m$^2$ and $r=1.5\times10^{11}$ m again, I get that the exchanged momentum per second ($F_{sail}$) is $5.8$ Newtons, as before.

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A similar problem is about a flux of bullets hitting superman. By the law of conservation of momentum, it's true that the change in momentum per bullet is $\Delta p = 2p$. If the flux is given in $\frac{n \text{ bullets}}{\text{minutes}}$, I can calculate the change in momentum per bullet then multiply by n bullets and divide by the minutes (converted to seconds) given to find F.

In the solar sail case I had two pieces of information: flux in photons per second and the energy of a photon. But $F = \frac{\Delta p}{\Delta t}$ was yielding an F with an order of magnitude of $10^{-3}$, which is too low. I suspect that the data was wrong, photons per second should had been photons per minute or something.

In either case, the problem is one dimensional, so no area of impact is discussed.

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I don't know that this is the correct way to do this, but here it goes.

I would start by using the equation $$p=\frac{E}{c}$$ to find the momentum of the photon, and then apply the law of conservation of momentum to find the resulting velocity of your sail...

Example:

$p_{photon}=1kg*m/s$ (for simplicity, not a realistic value)

$m_{sail} = 10kg$

then applying the law of conservation of momentum:

$$p_{photon}= p_{sail}$$ $$p_{sail}=mv$$ $$v=\frac{p_{sail}}{m}$$ $$v=\frac{1kg*m/s}{10kg}$$ $$v=0.1m/s$$

and then figure out the rate at which photons are hitting the sail per time period to work out the total number of photon impacts...

$t=10s$

$r=10,000photons/second$ (again, not a realistic value) Edit: This is across the entire area of the sail. Obviously, a realistic sail with a realistic number of photon impacts would have a much higher value here.

$$N_{photons}=photons/second*seconds$$ $$N_{photons}=100,000photons$$

figure out the total velocity increase over a time period, and then use $F=ma$ to solve for force. $$a_{sail}=\frac{10,000m/s}{10s}$$ $$a_{sail}=1,000m/s^2$$ $$F_{photons}=m_{sail}a_{sail}$$ $$F_{photons}=10,000N$$ Realistically, the momentum of the photon should be orders of magnitude smaller, and therefore the force should be smaller as well. Hope that helps!

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  • $\begingroup$ Some quick calculating using rough estimates and photon energies from the blue band, I get something like $5x10^-28N$ for a 10kg sail with a photon impact rate of a billion photons per second. $\endgroup$ – CoilKid May 25 '15 at 4:07
  • $\begingroup$ C^2 is incorrect $\endgroup$ – Jimmy360 May 25 '15 at 5:38
  • $\begingroup$ You can use a photon energy of approx. 1.5eV for the green part of the spectrum and then use 1eV = 1.6e-19J. Your estimates for the photon flux on a small sail are 15 orders of magnitude off. :-) $\endgroup$ – CuriousOne May 25 '15 at 5:41
  • $\begingroup$ $E = pc^2$ is dimensionally incorrect. $\endgroup$ – Rob Jeffries May 25 '15 at 14:54
  • $\begingroup$ @RobJeffries Isn't the one taking momentum into account $E^2=m^2c^4+p^2c^4$? Assuming no mass, that should become $E=pc^2$? $\endgroup$ – CoilKid May 25 '15 at 15:09

protected by AccidentalFourierTransform Dec 8 '17 at 15:42

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