# Question Video: Exploring the Use of Photon’s Momentum as a Source of Thrust Physics • 9th Grade

A solar sail is a proposed method of spacecraft propulsion that uses the momentum of photons as a source of thrust. When photons hit the sail, they are absorbed and their momentum is transferred to the sail. If laser light with a wavelength of 200 nm is used to propel the sail, how many photons must hit the sail for it to gain 1 kg⋅m/s of momentum? Use a value of 6.63 × 10⁻³⁴ J⋅s for the Planck constant. Give your answer to 3 significant figures.

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### Video Transcript

A solar sail is a proposed method of spacecraft propulsion that uses the momentum of photons as a source of thrust. When photons hit the sail, they are absorbed and their momentum is transferred to the sail. If laser light with a wavelength of 200 nanometers is used to propel the sail, how many photons must hit the sail for it to gain one kilogram meter per second of momentum? Use a value of 6.63 times 10 to the negative 34th joule seconds for the Planck constant. Give your answer to three significant figures.

All right, so in this example, we have this structure called a solar sail. A solar sail is a surface, we could sketch it like this, that is designed to absorb or reflect incoming photons. The idea is that every incoming photon has some amount of momentum. And when those photons are, say, absorbed by the sail, that momentum is transferred to the sail. The effect of this is to create an overall force on the sail and the spacecraft it’s attached to. This force can move spacecraft through space. And so, indeed, these incoming photons are a source of thrust.

Now, in our example, these incoming photons that land on the sail are provided by a laser. The wavelength of these photons, we can call it 𝜆, is equal to 200 nanometers. We want to know just how many photons at this wavelength, and we’ll call that number capital 𝑁, would be needed to deliver a gain in momentum to the solar sail of one kilogram meter per second. We can refer symbolically to that gain in momentum as Δ𝑃.

To figure out how many photons are needed to create this change in momentum in our sail, we’ll need to know how much momentum each one of our individual photons with a wavelength of 200 nanometers possesses. Written as an equation, this amount of momentum for a single photon is given as Planck’s constant divided by the wavelength of that photon. In our problem statement, we’re told to treat Planck’s constant as exactly 6.63 times 10 to the negative 34th joule seconds.

So knowing ℎ and knowing 𝜆, we’ll be able to calculate the momentum of a single photon emitted by our laser. But recall that what we want to do is figure out how many of these photons will be necessary to impart a gain in momentum of one kilogram meter per second to the sail. To see how to calculate that, let’s clear a bit of space on screen.

Okay, much better. Now, like we said, we want to solve for capital 𝑁, the number of photons emitted by our laser in order to impart this change in momentum to our sail. Since Δ𝑃 is the total momentum change of our sail, we can say that it’s equal to the momentum of one of the photons emitted by the laser multiplied by the total number of those photons. We can write this because each one of our individual photons all have the same wavelength, 200 nanometers. Therefore, each one provides the same amount of momentum to the sail.

Now, just as an interesting side note, notice that our photons are being absorbed by the sail. This is in contrast to being reflected. If the photons were reflected back, then that means they would impart twice as much momentum change to our sail. And that’s because rather than simply having their forward momentum being stopped, like it is when they’re absorbed, reflection would indicate that that momentum is reversed in the opposite direction. This would double the change in momentum experienced by the sail.

And therefore, if our sail reflected photons, then over here at our equation from the momentum of a single photon, we would multiply this by two. That’s because ℎ over 𝜆 of momentum would be transmitted to the sail when a photon is stopped. And then that same amount would be transmitted when the photon is sent back the way it came. This is why we have a factor of two whenever our sail is reflective material.

But since our sail in this example is not reflective, but it absorbs photons, we won’t use this factor of two in our equation. In this case, the momentum transmitted to this sail by an individual photon truly is equal to ℎ over 𝜆. Plugging that expression in for 𝑃 sub p, we now have this equation where it’s capital 𝑁 that we want to solve for.

To do that, we can multiply both sides of the equation by 𝜆 over ℎ. This cancels out both 𝜆 and ℎ on the right-hand side. And we find that the wavelength of our photons divided by Planck’s constant all multiplied by the total gain in momentum of our sail is equal to the number of photons incident on it.

Our next step is to substitute in the values in the left-hand side of this equation. But before we plug in 200 nanometers for our wavelength, let’s change the units here from nanometers into meters. We can do this by recalling that one nanometer is equal to 10 to the negative ninth meters. Which means that we can reexpress 𝜆 as 200 times 10 to the negative ninth meters. So we’ll substitute that value in for 𝜆 in our equation. And we’ll replace ℎ with 6.63 times 10 to the negative 34th joule seconds. And Δ𝑃, we know, is equal to one kilogram meter per second.

Looking at this expression, we can note that a joule is equal to a newton times a meter. And that a newton is equal to a kilogram meter per second squared. Which means we can replace the unit joule with a kilogram meter squared per second squared. The value in doing this is that now we see that the units on the left-hand side of this expression entirely cancel out. That is, kilograms cancel out with kilograms. Meters squared in the numerator cancels out with meters squared in the denominator. And after canceling one factor of seconds in the denominator, all that’s left is one over seconds in the numerator and denominator. So that cancels as well. We’re left with a unitless result, which is good because we’re looking for a pure number.

To three significant figures, 𝑁 is 3.02 times 10 to the 26th. That’s the number of photons that would need to land on the sail in order to impart one kilogram meter per second of momentum gain to it.