### Video Transcript

The diagram shows an artist’s conception of a solar sail. A solar sail is a thin, light, highly reflective sheet that can be used as a method of propulsion for spacecraft. The solar sail reflects light from a nearby star. A net force is exerted on the sail due to radiation pressure. If a spacecraft powered by a solar sail sets sail from Earth, traveling away from the Sun, what area would the solar sail have to have in order for the force exerted on it due to radiation pressure to be 1.00 newtons? The intensity of light from the Sun on Earth is 1350 watts per meter squared. Assume that all of the light that hits the sail is reflected. Use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum.

Okay, so we can see that we’ve got a picture showing a solar sail. That’s this square here. The question tells us that a solar sail acts as a method of propulsion for spacecraft and that it works by reflecting a light from a nearby star. When the light reflects, a net force is exerted on the sail as a result of radiation pressure. We are being asked to work out what area the sail needs to have in order for this force to have a magnitude of 1.00 newtons. Let’s begin by clearing some space and summarizing the key information given to us in the question.

Okay, first off, we know that a force of 1.00 newtons is exerted on the sail due to radiation pressure. The question told us that the solar sail set sail from Earth and that it was traveling away from the Sun. We were also told that the intensity of the light from the Sun at the location of Earth was equal to 1350 watts per meter squared. And since the solar sail is moving in the direction away from the Sun, starting at the location of Earth, we can assume that the light with this intensity is directed at the sail. In other words, if this here is the solar sail, then the light waves are coming in like this, perpendicular to the surface. We are also told that we can assume 100 percent reflection of light that’s incident on the solar sail. And finally, we are told to use a value for the speed of light in vacuum of 3.00 times 10 to the eight meters per second.

We are being asked to find the area of the solar sail, which we’ll label as 𝐴. To do this, let’s begin by recalling why it is that light incident on the sail exerts a force on it. We can recall that even though light waves don’t have mass, they can still transfer momentum. So if we have a load of light waves like this one colliding with and reflecting off a surface, then those light waves experience a change in momentum. This momentum change means that there is a force, since whenever something’s momentum changes by an amount Δ𝑝 over a time Δ𝑡, then there’s a force 𝐹 equal to Δ𝑝 divided by Δ𝑡. So when the light waves reflect off the solar sail, they must be exerting a force on its surface.

We are trying to work out the area 𝐴 of the solar sail. We can recall that whenever a force 𝐹 acts over a surface with an area 𝐴, then there’s a pressure on that surface, capital 𝑃, equal to the force 𝐹 divided by the area 𝐴. This pressure is precisely the radiation pressure that we are told about in the question. We can recall that for a surface which reflects 100 percent of the light incident on it, the radiation pressure 𝑃 exerted by the light on the surface is equal to two times the intensity of the light, 𝐼, divided by the speed of light, 𝑐. Since we were told in the question that we can assume 100 percent reflection of light, that means that this equation does apply to our solar sail.

We know that the force exerted by the light is 1.00 newtons. So in this equation, that value of 1.00 newtons is our value for 𝐹. We also know that the light’s intensity is 1350 watts per meter squared. So that’s our value for the quantity 𝐼 in this equation for the radiation pressure. Finally, we are told to use a value of 3.00 times 10 to the eight meters per second for the speed of light 𝑐. We are trying to calculate the value of 𝐴 in this equation. So to do that, we need the values of the quantities 𝑃 and 𝐹. We already know the value of the force 𝐹. At the moment, we don’t know the value of the radiation pressure 𝑃. However, we do know the value of the intensity 𝐼 and the speed of light 𝑐. So we can use those values in this equation in order to calculate the radiation pressure.

Let’s go ahead and sub our values for 𝐼 and 𝑐 into this equation. When we do this, we get that 𝑃 is equal to two times 1350 watts per meter squared divided by 3.00 times 10 to the eight meters per second. We can notice that the light intensity is given in units of watts per meter squared, which is the SI base unit for intensity. And the speed of light is in units of meters per second, the SI base unit for speed. This means that the radiation pressure 𝑃 that we’ll calculate will be in the SI base units for pressure. So that’s units of newtons per meter squared. When we evaluate this expression, we find that the radiation pressure 𝑃 is equal to nine times 10 to the negative six newtons per meter squared. So this value for 𝑃 is the radiation pressure that’s exerted on the solar sail by light from the Sun.

We now want to calculate what area 𝐴 that sail must have in order for this pressure to result in a force 𝐹 of 1.00 newtons. Let’s take this equation here and rearrange it to make the area 𝐴 the subject. So we’ll start with our equation 𝑃 is equal to 𝐹 divided by 𝐴. The first step is to multiply both sides of the equation by the area 𝐴. Then, on the right-hand side, there’s an 𝐴 in the numerator which cancels with the 𝐴 in the denominator, leaving us with 𝐴 multiplied by 𝑃 is equal to 𝐹. Then, we divide both sides of the equation by the radiation pressure 𝑃. On the left-hand side, the 𝑃 in the numerator cancels with the 𝑃 in the denominator. This gives us an equation that says the area 𝐴 is equal to the force 𝐹 divided by the radiation pressure 𝑃.

We can now take our values for the quantities 𝐹 and 𝑃 and sub them into this equation in order to calculate the area 𝐴 of the solar sail. When we do this, we get that the area 𝐴 is equal to 1.00 newtons divided by nine times 10 to the negative six newtons per meter squared. Evaluating this expression, we calculate an area 𝐴 of 1.1 recurring times 10 to the five meters squared. Let’s round this result to two decimal places matching the values for the force and the speed of light given to us in the question. This gives us our answer to the question that the area of the solar sail must be equal to 1.11 times 10 to the five meters squared.