### Video Transcript

Radiation known as the solar wind is incident on the top of Earth’s atmosphere with an average intensity of 1.30 kilowatts per meter squared. Suppose that you are building a solar sail that will use the radiation pressure of the solar wind to propel a small toy spaceship with a mass of 0.100 kilograms. The sail is made from a material which perfectly reflects radiation of the solar wind. To assess whether such a project is feasible, answer the following questions, assuming that radiation photons are incident only in normal direction to the sail reflecting surface. What is the radiation pressure of the solar wind on the solar sail? How much acceleration of the toy spaceship would the solar sail produce if the sail’s area is 10.00 meters squared? How much velocity increase of the toy spacecraft would the sail produce in one day’s acceleration? Consider one day as 86.4 times 10 to the third seconds.

In this three-part problem, we’ll call the radiation pressure in part one we want to solve for capital 𝑃. In part two, we’ll call the acceleration of the toy spaceship 𝑎. And in part three, we’ll call the toy spacecraft’s increase in velocity Δ𝑣.

To begin solving for these values, let’s draw a diagram of this scenario. Say we have a picture of Earth and then a dotted line representing the outer edge of Earth’s atmosphere. This is where our solar sail and the payload of the toy spaceship exist. This sail is designed to receive radiation from the sun, called the solar wind, that powers the spaceship and moves it through space.

In the problem statement, we’re told that, at this point in Earth’s atmosphere, the radiation from the sun has an intensity 𝐼 of 1.30 kilowatts per meter squared. To solve for the radiation pressure this solar wind exerts on the sail, we can recall the mathematical relationship for this type of pressure.

Pressure in general is defined as a force spread over an area 𝐴. And with regard to radiation, this pressure is equal to the intensity 𝐼 of that radiation over the speed of light. This particular formulation — radiation pressure equals 𝐼 over 𝑐 — is true when the material receiving radiation absorbs it.

On the other hand, if the material reflects the radiation, a factor of two appears in this equation, because now instead of photon momentum simply being stopped as in the case of an absorbent material, it’s reversed. So the pressure felt by the material is twice as much.

In the case of our solar sail, we’re told in the problem statement that it perfectly reflects radiation from the sun. So when we write an equation for radiation pressure on our solar sail, we include the factor of two. Looking at this equation, we see that 𝐼, the intensity of the sunlight, is given in the problem statement. The speed of light we’ll treat in this problem as exactly 3.00 times 10 to the eighth meters per second.

So we’re now ready to plug in and solve for 𝑃, the radiation pressure. When we do, we’re careful to use a value in units of watts per meter squared for our intensity, SI units. Entering these values on our calculator, we find that the radiation pressure experienced by the sail is 8.67 times 10 to the negative six newtons per meter squared. Notice that this value has units of force per area consistent with our definition of pressure.

We can now move on to solving for the acceleration that the solar sail and toy spaceship together experience due to this radiation pressure. If we look back at our equation for radiation pressure, we see that force equals pressure times area. And we can recall we’re told the area of the sail in the problem statement. Calling that area 𝐴, it’s equal to 10.00 meters squared.

Now we can recall an equation that relates force with acceleration. Newton’s second law says that the net force acting on an object equals its mass times its acceleration. If we assume that the force due to the solar wind is the only force acting on our sail and toy spaceship, then we can write that that force which equals pressure times area also equals mass times acceleration. If we rearrange this equation to solve for 𝑎, we see that it’s pressure times area over mass.

In the statement of the problem, we’re told that the mass of the toy spaceship is 0.100 kilograms. We’re also given 𝐴 in the problem statement. And we’ve solved for the pressure 𝑃. So we’re now ready to plug in and solve for acceleration 𝑎.

When we enter these values on our calculator, we find that, to three significant figures, the acceleration is 8.67 times 10 to the negative fourth meters per second squared. That’s how much the toy spaceship accelerates due to the solar wind.

We now lastly want to solve for the change in velocity that the spaceship experiences over the course of one day, 𝑡 equals 86.3 times 10 to the third seconds, due to the acceleration that it undergoes, caused by the radiation pressure. We can recall that acceleration 𝑎 is defined as a change in velocity Δ𝑣 over change in time.

Rearranging this equation and expressing it in terms of our variables, that means that Δ𝑣 equals 𝑎 times 𝑡. Since we’ve solved for 𝑎 and we’re given 𝑡, we’re now ready to plug in and solve for Δ𝑣. When we do, as we look at the units, notice that a factor of seconds cancels out. And our final answer will have units of meters per second, the units of velocity.

Calculating this value, we find that it’s equal to 74.9 meters per second. That’s how fast the spaceship would be moving after one day of being accelerated from rest by the solar wind.