### Video Transcript

A large asteroid is on a collision course with Earth. The asteroid has a mass of 300000 kilograms. The asteroid can be deflected if a 0.010-newton force can be applied to it for 2.0 times 10 to the six seconds. If a powerful laser is used to shine light on a 12-meter-squared area of the asteroid, what is the intensity of light needed in order to deflect the asteroid using radiation pressure. Assume that the surface of the asteroid is 100 percent reflective. Use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum. Give your answer in scientific notation to one decimal place.

In this question, we have an asteroid which is headed toward Earth. And the goal is to deflect that asteroid using radiation pressure. We can draw a sketch of this situation showing Earth and the asteroid which is headed toward it. Since the aim is to deflect this asteroid using radiation pressure, let’s recall how radiation pressure is defined.

The radiation pressure exerted by light on a surface is typically labeled as capital 𝑃. And this radiation pressure 𝑃 is equal to two 𝐼 divided by 𝐶, where 𝐼 is the intensity of the light and 𝐶 is the speed of light. It’s important to note that this equation only applies for perfectly reflective surfaces. Luckily for us, we’re told in the question that we can assume that the surface of the asteroid is 100 percent reflective, which means we can indeed use this equation.

We’re asked to work out the intensity of light needed in order to deflect the asteroid. So what we’re trying to do is find the value of 𝐼. We’re not directly told anything about the value of the pressure 𝑃. However, we are told that we need a force with a magnitude of 0.010 newtons in order to deflect the asteroid. And we’re also told that our laser can be used to shine light over a 12-meter-squared area of the asteroid.

We have indicated this area on the asteroid in our sketch and labeled the area as capital 𝐴, so we have capital 𝐴 is equal to 12 meters squared. We’ll also label the required force as capital 𝐹 so that we have capital F is equal to 0.010 newtons. We can recall that pressure is defined as force per unit area. We can express this statement mathematically as pressure 𝑃 is equal to force F divided by area 𝐴. Since we know the force F required in order to deflect the asteroid and we know the area 𝐴 over which we can apply this force, then we can substitute those values into this equation in order to calculate the required value of the radiation pressure 𝑃.

Doing these substitution gives us that 𝑃 is equal to 0.010 newtons divided by 12 meters squared. Since the force is given in its SI base unit of newtons and the area is given in its SI base unit of meters squared, then the pressure that we calculate will also be expressed in its own SI base unit. The SI base unit for pressure is the pascal. And doing the division gives us a pressure of 0.00083 pascals where the bar over the three is used to indicate that this digit is recurring.

If we now look back at our equation for the radiation pressure in terms of the intensity of the light, we see that we now know the value of the radiation pressure 𝑃. And we’re told to use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum. So that’s our value of 𝐶. What this means is that if we rearrange this equation in order to make the intensity 𝐼 the subject, then we now have all of the information that we need in order to calculate this intensity.

So let’s take this equation and rearrange it to make 𝐼 the subject. First, we’ll multiply both sides of the equation by 𝐶, the speed of light. On the right-hand side, the 𝐶 in the numerator cancels with the 𝐶 in the denominator. And so we have that 𝐶 multiplied by 𝑃 is equal to two 𝐼. Then we divide both sides of the equation by two. Again, on the right-hand side, the two in the numerator cancels with the two in the denominator. And so we have that 𝐶 multiplied by 𝑃 divided by two is equal to 𝐼. We can also write this equation as 𝐼 is equal to 𝐶 multiplied by 𝑃 divided by two.

All that’s left to do is to substitute in our values for 𝐶 and 𝑃 into this equation. Doing this gives us that the intensity 𝐼 is equal to 3.00 times 10 to the eight meters per second, so that’s our value for the speed of light 𝐶, multiplied by 0.00083 pascals, that’s our radiation pressure 𝑃, divided by two. Evaluating this expression gives us that 𝐼 is equal to 1.25 times 10 to the seven watts per meters squared. Notice that the speed 𝐶 was given in its SI base unit and we had a pressure 𝑃 also in its SI base unit. And so we got an intensity 𝐼 with units of watts per meter squared, which is the SI base unit for intensity.

The last thing to notice is that we were told to give our answer in scientific notation to one decimal place. We already have the value in scientific notation, so we just need to round it to one decimal place of precision. Doing this gives us our final answer to the question that the intensity of light needed in order to be able to deflect this asteroid using radiation pressure is equal to 1.3 times 10 to the seven watts per meter squared.

Now that we’ve reached this answer, it’s worth pointing out that the question actually gave us some extra information that we didn’t even need. Namely, we were told the mass of the asteroid and the time for which the force was applied. This information was simply given to us in order to distract us and to make us think that perhaps we needed to use it.