Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Video: Force Exerted by Radiation Pressure

Ed Burdette

Sunlight falls perpendicularly on a kite with a reflecting surface of area 0.75 m². The average intensity of the sunlight is 900 W/m². What is the average force on the kite due to radiation pressure? If the kite material is black and absorbs all sunlight, what is the average force on the kite due to radiation pressure?

03:55

Video Transcript

Sunlight falls perpendicularly on a kite with a reflecting surface of area 0.75 metre squared. The average intensity of the sunlight is 900 watts per metre squared. What is the average force on the kite due to radiation pressure? If the kite material is black and absorbs all sunlight, what is the average force on the kite due to radiation pressure?

In this statement, weโ€™re told that the kite has an area of 0.75 metre squared; we will call that ๐ด. Weโ€™re also told that in this case the intensity of the sunlight is 900 watts per metre squared; we will call that value ๐ผ. We want to know the average force on the kite due to the radiation pressure when the kite reflects the light; we will call that ๐น sub ๐‘Ÿ. And then if instead of reflecting, the kite absorbs the sunlight, we want to know that average force on the kite again due to the radiation pressure; we will call that value ๐น sub ๐‘Ž.

Letโ€™s start our solution by drawing a diagram of this situation. In this first scenario, we have sunlight hitting a kite perpendicularly and then being reflected back by the kite. This sunlight exerts a pressure on the kite called radiation pressure, which is due to the fact that photons have momentum. This pressure which is equal to a force spread over an area is equal to the intensity of the incident light ๐ผ divided by the speed of light ๐‘.

Thatโ€™s true when the material is an absorbing material. If the material is perfectly reflecting instead, we add a factor of two to the equation for radiation pressure; thatโ€™s because the photons incident on a surface experience two times the change in momentum when theyโ€™re reflected rather than when they are absorbed. So when we apply this relationship to our scenario, in the first case since we have a reflecting material on our kite, we include the factor of two in our equation.

When we rearrange to solve for ๐น sub ๐‘Ÿ, the force the kite experiences when it reflects incoming radiation, that force is two times the incident intensity of radiation ๐ผ times the kite area ๐ด all over the speed of light ๐‘, which weโ€™ll assume is exactly 3.00 times 10 to the eighth metres per second. When we plug these values in to this equation for ๐น sub ๐‘Ÿ and enter these values on our calculator, we find that ๐น sub ๐‘Ÿ is equal to 4.5 times 10 to the negative sixth newtons. Thatโ€™s the force the sunlight exerts on the kite when the kite reflects the sunlight back.

Next, we turn to considering what force the light would exert on the kite if instead of reflecting the light back, the kite merely obsorbs the light. In this case, our radiation pressure equation would not include the factor of two. And when we rearrange for the force ๐น sub ๐‘Ž, it would be equal simply to ๐ผ times ๐ด, the area, divided by the speed of light ๐‘.

When we plug in for these three values and enter the values on our calculator, we find that ๐น sub ๐‘Ž equals 2.3 times 10 to the negative sixth newtons. So the force when the photons are absorbed is half as much as it is when theyโ€™re perfectly reflected.