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  1. Problems
  2. /A sail has an area of 4000 m² and a total mass of 18 kg…
MediumPhysicsHigh School (11–12)Waves & Sound › Electromagnetic WavesNumericalNEETBoard ExamAPA-LevelIBAEEE (Amrita)
+19 more exams
AMUEEEAP EAPCETBCECECOMEDK UGETCUET UGGUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEE

A sail has an area of 4000 m² and a total mass of 18 kg…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A sail has an area of 400040004000 m² and a total mass of 181818 kg. It faces the Sun straight on at 0.80.80.8 of the Earth's distance from the Sun. It reflects a fraction 0.850.850.85 of the light and absorbs the rest. Near the Earth the intensity of sunlight is 136113611361 W/m², and it falls as 1/d21/d^21/d2. Use c=3.00×108c = 3.00 \times 10^{8}c=3.00×108 m/s. Find:

  1. the radiation pressure on the sail,
  2. the force on it,
  3. its acceleration,
  4. its speed after 202020 days, if the push stays constant.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

µPa
mN
mm/s²
km/s
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The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The intensity there. I=1361 W/m2(0.8)2=2126.6 W/m2I = \dfrac{1361\ \text{W/m}^2}{(0.8)^2} = 2126.6\ \text{W/m}^2I=(0.8)21361 W/m2​=2126.6 W/m2.

Step 2. The pressure. p=(1+ρ)Ic=(1+0.85)×2126.6 W/m23.00×108 m/s=13.1 μPap = \dfrac{(1 + \rho)I}{c} = \dfrac{(1 + 0.85) \times 2126.6\ \text{W/m}^2}{3.00 \times 10^{8}\ \text{m/s}} = 13.1\ \mu\text{Pa}p=c(1+ρ)I​=3.00×108 m/s(1+0.85)×2126.6 W/m2​=13.1 μPa.

Step 3. The force. F=pA=13.1×10−6 Pa×4000 m2=52.5 mNF = pA = 13.1 \times 10^{-6}\ \text{Pa} \times 4000\ \text{m}^2 = 52.5\ \text{mN}F=pA=13.1×10−6 Pa×4000 m2=52.5 mN.

Step 4. The acceleration. We want aaa, so a=Fm=0.0525 N18 kg=2.91 mm/s2a = \dfrac{F}{m} = \dfrac{0.0525\ \text{N}}{18\ \text{kg}} = 2.91\ \text{mm/s}^2a=mF​=18 kg0.0525 N​=2.91 mm/s2.

Step 5. The speed. The time is 20×86 400 s=1.728×106 s20 \times 86\,400\ \text{s} = 1.728 \times 10^{6}\ \text{s}20×86400 s=1.728×106 s. So v=at=5.04 km/sv = at = 5.04\ \text{km/s}v=at=5.04 km/s.

Answer: 13.1 μPa13.1\ \mu\text{Pa}13.1 μPa; 52.552.552.5 mN; 2.912.912.91 mm/s²; 5.045.045.04 km/s.

Watch out. The push stays constant only if the sail stays at the same distance. In truth, the sail moves, and the push changes.

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  1. Momentum and radiation pressureEnergy, Intensity and Momentum of Electromagnetic Waves

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