Force on Earth by the radiation pressure of Sun

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SUMMARY

The force on Earth due to solar radiation pressure is estimated using the intensity of sunlight at Earth's orbit, which is 1.37 kW/m². The radiation pressure is calculated as 4.5667e-6 N/m². The most accurate approach to determine the force involves using the effective area of the Earth, approximated as πr², rather than the total surface area. This method yields a force of approximately 5.823e8 N, assuming that the radiation is absorbed rather than reflected.

PREREQUISITES
  • Understanding of radiation pressure and its calculation (I/c)
  • Familiarity with basic physics equations (Pressure=F/A)
  • Knowledge of Earth's surface area (510,072,000 km²)
  • Concept of effective area in relation to solar radiation (πr²)
NEXT STEPS
  • Research the calculation of radiation pressure in different contexts
  • Study the effects of solar radiation on planetary bodies
  • Explore the implications of radiation pressure in astrophysics
  • Learn about the absorption and reflection of solar radiation by Earth
USEFUL FOR

Students in physics, astrophysics researchers, and anyone interested in understanding the effects of solar radiation on planetary dynamics.

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Homework Statement


Estimate the force on Earth due to the pressure of the radiation on Earth by the Sun. (At Earth's orbit, the intensity of sunlight is 1.37 kW/m^2.)

Homework Equations


Radiation Pressure=I/c
Pressure=F/A
Using SA of earth=510,072,000 km2

The Attempt at a Solution


So I have been working at this problem for a while now and can't seem to get the correct answer. I found the radiation pressure to be 4.5667e-6 N/m^2. I have since then tried taking this and multiplying it by the surface area of the Earth in many different forms to no avail. What am I missing? I have tried these coefficients to Prad*SA, 1 (acting on whole surface), 1/2 (half the surface since sun is on one side), 1/(2sqr2) (thought possibly rms), 1/sqr2 (similar to before, just trying anything), and 1/(4sqr2), 1/4. My force for Prad*SA=2.329e9 N.
 
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Welcome to PF!
I agree with 4.5667e-6 N/m^2.
I'm puzzled with what area to use. The sun only shines on one side of the Earth at a time so you certainly wouldn't use the whole surface area. Half of it doesn't seem right either because the sun hits a good part of it at a very high angle, drastically reducing the intensity. I think taking the area to be πr² is the best first estimate because it counts all of the radiation hitting the Earth. Of course all this assumes that the radiation is absorbed rather than reflected.
 
Thank you for welcoming me and thank you for the help, using just pi*r^2 worked. 5.823e8 N is the answer.
 

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