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Question: The average intensity of the solar radiation that strikes normally on a surface just outside Earth’s atmosphere is 1.4kw/m2.(a) What radiation pressure Pr is exerted on this surface, assuming complete absorption?(b) For comparison, find the ratio of Pr to Earth’s sea-level atmospheric pressure, which is1.0×105 P²¹

Short Answer

Expert verified
  1. pr=4.7×10−6 N/m2
  2. prPe=4.7×10−11

Step by step solution

01

Step 1: Given data

The average intensity of solar radiation is,

I=1.4×103 W/³¾2

Earth’s sea level atmospheric pressure is,

Pe=1.0×105 P²¹

02

Determining the concept

Radiation pressure depends upon the intensity of radiation and the speed of light. Radiation pressure is the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field. This includes the momentum of light or electromagnetic radiation of any wavelength that is absorbed, reflected, or otherwise emitted by matter on any scale.

Formulae are as follows:

Radiation pressure (Pr):

pr=Ic

Where,I−intenisity(Wm2), Pr is the radiation pressure, c is the speed of light.

03

(a) Determining theradiation pressure Pr is exerted on this surface, assuming complete absorption

To find the radiation pressure Pr, we can use the formula as,

pr=Ic

Substitute the values in the above expression, and we get,

pr=1400 W/³¾23×108″¾/²õpr=4.7×10−6â‹…(1 W/³¾21″¾/²õ×1 Nâ‹…m/s1 W)pr=4.7×10−6 N/³¾2

Therefore, the radiation pressure is

pr=4.7×10−6 N/³¾2

04

(b) Determining theratio of Pr to Earth’s sea-level atmospheric pressure, which is 1.0×105 Pa for comparison. 

The ratio of pressure Pr to Earth’s sea level atmospheric pressure can be calculated as,

prPe=4.7×10−61×105prPe=4.7×10−11

Therefore, the ratio of pressure Pr to Earth’s sea level atmospheric pressure is prPe=4.7×10-11.

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