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the intensity of sunlight reaching the earth is 1360 w/m2 . part a assuming all the sunlight is absorbed, what is the radiation-pressure force on the earth? give your answer in newtons. express your answer with the appropriate units.

2 Answers

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Final answer:

The radiation-pressure force on the Earth is approximately 2.67 x 10^6 N (newtons).

Step-by-step explanation:

The radiation-pressure force on the Earth can be calculated using the formula:
Force = Intensity/Area

The intensity of sunlight reaching the Earth is given as 1360 W/m². Assuming all the sunlight is absorbed, the force can be calculated by dividing the intensity by the area of the Earth. The area of the Earth is approximately 5.10 x 10^14 m². Plugging the values into the formula:
Force = 1360 W/m² / 5.10 x 10^14 m².

The force exerted by the radiation pressure on the Earth is approximately 2.67 x 10^6 N (newtons).

User Mike Perham
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6 votes

Final answer:

To calculate the radiation-pressure force on Earth assuming all sunlight is absorbed, the radiation pressure is found using the intensity of sunlight and the speed of light in vacuum. Given the cross-sectional area of Earth, the force can be found by multiplying the pressure with the area.

Step-by-step explanation:

The intensity of sunlight reaching Earth is 1360 W/m². To calculate the radiation-pressure force on Earth, we must first assume that the sunlight is fully absorbed. The pressure due to radiation can be found using the formula p = I/c, where I is the intensity of radiation and c is the speed of light in a vacuum.

The pressure p is then 1360 W/m² / (3 x 10⁸ m/s), which gives us 4.533 x 10⁶ Pa (Pascals). To find the force, we use the formula F = p × A, where A is the cross-sectional area of the Earth perpendicular to the sunlight. The average radius of the Earth is approximately 6.371 x 10⁶ m, giving us an area A = π × (6.371 x 10⁶ m)². Multiplying the pressure by this area will provide the radiation-pressure force in newtons.

However, upon calculating the above, it is evident that this formulation does not include the momentum transfer for reflection, and since Earth reflects about 30% of the incident sunlight, an additional factor accounting for perfect absorption would be needed, and as such, the provided calculation is only partially accurate without further refinement.

User Jbyen
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