Calculating Intensity of Sunlight Reaching Jupiter

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Homework Help Overview

The problem involves calculating the intensity of sunlight reaching Jupiter, given that it is 5.4 times farther from the Sun than Earth, with an initial intensity value provided for Earth. The context is based on the principles of light intensity and distance from a source, treating the Sun as an isotropic emitter.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between intensity and distance from the Sun, questioning whether the area of Earth or Jupiter is relevant to the calculation. There are attempts to clarify the concept of the area involved in the problem, particularly the surface area of a sphere at a given distance from the Sun.

Discussion Status

Some participants have provided hints regarding the conservation of energy and the spherical nature of light propagation. There is an ongoing exploration of the assumptions regarding the areas involved, with some clarification offered about focusing on the area of the sphere at the distance of the planets rather than the planets themselves.

Contextual Notes

Participants note the lack of specific information about the distances from the Earth and Jupiter to the Sun, which may be necessary for calculations. There is also mention of potential confusion regarding the mathematical approach to the problem.

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Homework Statement



The intensity of sunlight that reaches the Earth's atmosphere is about 1500 W/m2. What is the intensity of the sunlight that reaches Jupiter? Jupiter is 5.4 times as far from the sun as Earth. [Hint: Treat the Sun as an isotropic source of light waves.]
____ W/m2

Homework Equations



I = power/area?


The Attempt at a Solution



How do you do this? they don't give you that much information. Are you suppose to assume Earth's area? or jupiter's area? or does the area even matter?
 
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While the intensity will change depending on the distance from the sun, the power will not due to conservation of energy.

HINT: Find the power of the light hitting everything at the distance of Earth. This is the same as the power of the light hitting everything at the distance of Jupiter.

HINT 2: The light radiates away spherically, so the light at a distance, r, from the sun, impinges on a spherical surface of radius, r. The area you will be concerned with is the area of that surface.
 
Last edited:
does this depend on the area of Earth and jupiter?
 
Jtappan said:
does this depend on the area of Earth and jupiter?

I added an extra hint above about the area. Does that help, or are you still confused about the area?
 
so it is the area of jupiter? and is it assumed that we know the radius of Jupiter to figure out the problem?
 
Jtappan said:
so it is the area of jupiter? and is it assumed that we know the radius of Jupiter to figure out the problem?

No, we are not talking about the area of either one of the planets, but rather the area of the "imaginary sphere" centered at the sun with a radius equal to the radius of orbit of the planets.

Think about it like this:

The light that is hitting the Earth is also hitting every other point that far away from the sun, i.e. every point on a sphere of radius equal to the distance from the Earth to the sun. The area you are concerned with is the surface area of this sphere. The same goes for when the light gets to Jupiter's distance.

You should be able to look up the distance from the Earth to the Sun and from Jupiter to the Sun.

Does this help?
 
Yea thank you so much for your help! I was doing my math wrong a few times and that's where i got confused. I was doing 2piR instead of pi R^2.
 
I have another quick question that you may be able to answer...

The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at the Earth's upper atmosphere is 1.4 kW/m2. At what rate does the Sun emit electromagnetic waves? (In other words, what is the power output?)
____ WIs this the same type of problem??
 
Yes, this question follows along the same lines.
 

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