Astronomy/energy from stars

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Question
I was debating with someone the other day about where all the aggregate energy from the stars goes.  We were hung up on the question - if there are so many stars in the universe, and each emits energy, why aren't we all fried on earth?  This is based on the assumption that there must be enough energy emitted from all the stars to make this plausible, which may or may not be correct.  If it is, we concluded, than most of the energy goes to expanding the universe AND/OR the aggregate energy's intensity is lessened as it encounters black holes, particles in space, etc.  Is this true?  Are there any other explanations?  

Thanks,

ML: Washington DC  

Answer
Well, the DISTANCE is a big factor in this.  The Sun, which is only 93 million miles away, doesn't even "fry" us, so the energy coming from stars hundreds, thousands, and millions of LIGHT YEARS away, has no effect on us.  Their energy is strong, but not from these distances.

Steve

Astronomy

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Steven LJ Russo

Expertise

As a space science educator in a planetarium, my work centers around teaching people about the night sky and the solar system. I have a strong background in those areas, plus the history of NASA and spaceflight, and meteorology.

Experience

Experience in the area. I have been an amateur astronomer for 47 years, and have been teachng space science in planetariums for 34 years. For 15 years I was a radio and television meteorologist, and for the past 20 years I have been a space science writer for two newspapers in New York State. I am a member of the Middle Atlantic Planetarium Society and the International Planetarium Society. I have had a number of articles published in several astronomy journals, including "The Constellation" and the "Planetarian". Education/Credentials. I hold a B. S. from Wagner College and an M. S. from State University of New York at Oswego. Awards and Honors. I have been awarded the "Fellows" award from the International Planetarium Society for more than 20 years of continuous service in the planetarium field.

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