Asked by jean
the rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50kW/m2. the distance from the earth to the sun is 1.50*10^11m, and the radius of the sun is 6.96 * 10^8m. if the sun radiates as an ideal black body (e=1), what is the temperature of its surface?
Answers
Answered by
Anonymous
Radiant emittance is R=σT⁴
The solar radiation power is N=RS₁=σT⁴4π(r₁)²,
where
S₁ is the Sun surface, and r₁ is the Sun radius.
The power radiated by the Sun falls on the inner surface, which radius is equal to the distance from the Sun to the Earth - r₂. This area is equal to 4π(r ₂)².
The rate of radiant energy is K = 1500 W/m².
K=σT⁴4π(r₁)²/4π(r ₂)² =
= σT⁴(r₁/r ₂)²;
T=∜[K(r₂/r₁)²/σ]= ∜[1500(1.5•10¹¹/6.96•10⁸)²/5.67•10⁻⁸]= 5921 K
The solar radiation power is N=RS₁=σT⁴4π(r₁)²,
where
S₁ is the Sun surface, and r₁ is the Sun radius.
The power radiated by the Sun falls on the inner surface, which radius is equal to the distance from the Sun to the Earth - r₂. This area is equal to 4π(r ₂)².
The rate of radiant energy is K = 1500 W/m².
K=σT⁴4π(r₁)²/4π(r ₂)² =
= σT⁴(r₁/r ₂)²;
T=∜[K(r₂/r₁)²/σ]= ∜[1500(1.5•10¹¹/6.96•10⁸)²/5.67•10⁻⁸]= 5921 K
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