Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
A small satellite is in orbit around the Earth at a height of 1000 km above the surface. It has a mass of 100 kg, a radius of 1 m, an absorptivity of 0.9 for IR radiation and 0.5 for visible radiation. a) What solid angle does the Earth subtend when viewed from the satellite? b) Calculate the radiative equilibrium temperature of the satellite when it is in the shadow of the Earth. c) What is the radiative equilibrium temperature of the satellite at the moment when it passes completely out of the shadow of the Earth? (Note that at this time the Earth is still completely dark when viewed from the satellite.)

A small satellite is in orbit around the Earth at a height of 1000 km above the surface. It has a mass of 100 kg, a radius of 1 m, an absorptivity of 0.9 for IR radiation and 0.5 for visible radiation. a) What solid angle does the Earth subtend when viewed from the satellite? b) Calculate the radiative equilibrium temperature of the satellite when it is in the shadow of the Earth. c) What is the radiative equilibrium temperature of the satellite at the moment when it passes completely out of the shadow of the Earth? (Note that at this time the Earth is still completely dark when viewed from the satellite.)

icon
Related questions
Question

Can you try doing Part-A of this question I have included formula sheet also. 

A small satellite is in orbit around the Earth at a height of 1000 km above the surface. It has a mass of 100 kg, a radius of 1 m, an absorptivity of 0.9 for IR radiation and 0.5 for visible radiation. a) What solid angle does the Earth subtend when viewed from the satellite? b) Calculate the radiative equilibrium temperature of the satellite when it is in the shadow of the Earth. c) What is the radiative equilibrium temperature of the satellite at the moment when it passes completely out of the shadow of the Earth? (Note that at this time the Earth is still completely dark when viewed from the satellite.)
Transcribed Image Text:A small satellite is in orbit around the Earth at a height of 1000 km above the surface. It has a mass of 100 kg, a radius of 1 m, an absorptivity of 0.9 for IR radiation and 0.5 for visible radiation. a) What solid angle does the Earth subtend when viewed from the satellite? b) Calculate the radiative equilibrium temperature of the satellite when it is in the shadow of the Earth. c) What is the radiative equilibrium temperature of the satellite at the moment when it passes completely out of the shadow of the Earth? (Note that at this time the Earth is still completely dark when viewed from the satellite.)
Earth-Sun mean distance: 149.598 x 10⁰ m Radius of Sun: 6.96 x 10⁰ m Radius of Earth: 6371 x 10³ m Effective temperature of Sun: 5770 K Cross sectional area of a sphere: R² Surface area of a sphere: 4+R² Solid Angle: = Area on sphere/R²; d = sin0d0 dø Albedo of the Earth: A = 0.3 Plank Function: B(λ,T) = 2hc² hc 25 ekλT-1 Plank's constant: h = 6.626 × 10-34 Js Boltzmann's constant: k = 1.381× 10-23 J/K Speed of light: c = 3 × 108 m/s Flux: F = I cose d Watts Stephan-Boltzmann Law: F = 6T4 m² Stephan's constant: = 5.67 × 108 W m²2 K-4 Net flux upward or downward for isotropic radiation: F=nl Wien's displacement Law: Ap=2898/T µm Kirchoff's Law: absorptivity = emissivity Watts m².um.sr Run Flux of solar Radiation at Earth: F = Tsun = 1370 Watts/m² D²-s Optical cross section: k in Effective Radiating Temperature for the Earth (current climate): T₂ = 255 K Beer's Law: 1(X) = 1(0) exp[-x]; x = · S² kpdx = fondx m² kg or o in m² molecule Ideal Gas Law: P = pRT; R = 287 J Kg¹ K-¹ Ideal Gas Law: P = n k T; k = 1.381 × 10-²³ J/K Ideal Gas Law for Water Vapour: e=py Ry T; R = 461.5 J Kg¹ K-¹ Hydrostatic Equation: Barometric Law: P(z) = P(0) exp (− ²); H = RT/g; P₁ = 100 × 10³ Pa ; g = 9,81 m/s² dP dz ,(Po-P). Force of buoyancy: FB = 9- ·=g₁ P =-pg Adiabatic Lapse! dT Rate: =- dz g Cp First Law of Thermodynamics: dq = c₂dT + P da or Sq = ₁₂ dT - ² dp Joules kg-deg C Specific Heat Capacity for Air: Potential Temperature: 0 = T (T-To) To = -9.8 °C/km =1005 Cp Brunt-Vaisala Frequency, or Buoyancy Frequency: N² = g de 0 dz R Latent Heat of Condensation for water: 2535 J/g Latent Heat of Sublimation for water: 2834 J/g CAPE = -R des Les Clausius-Clapeyron Equation: dT R₂T² Solution to C-C equation: - = exp es(T) eso LFC dT Saturated (Wet) Adiabatic Lapse Rate: T = -- = dz EL [(T-To) dinPo Ry To LWS Equivalent Potential Temperature: 0 ~ 0 exp CpT - })}; at To = 0 °C (273 K), es = 611 Pa Velocity: v²v² = 2a(z-z₁); a is acceleration - 1+ Id Adiabatic liquid/ice water content: x = − 4w₁ (2); or x = − 4ps (1) - L dws T dz Cp đồ
Transcribed Image Text:Earth-Sun mean distance: 149.598 x 10⁰ m Radius of Sun: 6.96 x 10⁰ m Radius of Earth: 6371 x 10³ m Effective temperature of Sun: 5770 K Cross sectional area of a sphere: R² Surface area of a sphere: 4+R² Solid Angle: = Area on sphere/R²; d = sin0d0 dø Albedo of the Earth: A = 0.3 Plank Function: B(λ,T) = 2hc² hc 25 ekλT-1 Plank's constant: h = 6.626 × 10-34 Js Boltzmann's constant: k = 1.381× 10-23 J/K Speed of light: c = 3 × 108 m/s Flux: F = I cose d Watts Stephan-Boltzmann Law: F = 6T4 m² Stephan's constant: = 5.67 × 108 W m²2 K-4 Net flux upward or downward for isotropic radiation: F=nl Wien's displacement Law: Ap=2898/T µm Kirchoff's Law: absorptivity = emissivity Watts m².um.sr Run Flux of solar Radiation at Earth: F = Tsun = 1370 Watts/m² D²-s Optical cross section: k in Effective Radiating Temperature for the Earth (current climate): T₂ = 255 K Beer's Law: 1(X) = 1(0) exp[-x]; x = · S² kpdx = fondx m² kg or o in m² molecule Ideal Gas Law: P = pRT; R = 287 J Kg¹ K-¹ Ideal Gas Law: P = n k T; k = 1.381 × 10-²³ J/K Ideal Gas Law for Water Vapour: e=py Ry T; R = 461.5 J Kg¹ K-¹ Hydrostatic Equation: Barometric Law: P(z) = P(0) exp (− ²); H = RT/g; P₁ = 100 × 10³ Pa ; g = 9,81 m/s² dP dz ,(Po-P). Force of buoyancy: FB = 9- ·=g₁ P =-pg Adiabatic Lapse! dT Rate: =- dz g Cp First Law of Thermodynamics: dq = c₂dT + P da or Sq = ₁₂ dT - ² dp Joules kg-deg C Specific Heat Capacity for Air: Potential Temperature: 0 = T (T-To) To = -9.8 °C/km =1005 Cp Brunt-Vaisala Frequency, or Buoyancy Frequency: N² = g de 0 dz R Latent Heat of Condensation for water: 2535 J/g Latent Heat of Sublimation for water: 2834 J/g CAPE = -R des Les Clausius-Clapeyron Equation: dT R₂T² Solution to C-C equation: - = exp es(T) eso LFC dT Saturated (Wet) Adiabatic Lapse Rate: T = -- = dz EL [(T-To) dinPo Ry To LWS Equivalent Potential Temperature: 0 ~ 0 exp CpT - })}; at To = 0 °C (273 K), es = 611 Pa Velocity: v²v² = 2a(z-z₁); a is acceleration - 1+ Id Adiabatic liquid/ice water content: x = − 4w₁ (2); or x = − 4ps (1) - L dws T dz Cp đồ
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer