Question

QUESTION 1

1. Estimate The Temperature For A Planet In Other Solar System

   Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 8000 Kelvins and a radius Sr = 1×10⁹ meters. Scientists also measured the distance (D) between the star and the new-discovered planet: D = 1 AU ~ 1.5× 10¹¹ meters.

   The solar power over unit area from the star’s surface (Ps) can be calculated from the star’s surface temperature Ts (8000 Kelvins) by the Stefen-Boltzman law Ps=σ×(Ts)⁴, where σ is Stefen-Boltzman constant (5.67 × 10^-8 Watt/meter²/Kelvin⁴ ). What is the solar power over unit area from the star’s surface (Ps)?

   	A.	Ps ~ 5.67 × 108 Watt/meter2
   	B.	Ps ~ 2.32 × 108 Watt/meter2
   	C.	Ps ~ 5.67 × 109 Watt/meter2
   	D.	Ps ~ 2.32 × 109 Watt/meter2

1 points   

QUESTION 2

1. The solar power (Ps) decreases from the star’s surface to the distance at the planet. Assuming the solar power over unit area at the distance of the planet as Pp, we have Pp=Ps×(Sr/D)², where Sr is the radius of the star (1 × 10⁹ meters) and D is the distance between the star and the planet (~1.5× 10¹¹ meters). With the calculated solar power over unit area from the star’s surface (Ps) in Question 1, please estimate the solar power over unit area at the distance of the planet (Pp).

   	A.	Pp ~ 2.52  × 103 Watt/meter2
   	B.	Pp ~ 1.03  × 103 Watt/meter2
   	C.	Pp ~ 2.52  × 104 Watt/meter2
   	D.	Pp ~ 1.03 × 104 Watt/meter2

1 points   

QUESTION 3

1. Likewise, the emitted power over unit area from the planet (EP) can be calculated by the Stefen-Boltzman law with EP=σ×(Tp)⁴. The parameter Tp is the equilibrium temperature of the planet, which is close to the actual temperature of the planet. In addition, scientists measured the global albedo (A) of the planet, which is the ratio between the reflected solar power and the total solar power. The global albedo of the planet A=0.4. With the albedo A, the absorbed solar power over unit area by the planet (AP) can be expressed by AP=Pp×(1−A), where Pp is the solar power over unit area at the distance of the planet (refer to Question 2). Note: AP and EP are powers over unit area. The planet absorbs the solar power through an imaginary disc that has a radius equal to the radius of the planet (Pr). The area of the imaginary disc for the absorbed solar power (S_AP) can be expressed as S_AP = π×(Pr)². Therefore, the total absorbed solar power (AP_total) by the planet can be calculated by AP_total=AP×S_AP=AP×π×(Pr)²= Pp×(1−A)×π×(Pr)². At the same time, the planet emits power from its whole sphere. The surface area of spherical planet (S_EP) can be calculated by S_EP=4π×(Pr)², and hence the total emitted power (EP_total) by the planet can be expressed as EP_total=EP×S_EP=EP×4π×(Pr)²=σ×(Tp)⁴×4π×(Pr)². If we assume a balance between the total absorbed solar power and the total emitted power for the planet, we have AP_total = EP_total. Based on the balance AP_total = EP_total, please estimate the equilibrium temperature of the planet (Tp).   

   	A.	Tp ~ 306 Kelvins
   	B.	Tp ~ 406 Kelvins
   	C.	Tp ~ 506 Kelvins
   	D.	Tp ~ 1000 Kelvins

Answer 1

Thank you for asking. Let's try to answer each questions one by one abs within the stipulated time frame.

1.) We are given -

Stefan boltzman constant = 5.67 × 10-8 Watt/meter2/Kelvin4

Star surface temperature = 8000 kelvin

Now we have to calculate the solar power generated per unit area(Ps)= Ts4

= Solar power over unit area from the star’s surface (Ps)=>

5.67*10-8* 4096*1012=23,224.32*104= 2.32*108watt/meter2 Answer

2.) Here we are given -

Sr is the radius of the star 1 × 109 meters and D is the distance between the star and the planet ~1.5× 1011 meters

Ps= 2.32*108watt/ meter2

So, Pp=Ps×(Sr/D)2

= Pp= 2.32*108*(1*109/1.5*1011)2

  = 2.32*108*0.4356*10-4 = 1.03*104watt/ meter2 Answer

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