Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
QUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D=2 AU - 3.0x10¹1 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where a is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? Ps~ 2.87 x 108 Watt/meter² O Ps~ 5.67 x 108 Watt/meter² O Ps~ 2.87 x 10 Watt/meter2 O Ps~ 5.67 x 10 Watt/meter2 QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per 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 x meters) and D is the distance between the star and the planet (-3.0x 1011 meters). With the calculated solar power per unit area from the star's surface (Ps) in Question 1, please estimate the solar power per unit area at distance of the planet (Pp). 0 Pp 3.15 x 102 Watt/meter² O Pp~ 6.30 x 102 Watt/meter2 Pp~3.15 x 103 Watt/meter² 0 Pp~ 6.30 x 103 Watt/meter2

QUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D=2 AU - 3.0x10¹1 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where a is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? Ps~ 2.87 x 108 Watt/meter² O Ps~ 5.67 x 108 Watt/meter² O Ps~ 2.87 x 10 Watt/meter2 O Ps~ 5.67 x 10 Watt/meter2 QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per 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 x meters) and D is the distance between the star and the planet (-3.0x 1011 meters). With the calculated solar power per unit area from the star's surface (Ps) in Question 1, please estimate the solar power per unit area at distance of the planet (Pp). 0 Pp 3.15 x 102 Watt/meter² O Pp~ 6.30 x 102 Watt/meter2 Pp~3.15 x 103 Watt/meter² 0 Pp~ 6.30 x 103 Watt/meter2

icon
Related questions
Question
QUESTION 3 Likewise, the emitted power per unit area from the planet (EP) can be calculated by the Stefen-Boltzman law with EP-OT4. The parameter T 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=0.5. With the albedo A, the absorbed solar power per unit area by the planet (AP) can be expressed by AP-Pp(1-A), where Pp is the solar power per unit area at the distance of the planet (see Question 2). Note: AP and EP are powers per unit area. The planet absorbs the solar power through an imaginary disc that has a radius equal to the radius of the planet (r). The area of the imaginary disc for the absorbed solar power (SAP) can be expressed as SAP = Tr². Therefore, the total absorbed solar power (APtotal) by the planet can be calculated by APtotal-AP x SAP-AP T²= Pp (1-A) r². At the same time, the planet emits power from its whole sphere. The surface area of a spherical planet (SEP) can be calculated by SEP=4Tr², and hence the total emitted power (EP total) by the planet can be expressed as EP total-EPXSEP-EPX4T₁²=OT44πr². If we assume a balance between the total absorbed solar power and the total emitted power for the planet, we have APtotal = EPtotal- Based on the balance AP total = EP total and the estimated Pp in Question 2, please estimate the equilibrium temperature of the planet (T). 0 T~ 343 Kelvins OT-443 Kelvins OT-543 Kelvins 0 T~ 643 Kelvins
Transcribed Image Text:QUESTION 3 Likewise, the emitted power per unit area from the planet (EP) can be calculated by the Stefen-Boltzman law with EP-OT4. The parameter T 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=0.5. With the albedo A, the absorbed solar power per unit area by the planet (AP) can be expressed by AP-Pp(1-A), where Pp is the solar power per unit area at the distance of the planet (see Question 2). Note: AP and EP are powers per unit area. The planet absorbs the solar power through an imaginary disc that has a radius equal to the radius of the planet (r). The area of the imaginary disc for the absorbed solar power (SAP) can be expressed as SAP = Tr². Therefore, the total absorbed solar power (APtotal) by the planet can be calculated by APtotal-AP x SAP-AP T²= Pp (1-A) r². At the same time, the planet emits power from its whole sphere. The surface area of a spherical planet (SEP) can be calculated by SEP=4Tr², and hence the total emitted power (EP total) by the planet can be expressed as EP total-EPXSEP-EPX4T₁²=OT44πr². If we assume a balance between the total absorbed solar power and the total emitted power for the planet, we have APtotal = EPtotal- Based on the balance AP total = EP total and the estimated Pp in Question 2, please estimate the equilibrium temperature of the planet (T). 0 T~ 343 Kelvins OT-443 Kelvins OT-543 Kelvins 0 T~ 643 Kelvins
QUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D = 2 AU - 3.0x1011 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where o is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? O Ps ~ 2.87 x 108 Watt/meter2 O Ps ~ 5.67 x 108 Watt/meter2 O O Ps ~ 2.87 x 10 Watt/meter2 Watt/meter² Ps ~ 5.67 x 10⁹ QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per 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 x 10⁹ meters) and D is the distance between the star and the planet (-3.0x 1011 meters). With the calculated solar power per unit area from the star's surface (Ps) in Question 1, please estimate the solar power per unit area at the distance of the planet (Pp). 0 Pp~ 3.15 x 102 Watt/meter² 0 Pp ~ 6.30 x 102 Watt/meter² Pp~ 3.15 x 10³ Watt/meter2 0 Pp ~ 6.30 x 10³ Watt/meter2
Transcribed Image Text:QUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D = 2 AU - 3.0x1011 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where o is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? O Ps ~ 2.87 x 108 Watt/meter2 O Ps ~ 5.67 x 108 Watt/meter2 O O Ps ~ 2.87 x 10 Watt/meter2 Watt/meter² Ps ~ 5.67 x 10⁹ QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per 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 x 10⁹ meters) and D is the distance between the star and the planet (-3.0x 1011 meters). With the calculated solar power per unit area from the star's surface (Ps) in Question 1, please estimate the solar power per unit area at the distance of the planet (Pp). 0 Pp~ 3.15 x 102 Watt/meter² 0 Pp ~ 6.30 x 102 Watt/meter² Pp~ 3.15 x 10³ Watt/meter2 0 Pp ~ 6.30 x 10³ Watt/meter2
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer