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fomenos
3 years ago
8

A planet orbits a star in an elliptical orbit. At a particular instant the momentum of the planet is

Physics
1 answer:
Masja [62]3 years ago
7 0

Answer:

L = m v r (The momentum remains constant)

Explanation:

Even in an ellipsoidal orbit, the law of conservation of angular momentum always apply. When the plant approached the perihelion, the radius of the orbit decreases and the speed of the star increases to conserve the momentum. Similarly, when the planet approaches the aphelion, the speed of the star decreases as the radius increases to conserve the momentum. So, the momentum at a particular instant can be calculated by L = m v r

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Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

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3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

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      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

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range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
3 years ago
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