JWST’s FASCINATING find on EXOPLANET K2-18 b! #space #universe #exoplanets #k218b #science #cosmos

1 0 1 minute read

The James Webb Space Telescope (JWST) has made a groundbreaking discovery on the exoplanet K2-18b, a distant world that could hold the key to understanding the potential for life beyond Earth. In this video, we dive deep into the latest findings, exploring what makes K2-18b so special and why this discovery could change everything we know about exoplanets. Could this be the next step in our search for habitable worlds? Join us as we uncover the promise of exoplanets and what JWST’s discoveries mean for the future of space exploration. Whether you’re a space enthusiast or new to the wonders of the universe, this journey into the unknown is not to be missed!

#JWST #Exoplanets #K218b #SpaceExploration #Astronomy #Science #HabitableZone #Cosmos #NASA #ExoplanetDiscovery #SpaceScience #Universe #Astrobiology #NewWorlds #SpaceMission #JWSTFindings #LifeBeyondEarth #CosmicDiscovery #SpaceExploration2024 #planetaryscience #astrobitsdaily

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Dr Woblok 5 days ago

1 0 1 minute read

Let's go ahead and calculate the estimated temperature on the side of K2-18b facing its star, assuming it's tidally locked.

### Given:
– *Luminosity of K2-18 (L*)**: K2-18 is a red dwarf with an estimated luminosity of about 0.022 times that of the Sun (L ≈ 0.022 L☉).
– **Distance of K2-18b from K2-18 (d)**: Approximately 0.1429 AU.
– **Stefan-Boltzmann constant (σ)**: ( 5.67 times 10^{-8} , text{W m}^{-2} text{K}^{-4} )
– **Albedo (A)**: Assume an albedo of 0.3 (a typical estimate for a planet with a reflective atmosphere).

### Calculation:

1. **Convert the distance to meters**:
– 1 AU ≈ ( 1.496 times 10^{11} ) meters.
– Distance ( d ) = 0.1429 AU ≈ ( 0.1429 times 1.496 times 10^{11} ) meters ≈ ( 2.137 times 10^{10} ) meters.

2. **Calculate the equilibrium temperature**:
[
T_{eq} = left( frac{L_* (1 – A)}{16 pi sigma d^2} right)^{1/4}
]

– ( L_* ) = ( 0.022 times 3.828 times 10^{26} , text{W} ) (since L☉ ≈ ( 3.828 times 10^{26} ) W).
– ( L_* ) ≈ ( 8.42 times 10^{24} , text{W} ).
– ( T_{eq} ) ≈ ( left( frac{8.42 times 10^{24} times (1 – 0.3)}{16 pi times 5.67 times 10^{-8} times (2.137 times 10^{10})^2} right)^{1/4} ).

3. **Simplify the calculation**:

– Numerator: ( 8.42 times 10^{24} times 0.7 ) ≈ ( 5.894 times 10^{24} , text{W} ).
– Denominator: ( 16 pi times 5.67 times 10^{-8} times 4.57 times 10^{20} ) ≈ ( 1.31 times 10^{15} , text{W m}^{-2} ).

[
T_{eq} approx left( frac{5.894 times 10^{24}}{1.31 times 10^{15}} right)^{1/4} approx left( 4.5 times 10^9 right)^{1/4} text{K}
]

[
T_{eq} approx 316 , text{K}
]

### Conclusion:
The estimated equilibrium temperature on the side of K2-18b facing its star, assuming tidal locking, is approximately *316 K* (around 43°C or 109°F). This is a rough estimate and does not take into account atmospheric effects or heat redistribution across the planet's surface.

One Comment

@samiamgreeneggaandham says:
September 12, 2024 at 2:48 pm
Let's go ahead and calculate the estimated temperature on the side of K2-18b facing its star, assuming it's tidally locked.
### Given:
– *Luminosity of K2-18 (L*)**: K2-18 is a red dwarf with an estimated luminosity of about 0.022 times that of the Sun (L ≈ 0.022 L☉).
– **Distance of K2-18b from K2-18 (d)**: Approximately 0.1429 AU.
– **Stefan-Boltzmann constant (σ)**: ( 5.67 times 10^{-8} , text{W m}^{-2} text{K}^{-4} )
– **Albedo (A)**: Assume an albedo of 0.3 (a typical estimate for a planet with a reflective atmosphere).
### Calculation:
1. **Convert the distance to meters**:
– 1 AU ≈ ( 1.496 times 10^{11} ) meters.
– Distance ( d ) = 0.1429 AU ≈ ( 0.1429 times 1.496 times 10^{11} ) meters ≈ ( 2.137 times 10^{10} ) meters.
2. **Calculate the equilibrium temperature**:
[
T_{eq} = left( frac{L_* (1 – A)}{16 pi sigma d^2} right)^{1/4}
]
– ( L_* ) = ( 0.022 times 3.828 times 10^{26} , text{W} ) (since L☉ ≈ ( 3.828 times 10^{26} ) W).
– ( L_* ) ≈ ( 8.42 times 10^{24} , text{W} ).
– ( T_{eq} ) ≈ ( left( frac{8.42 times 10^{24} times (1 – 0.3)}{16 pi times 5.67 times 10^{-8} times (2.137 times 10^{10})^2} right)^{1/4} ).
3. **Simplify the calculation**:
– Numerator: ( 8.42 times 10^{24} times 0.7 ) ≈ ( 5.894 times 10^{24} , text{W} ).
– Denominator: ( 16 pi times 5.67 times 10^{-8} times 4.57 times 10^{20} ) ≈ ( 1.31 times 10^{15} , text{W m}^{-2} ).
[
T_{eq} approx left( frac{5.894 times 10^{24}}{1.31 times 10^{15}} right)^{1/4} approx left( 4.5 times 10^9 right)^{1/4} text{K}
]
[
T_{eq} approx 316 , text{K}
]
### Conclusion:
The estimated equilibrium temperature on the side of K2-18b facing its star, assuming tidal locking, is approximately *316 K* (around 43°C or 109°F). This is a rough estimate and does not take into account atmospheric effects or heat redistribution across the planet's surface.