First signs of life may have been detected on an exoplanet...

originally posted by: ArMaP

originally posted by: beyondknowledge2
Two times size means eight times gravity.

Not really.

Gravity is related to the mass of the planet, and although K2-18b is 2.6 times larger than Earth, its mean density is only 2.67, while Earth's mean density is 5.51.
Also, as K2-18b has a bigger radius, gravity at the surface is not as big, as the distance from the surface to the centre of gravity is bigger, resulting in gravity at the surface of 12.43 m/s2, only 1.27 times Earth's gravity at the surface.

Exactly the point I was trying to relate earlier.

a reply to: ArMaP
Okay so they know the planet mass and density already?
I didn't know the James webb could do that ?

a reply to: 0bserver1

I suppose mass is calculated based on the planet's orbit and density based on mass and size.

originally posted by: 0bserver1
a reply to: ArMaP
Okay so they know the planet mass and density already?
I didn't know the James webb could do that ?

It appears the mass and density was determined a while ago.

en.wikipedia.org...#:~:text=K2%2D18b%20has%20a%20mass,like%20Mercury%20is%20also%20possible.

originally posted by: quintessentone

originally posted by: ArMaP

originally posted by: beyondknowledge2
Two times size means eight times gravity.

Not really.

Gravity is related to the mass of the planet, and although K2-18b is 2.6 times larger than Earth, its mean density is only 2.67, while Earth's mean density is 5.51.
Also, as K2-18b has a bigger radius, gravity at the surface is not as big, as the distance from the surface to the centre of gravity is bigger, resulting in gravity at the surface of 12.43 m/s2, only 1.27 times Earth's gravity at the surface.

Exactly the point I was trying to relate earlier.

Mass approximately 9 times the mass of Earth.
Radius approximately 3 times the radius of Earth. That gives you the same gravitational field strength because of the inverse square law.

originally posted by: Muldar

originally posted by: quintessentone

originally posted by: ArMaP

originally posted by: beyondknowledge2
Two times size means eight times gravity.

Not really.

Gravity is related to the mass of the planet, and although K2-18b is 2.6 times larger than Earth, its mean density is only 2.67, while Earth's mean density is 5.51.
Also, as K2-18b has a bigger radius, gravity at the surface is not as big, as the distance from the surface to the centre of gravity is bigger, resulting in gravity at the surface of 12.43 m/s2, only 1.27 times Earth's gravity at the surface.

Exactly the point I was trying to relate earlier.

Mass approximately 9 times the mass of Earth.
Radius approximately 3 times the radius of Earth. That gives you the same gravitational field strength because of the inverse square law.

Not according to scientists.

It is unlikely for the exoplanet K2-18b to have land due to its low density, and if it is habitable with an ocean, there cannot be land masses exposed to the thin atmosphere.
The possibility of tidal interaction between the planets in the K2-18 system could be influencing the convection and tides on K2-18b, given their close proximity and the unknown mass of the second planet.
The mass of K2-18b is only known to be a minimum because there is no inclination measurement or transit of the star to measure its radius.

eightify.app...

originally posted by: quintessentone

originally posted by: Muldar

originally posted by: quintessentone

originally posted by: ArMaP

originally posted by: beyondknowledge2
Two times size means eight times gravity.

Not really.

Gravity is related to the mass of the planet, and although K2-18b is 2.6 times larger than Earth, its mean density is only 2.67, while Earth's mean density is 5.51.
Also, as K2-18b has a bigger radius, gravity at the surface is not as big, as the distance from the surface to the centre of gravity is bigger, resulting in gravity at the surface of 12.43 m/s2, only 1.27 times Earth's gravity at the surface.

Exactly the point I was trying to relate earlier.

Mass approximately 9 times the mass of Earth.
Radius approximately 3 times the radius of Earth. That gives you the same gravitational field strength because of the inverse square law.

Not according to scientists.

It is unlikely for the exoplanet K2-18b to have land due to its low density, and if it is habitable with an ocean, there cannot be land masses exposed to the thin atmosphere.
The possibility of tidal interaction between the planets in the K2-18 system could be influencing the convection and tides on K2-18b, given their close proximity and the unknown mass of the second planet.
The mass of K2-18b is only known to be a minimum because there is no inclination measurement or transit of the star to measure its radius.

eightify.app...

You've linked something different.
The density of a planet isn't the gravitational field strength. The gravitational field strength depends on the mass and radius and if you know them you can find its value at the surface of the planet. The numbers I gave you above are more or less correct.

New observations by JWST suggest that they do. The planet in question is K2-18 b. It has a mass nearly 9 times that of Earth, and a radius almost 3 times that of Earth. Its orbit is in the habitable zone of a red dwarf star just 120 light-years from Earth

www.universetoday.com... ater-ocean/

What if they are like, the Borg?

Shouldn't we be keeping our heads down?

As Hawking warned us.

Who knows? The Borg may already be here, among us?

Scary.

originally posted by: Oldcarpy2
What if they are like, the Borg?

Shouldn't we be keeping our heads down?

As Hawking warned us.

Who knows? The Borg may already be here, among us?

Scary.

The Borg take no prisoners.
Resistance is futile and everyone gets assimilated into the collective.

On the issue of aliens on that planet and how gravity may have affected life, the gravity is pretty much the same as here on earth.

a reply to: Muldar

We know.

originally posted by: Oldcarpy2
a reply to: Muldar

We know.

There have been discussions that gravity is very different on that planet. But that's not correct. It's almost the same due to the inverse square law we use to calculate g

originally posted by: Muldar

originally posted by: Oldcarpy2
a reply to: Muldar

We know.

There have been discussions that gravity is very different on that planet. But that's not correct. It's almost the same due to the inverse square law we use to calculate g

Watch the video I posted, they are claiming the gravity on that planet is different. It's a trippy video worth the watch.

originally posted by: quintessentone

originally posted by: Muldar

originally posted by: Oldcarpy2
a reply to: Muldar

We know.

There have been discussions that gravity is very different on that planet. But that's not correct. It's almost the same due to the inverse square law we use to calculate g

Watch the video I posted, they are claiming the gravity on that planet is different. It's a trippy video worth the watch.

I will watch it but how is the gravitational field strength very different? It could be slightly different if you have the exact values but if you take approximate values like 9 Earth masses and 3 Earth radii then the answer is pretty much the same. It's just basic high school physics

g = G M / R^2

That's how you calculate it at the surface of the planet.

originally posted by: Muldar

originally posted by: quintessentone

originally posted by: Muldar

originally posted by: Oldcarpy2
a reply to: Muldar

We know.

There have been discussions that gravity is very different on that planet. But that's not correct. It's almost the same due to the inverse square law we use to calculate g

Watch the video I posted, they are claiming the gravity on that planet is different. It's a trippy video worth the watch.

I will watch it but how is the gravitational field strength very different? It could be slightly different if you have the exact values but if you take approximate values like 9 Earth masses and 3 Earth radii then the answer is pretty much the same. It's just basic high school physics

g = G M / R^2

That's how you calculate it at the surface of the planet.

From Quora:

Well, according to The Extrasolar Planets Encyclopaedia, the exoplanet has a mass of 8.92 Earth masses and a radius of 2.37 Earth radius.

The gravitational acceleration formula is:

GM over r2

G is always 6.67408⋅10−11

Using this, we can determine that the gravitational acceleration is 15.6 m/s2, or almost 1.6 times Earth’s gravity.

originally posted by: quintessentone

originally posted by: Muldar

originally posted by: quintessentone

originally posted by: Muldar

originally posted by: Oldcarpy2
a reply to: Muldar

We know.

There have been discussions that gravity is very different on that planet. But that's not correct. It's almost the same due to the inverse square law we use to calculate g

Watch the video I posted, they are claiming the gravity on that planet is different. It's a trippy video worth the watch.

I will watch it but how is the gravitational field strength very different? It could be slightly different if you have the exact values but if you take approximate values like 9 Earth masses and 3 Earth radii then the answer is pretty much the same. It's just basic high school physics

g = G M / R^2

That's how you calculate it at the surface of the planet.

From Quora:

Well, according to The Extrasolar Planets Encyclopaedia, the exoplanet has a mass of 8.92 Earth masses and a radius of 2.37 Earth radius.

The gravitational acceleration formula is:

GM over r2

G is always 6.67408⋅10−11

Using this, we can determine that the gravitational acceleration is 15.6 m/s2, or almost 1.6 times Earth’s gravity.

Slightly different from what I said based on the approximate values.

The exact values can be found on the NASA website

exoplanets.nasa.gov...

Mass = 8.92 Earth masses
Radius = 0.211 × Jupiter's radius
which gives around 2.31537 × the radius of Earth

Which in turn gives 1.6634g = 16.32 m / s^2

This value is not that different to what we have here on Earth according to the data from NASA (assuming these are the exact values)

a reply to: Muldar

Why not choose the ATS handle 'Spooky Muldar" it would suit you well, signed Scully.

originally posted by: quintessentone
a reply to: Muldar

Why not choose the ATS handle 'Spooky Muldar" it would suit you well, signed Scully.

... but does it have a massive moon?

... internal tectonics?

... ozone layer?

... seasonal declination of 23 degrees+?

... sufficient biochemistry?

... insolation filters (orbit/revolution/van allen-equiv. belt)?

... adequate atmospheric circulation?

This is great news - a step in a hopeful direction! We just need to keep probing, analyzing. We can't count a chicken before the egg. Proof is still ahead of us - but the day is coming...

originally posted by: 0bserver1
a reply to: data5091
It's time they point that thing on Zeta Reticuli and tell us what they see?

probably not much