My boy-scout compass does not work on Mars.I have not gone to Mars yet but if I do I will get a special Mars compass.Decimal currency started in Australia on the 14th of February 1966.
@morphyms1817 I have also heard of studies proving that the sun can only strip away 0.001 bars of pressure from the atmosphere every billion years...
We just don't know Mars to its entirety...
We just don't know Mars to its entirety...
@SavageAntarctican
In all seriousness, please point me to such a study, I would be interested.
In all seriousness, please point me to such a study, I would be interested.
@morphyms1817 I don't remember where it was written. I just read it somewhere a few years ago...
Let me make my own estimate here:
I used a few resources, all of them being non-study websites (because I don't know how to use the Jeans Escape formula to calculate atmosphere mass loss and studies don't give me many useful tips on that nor the amount of solar wind/xray emission through time). I came up with this:
100g of mass is lost per second (a website told me this).
Assuming this, it can be calculated that 3.154*10^6 kg of mass is lost per year or 3.154*10^15 kg per billion years.
The sun started with a 100x stronger solar wind (compared to today) that went down to <10x by 3.8-4 billion years ago.
So, let's say it averaged 10x (they say averaged 10-20x so I'll take 10x) in the first billion years. I want to include that fact that Mars had a strong magnetic field until about 4 billion years ago.
3.154*10^15*10=3.154*10^16 kg of mass lost in the first billion years
Then after that, let's take a smaller number (say, 3x), so now, 9.46*10^15kg of mass lost
And then for the next 2.5 billion years, let's use current rates.
3.154*10^15*2.5= 7.88*10^15kg of mass lost.
This gives about 4.68 * 10^16 kg of mass loss.
The Earth's atmosphere has a mass of 5.15*10^18 kg.
Mars's atmosphere would have a mass of 1/2^2 (surface area because atmosphere wraps around the surface)*5.15*10^18 kg = 1.29*10^18 kg of atmosphere at 1 atm pressure.
4.68*10^16kg/1.29*10^18kg = 0.0363 atm lost.
0.0363 atm is lost in 4.6 billion years.
Which means that if Mars started with 0.5 bar - 2 bar (which is roughly 0.5 atm - 2 atm) of pressure, it wouldn't have lost much and would remain habitable...
But then, once again, this is just an estmation...
Let me make my own estimate here:
I used a few resources, all of them being non-study websites (because I don't know how to use the Jeans Escape formula to calculate atmosphere mass loss and studies don't give me many useful tips on that nor the amount of solar wind/xray emission through time). I came up with this:
100g of mass is lost per second (a website told me this).
Assuming this, it can be calculated that 3.154*10^6 kg of mass is lost per year or 3.154*10^15 kg per billion years.
The sun started with a 100x stronger solar wind (compared to today) that went down to <10x by 3.8-4 billion years ago.
So, let's say it averaged 10x (they say averaged 10-20x so I'll take 10x) in the first billion years. I want to include that fact that Mars had a strong magnetic field until about 4 billion years ago.
3.154*10^15*10=3.154*10^16 kg of mass lost in the first billion years
Then after that, let's take a smaller number (say, 3x), so now, 9.46*10^15kg of mass lost
And then for the next 2.5 billion years, let's use current rates.
3.154*10^15*2.5= 7.88*10^15kg of mass lost.
This gives about 4.68 * 10^16 kg of mass loss.
The Earth's atmosphere has a mass of 5.15*10^18 kg.
Mars's atmosphere would have a mass of 1/2^2 (surface area because atmosphere wraps around the surface)*5.15*10^18 kg = 1.29*10^18 kg of atmosphere at 1 atm pressure.
4.68*10^16kg/1.29*10^18kg = 0.0363 atm lost.
0.0363 atm is lost in 4.6 billion years.
Which means that if Mars started with 0.5 bar - 2 bar (which is roughly 0.5 atm - 2 atm) of pressure, it wouldn't have lost much and would remain habitable...
But then, once again, this is just an estmation...
@SavageAntarctican
I will review all carefully, your math skills are far ahead of me. Thank you, most helpful. Stellar and planetary science is of great interest to me.
On a tangent, I am waiting for a brown dwarf or red dwarf star to be discovered closer than Proxima Centauri.
On another tangent I am fascinated by the Kuiper belts and comet clouds of nearby stars.
I will review all carefully, your math skills are far ahead of me. Thank you, most helpful. Stellar and planetary science is of great interest to me.
On a tangent, I am waiting for a brown dwarf or red dwarf star to be discovered closer than Proxima Centauri.
On another tangent I am fascinated by the Kuiper belts and comet clouds of nearby stars.
@morphyms1817 If there was any star closer to the Sun than Proxima Centauri, we would know already.
The reason is heat signatures: every star has one, including brown dwarfs. So...
The reason is heat signatures: every star has one, including brown dwarfs. So...
@SavageAntarctican
That makes sense, the closest known brown dwarfs, a pair, are 7 LY distant. There is this search for a Super Earth at beyond 100 a.u. due to the strange orbits of many Kuiper Belt and other objects. Perhaps we can discuss it later. Thanks again for your guidance.
That makes sense, the closest known brown dwarfs, a pair, are 7 LY distant. There is this search for a Super Earth at beyond 100 a.u. due to the strange orbits of many Kuiper Belt and other objects. Perhaps we can discuss it later. Thanks again for your guidance.
The Kuiper Belt is where the earth-killer asteroid will come from..if indeed it is not already on it's way here.But don't start hoarding toilet paper just yet..we got a couple of billion yrs before it arrives.
The optometrist said I still have eyes.Tonight we will have chicken tikka masala for dinner.I will put some fresh lime juice from a lime from our lime tree in our backyard on it.
#349
7.9 [ optometrist determination (0.9) + lime orchard in S 3 (7.0) ]
7.9 [ optometrist determination (0.9) + lime orchard in S 3 (7.0) ]
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