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Yahoo! science articles? What, are you trying to send me back into a depressive cycle?Click here if you're having trouble sleeping.
"We confess our little faults to persuade people that we have no large ones." - François de La Rochefoucauld
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Hey Lori - are you as put out as I am about yet another "habitable" planet story where the planet -once again- turns out to be too big? What's the likely surface gravity of a planet 1.1 time the radius of Earth, assuming similar composition/density? I bet it comes out non-trivially more than 1.10 g...
It's almost as bad as every popular science story ever about binaries with a planet mentioning Tatooine in the headline.
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Not so. Volume is proportional to the cube of the radius. Mass is proportional to volume, so mass is proportional to the cube of the radius. Acceleration due to gravity is proportional to mass / r^2, which here means r^3 / r^2 which just equals r. So assuming a constant density, acceleration due to gravity is directly proportional to the radius, and you get 1.1 times the gravity. That said, we have absolutely no idea what the density (and mass) of this planet is, because the transit method alone doesn't allow for such a calculation.Click here if you're having trouble sleeping.
"We confess our little faults to persuade people that we have no large ones." - François de La Rochefoucauld
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My pathetic maths are only barely up to calculating the change in mass assuming same density, which is a poor assumption. I can't get a figure lower than 1.3g, but that's without allowing for the scaling of gravity with distance.
GeoModder is trying to tell me the same thing, 1.10g, and it's clear I'm in over my head. I still wouldn't want to live there.
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I notice that the margin for error at 59ly is decidedly significant for these purposes. We simply don't have good enough data to assume much more than it's nowhere like gas giant size.
However, given the previous assumptions, I'm getting around 33% greater mass. Is a mere 10% greater G consistent?
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What I think you may be forgetting is that while the strength of gravity may be ~1.3 times higher at any given point, the surface on this planet is 1.1 times farther away than it is on Earth. Gravity falls off with the square of the distance, so the equation you're looking at is 1.3/(1.1^2) = 1.07 gees. (I'm assuming that's lower than the 1.1 because 1.3 is slightly too low an approximation.)Click here if you're having trouble sleeping.
"We confess our little faults to persuade people that we have no large ones." - François de La Rochefoucauld
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Which comes out to pretty much exactly 1.1Click here if you're having trouble sleeping.
"We confess our little faults to persuade people that we have no large ones." - François de La Rochefoucauld
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Thanks.
I seem to have given you the impression that I'm an ascientific crank. I hope that turns out not to be the case. Nah, I think I do pretty well for someone with so near-zero formal schooling and sub-par talent for maths.
Is this a coincidence of the numbers? I mean, I shouldn't expect double the radius to turn out so near double the surface acceleration, should I?
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Nah, you're not a crank. You just don't know my handy dandy special physics student tricks.
For me, this isn't a math problem. I probably didn't explain myself very well, but my first explanation just looks at this in terms of proportionalities. You don't have to plug in any numbers or fiddle around with calculations; you just have to see how the variables in the problem relate to each other. When you learn physics, this is one of the tricks they teach you: be lazy when you can. Don't use real numbers unless you have to.
The takeaway for 1/r^2 forces is that distance is extremely important. For example, even though the Sun is ~300,000 times as massive as the Earth, the gravity at its surface is only ~30 times stronger.Click here if you're having trouble sleeping.
"We confess our little faults to persuade people that we have no large ones." - François de La Rochefoucauld
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