I would do the math myself, really ... if I knew where to start, and which numbers went where. Sorry, it's one of the costs of not being a scientist. But we've started hearing about the Kepler satellite and its apparent finds. Via BBC: A solar system including six planets around a star 2,000 light-years away has been spotted by astronomers. The planets range between two and four-and-a-half times the radius of Earth, and between two and 13 times its mass. Five of the planets orbit the star closer than Mercury orbits our Sun. (Boldface accent added) Is it possible ...? Could I appeal to my more scientific neighbors ...? There are a number of ifs about it, of course. But ... how does one calculate the gravitational acceleration at the surface of one of these planets? That is, when I was a kid, there were these fun little facts in childrens' books about how much you would weigh on the moon compared to how much you would weigh on Mars, &c. I mean, it's not so simple, is it, as twice the mass equals twice the acceleration? Thirteen times the Earth's mass? We're not really looking at 127.4 m/s[sup]2[/sup], are we? No, really, I can do the arithmetic, but I have no idea how to set it up. I'm not sure I would even know where to start looking. Or is it just the simple F=ma, leading to— Please Register or Log in to view the hidden image! Am I making too much of the question? I would only point out that it is curious to me that, no matter what planet he was on, Captain Kirk seemed to have little trouble moving around, suggesting that intelligent life only developed on planets of similar mass to Earth. I mean, I can suspend disbelief so Kirk, or Darth Vader, or whoever, can walk around on a spaceship. But it occurs to me that we rarely even notice the fact that even in some Martian stories, gravity isn't much of a consideration. It's a curiosity that struck me as I was reading the Beeb article. ____________________ Notes: Palmer, Justin. "Six exoplanets in close orbit around far-flung star". BBC News. February 2, 2011. BBC.co.uk. February 3, 2011. http://www.bbc.co.uk/news/science-environment-12333766 Wikipedia. "Newton (unit)". February 2, 2011. Wikipedia.org. February 3, 2011. http://en.wikipedia.org/wiki/Newton_(unit)
Hi Tiassa, You need Newton's Law of universal gravitation for the force between two masses with distance r between their centres: \(F = G \frac{m_1.m_2}{r^2}\) The force of gravity is proportional to each mass, and inversely proportional to the distance between their centres. So, doubling the mass of a planet without changing its size doubles its surface gravity. Doubling its radius without changing its mass quarters its surface gravity. Doubling the mass of the object on the surface also doubles the force, which is why the acceleration doesn't depend on the surface object's mass - assuming that the object is so much smaller than the planet that the planet doesn't detectably accelerate. Hmm. What a nasty wordy explanation. I hope it's at least semi-useful. If not, the Wiki-link might give you what you seek. But anyway, you are correct. Popular sci fi usually conveniently ignores such inconvenient physics (also biology and anthropology, but whatever makes a good story, right?)
Thank ye Actually, I remember that equation from high school. I usually apply it in a philosophical context, but that's beside the point. Thank you for the information. And, yes, it's whatever makes a good story. Unfortunately, good stories are harder to come by than Hollywood would have us believe.
If you want to compare ratios, the gravitational acceleration at a planet's surface is: \(g=GM/r^2\) where M is the planet's mass and r is its radius. So, for one of these new planets, compared to Earth: \(\frac{g_1}{g_{Earth}} = \frac{M}{M_{Earth}}\left(\frac{r_{Earth}}{r}\right)^2\) To put this another way, if the new planet is \(x\) times more massive than Earth, and \(y\) times the radius, then its surface gravity multiple, compared to the Earth is: \(g_{relative} = \frac{x}{y^2}\) For example, a planet is x=6 times the Earth's mass and y=2 times the Earth's radius would have a gravity equal to \(6/2^2 = 1.5\) times the Earth's gravity.
Thank ye Now that is nifty! Gorgeous. I mean, I read in one of the many articles that Kepler might have found at least one rocky planet. I don't recall that I saw any approximation of its dimensions, though. I'll have to go fish later today. At any rate, it gives me a chance to think about the part of planetary exploration I don't think about as often. Interstellar travel, to me, is all too often a question of speed limits. Suddenly I'm wondering, "Yes, but what do we do if we actually get there?" It's a fun change of pace. Thank ye, sir. ____________________ Notes: National Aeronautics and Space Administration. "Kepler: A Search for Habitable Planets". (n.d.) NASA.gov. February 3, 2011. http://www.nasa.gov/mission_pages/kepler/main/index.html
It's probably worth considering the extremes. If a planet had 2 times Earth's radius and 13 times its mass, its surface gravity would be 13/4 = 3.25 higher than Earth's. At the other end of the scale, if the planet had 4.5 times Earth's radius and twice Earth's mass, then its surface gravity would be 2/4.5^2 = 0.1 times Earth's. So, with these parameters we get planets that have surface gravities ranging from one-tenth of Earth's to three times. For comparison, the Moon's gravity is about one-sixth of Earth's, and Mar's is about one-third.
