Rocky Planet Discovered
Fraser Cain writes "Astronomers have discovered a rocky, terrestrial planet orbiting a nearby star, Gliese 876. The planet has approximately 7.5 times the mass of the Earth, double its radius, and orbits its parent star once every two days. This is the most Earthlike extrasolar planet discovered so far." Reader Karthik Narayanaswami points out that "the planet was discovered by the famed Berkeley astronomer Geoff Marcy," and adds a link to the news release from Berkeley.
With a new year every two days, everyone would be broke buying birthday cakes.
The thing has gotta be mighty close to the star. Mercury orbits in 60 days, right? This thing may not be a gas giant, but it must totally bake on the sunny side, and aren't there going to be some horrendous tidal forces with an orbit that close? It probably has no shortage of volcanism. Hey! It's Vulcan, maybe... if it can hold an atmosphere without having the stellar wind blow it all away. Whatever, it can't be Earth-like.
For all intensive purposes, "whom" is no longer a word. That begs the question, "who cares"?
which is only 15 light years away
So why not send some radio traffic which would obviously not be of natural origins. Surely 30ish years isn't that long to wait for a reply? (assuming the place has lifeforms which developed radio...)
Trolling is a art,
I am not an astronomer, but isn't mars more earthlike than that?
No.
A body moving in a circle of radius R at a uniform speed V experiences an acceleration a = (V*V)/R towards the center of the circle. In neither of the cases you mention does any centripetal acceleration come close to the local gravitational acceleration at the surface of the planet.
Case 1: The Earth: orbital speed V = 30 km/s, and R = 150 million km, so (V*V)/R is of order (10^8)/(10^11) m/s^2, or about 10^(-3) m/s^2. The local gravitational acceleration is about 10 m/s^2, of course. If you speak of the Earth's rotational motion at the equator, then very roughly V = 500 m/s and R = 6,400,000 m, so (V*V)/R has magnitude roughly (2.5 x 10^5) / 6.4 x 10^6 = 0.03 m/s^2; again, much less than 10 m/s^2 due to the gravitational pull of the Earth.
Case 2: The new planet. Its orbital radius is about 2 billion meters, so the circumference is about 7 billion meters; if it travels that distance in a period of 2 days = 170,000 seconds, then it speed is about V = 40,000 m/s. The orbital centripetal acceleration is therefore of order (16 x 10^8)/(2 x 10^9) = 0.8 m/s^2. That's much larger than the Earth's orbital centripetal acceleration, but still far less than the likely gravitational acceleration at the surface (or cloudtops) of this planet.
Michael Richmond "This is the heart that broke my finger."
mwrsps@rit.edu http://stupendous.rit.edu
This is pretty simple. Surface gravity for spherically-symmetrical masses scales linearly with mass and inverse-square with radius. The mass makes gravity 7.5 times higher, while the radius would make it 4 times lower, for a total surface gravity of about 1.9G.
Patrick Doyle
I mod down every jackass who puts his moderation policy in his sig. Oh, wait a sec....