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.

Comments

Berkeley Press Release

[metlin]

Here is the link to the Berkeley press release and information on Berkeley astronomer Geoff Marcy.

And oh, looks like Slashdot is continuing to mirror Boing Boing.

Re:minimum mass

[iostream_dot_h]

The planet has approximately 7.5 times the mass of the Earth, double its radius, and orbits its parent star once every two days.

g_newplanet = G(7.5M_Earth)/4(r_Earth^2), where g is the gravitational field strength and G is the gravitational constant. This is less than 2g_earth.

Seems close enough for life. Although I'm not sure why a stronger gravitational field would necessarily be a huge constraint on the development of life.

And to preempt the trolls, my little convenient formula is good enough. Relativistic effects can be ignored.

Re:minimum mass

[nacturation]

Is life possible at all under such gravity? Any examples?

And where do you propose we find such examples? In space perhaps? :) But seriously, check out this article.

Re:Doesn't sound very earthlike

[Pinefresh]

No you idiot, it said extra solar there, replied to myself.

Re:Orbital Velocity - significant acceleration?

[StupendousMan]

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.