4 ms·

Would gravity there be three Gs? I don't think we could handle that. The atmosphere might be too thick for us too. May it has a nice moon?

kqr2 16y ago
According to http://news.yahoo.com/s/afp/20100929/sc_afp/usastronomyplanet_20100929210707 http://news.yahoo.com/s/afp/20100929/sc_afp/usastronomyplane... , surface gravity might only be slightly higher than earth's.
- tocomment 16y ago
  How is that possible?
  - cryptoz 16y ago
    Density of a planet's interior is not uniform, and gravity tapers exponentially with distance. So, a planet with a dense core and a less-dense middle area could easily have less gravitational force than you might expect since the surface is so far away from the dense core.
    - hugh3 16y ago
      Surprisingly not so. If you have a body with a spherically symmetric mass distribution, from the outside it always has exactly the same gravitational field as a point mass at the centre. Proof: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell.html http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshel...
      - cryptoz 16y ago
        Wait, but didn't I specifically say I was referring to a body without a spherically symmetric mass distribution? Maybe I'm incorrect in my understanding of symmetric, but I was definitely referring to a non-symmetric case in my text.
        hugh3 16y ago
        Spherically symmetric means that it looks the same no matter which way you rotate it. You talked about a dense core, less dense middle-region planet, which is spherically symmetric. If the planet had an ultra-dense region off to one side then it wouldn't be spherically symmetric and it would have gravity slightly stronger on one side than on the other... it's hard to imagine such a planet existing though.
        cryptoz 16y ago
        Ah, gotcha. Thanks!
    - eru 16y ago
      > and gravity tapers exponentially with distance. Inverse square law.
  - imp 16y ago
    It's more massive, but the radius is bigger as well. It could be as low as 1.5 Gs. Modifying the gravitational formula (http://en.wikipedia.org/wiki/Earths_gravity http://en.wikipedia.org/wiki/Earths_gravity), it's m/r^2, so 3/(1.4^2) = 1.53 Gs. That's neglecting any local variations.

kqr2 16y ago

According to http://news.yahoo.com/s/afp/20100929/sc_afp/usastronomyplanet_20100929210707 http://news.yahoo.com/s/afp/20100929/sc_afp/usastronomyplane... , surface gravity might only be slightly higher than earth's.

tocomment 16y ago
How is that possible?
- cryptoz 16y ago
  Density of a planet's interior is not uniform, and gravity tapers exponentially with distance. So, a planet with a dense core and a less-dense middle area could easily have less gravitational force than you might expect since the surface is so far away from the dense core.
  - hugh3 16y ago
    Surprisingly not so. If you have a body with a spherically symmetric mass distribution, from the outside it always has exactly the same gravitational field as a point mass at the centre. Proof: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell.html http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshel...
    - cryptoz 16y ago
      Wait, but didn't I specifically say I was referring to a body without a spherically symmetric mass distribution? Maybe I'm incorrect in my understanding of symmetric, but I was definitely referring to a non-symmetric case in my text.
      - hugh3 16y ago
        Spherically symmetric means that it looks the same no matter which way you rotate it. You talked about a dense core, less dense middle-region planet, which is spherically symmetric. If the planet had an ultra-dense region off to one side then it wouldn't be spherically symmetric and it would have gravity slightly stronger on one side than on the other... it's hard to imagine such a planet existing though.
        cryptoz 16y ago
        Ah, gotcha. Thanks!
  - eru 16y ago
    > and gravity tapers exponentially with distance. Inverse square law.
- imp 16y ago
  It's more massive, but the radius is bigger as well. It could be as low as 1.5 Gs. Modifying the gravitational formula (http://en.wikipedia.org/wiki/Earths_gravity http://en.wikipedia.org/wiki/Earths_gravity), it's m/r^2, so 3/(1.4^2) = 1.53 Gs. That's neglecting any local variations.