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horseshoe7
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Possibly habitable planet Gliese 581 d is said to be (at a minimum) 7x the mass of the Earth... but what would the gravity be at the surface of Gliese 581 d?
7x the mass does not mean 7x the Gravity at the surface.
Gravitational force is directly proportional to mass and inversely proportional to the square of the distance.
On a SUPER EARTH like Gliese 581 d, the planet surface will be farther from the center of mass (given a similar composition of elements to the Earth and/or a similar density of the composition)... now, the distance (radius) would grow in relationship to the Volume(which grows due to the increased mass of the object), according to the formula (check my work folks - I'm an Engineer, but haven't had to do much "real math" in some time):
r = cube root of (3/4*V*pi)
So, gravity is inversely proportional to the square of distance, but distance(radius) is proportional to the cube root of Volume(~proportional to mass), so it does seem that, when calculating gravity of a 7x more massive Super Earth (vs. Earth), then the Super Earth would only have 7x - (square root of 7 or 2.65) or 4.35x more gravity than Earth... again, check my work folks - but, I believe my premise stands true in any event - 7x the mass does not equal 7x the gravity.
However, if my calculations are correct, my 200lb body would feel like about 867lbs on the surface of Gliese 581 d... even if I went on a crash diet to get to ~150lbs, I'd STILL weigh about 650 pounds there!... maybe we need to "continue the seach" for more suitable habitable planets? ... hopefully, there is a more Earth-sized planet Gliese 581 f that is right in the middle of the habitable zone of the Gliese 581 system?
7x the mass does not mean 7x the Gravity at the surface.
Gravitational force is directly proportional to mass and inversely proportional to the square of the distance.
On a SUPER EARTH like Gliese 581 d, the planet surface will be farther from the center of mass (given a similar composition of elements to the Earth and/or a similar density of the composition)... now, the distance (radius) would grow in relationship to the Volume(which grows due to the increased mass of the object), according to the formula (check my work folks - I'm an Engineer, but haven't had to do much "real math" in some time):
r = cube root of (3/4*V*pi)
So, gravity is inversely proportional to the square of distance, but distance(radius) is proportional to the cube root of Volume(~proportional to mass), so it does seem that, when calculating gravity of a 7x more massive Super Earth (vs. Earth), then the Super Earth would only have 7x - (square root of 7 or 2.65) or 4.35x more gravity than Earth... again, check my work folks - but, I believe my premise stands true in any event - 7x the mass does not equal 7x the gravity.
However, if my calculations are correct, my 200lb body would feel like about 867lbs on the surface of Gliese 581 d... even if I went on a crash diet to get to ~150lbs, I'd STILL weigh about 650 pounds there!... maybe we need to "continue the seach" for more suitable habitable planets? ... hopefully, there is a more Earth-sized planet Gliese 581 f that is right in the middle of the habitable zone of the Gliese 581 system?