Consider an Earthlike Planet. It has a similar composition and density to the Earth with an iron core and water oceans. What is the formula to determine the acceleration due to gravity at the planet's surface?

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    $\begingroup$ ...What? If your planet has the same mass than earth, then it would also have the same gravitational attraction. $\endgroup$ – AngelPray Jul 27 '17 at 10:33
  • $\begingroup$ If it's an earth-like planet with the mass of earth it will have the same gravity as Earth. And water oceans will influence the mass. After all adding water means adding mass. I am not sure I understand what you are asking here. $\endgroup$ – Sec SE - clear Monica's name Jul 27 '17 at 10:34
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    $\begingroup$ Not the same mass as earth , but proportionally the same density and composition , I guess Iexpressed poorly the question. $\endgroup$ – Naima Jul 27 '17 at 10:37
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    $\begingroup$ $F=G\frac{m_1 m_2}{r^2}$ $\endgroup$ – Fl.pf. Jul 27 '17 at 10:52
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    $\begingroup$ If you express yourself poorly, please edit your question to make it clear. $\endgroup$ – Mołot Jul 27 '17 at 12:15

Gravity at surface is easy enough: in SI, $g=\frac{GM}{r^2}$.

G is the gravitational constant, M is the planet's mass, r is its radius, g is the surface gravity.

If you're working in multiples of Earth rather than SI, $g_n=ga_r^{-2}b_m$ can be used, where $a_r$ is the scaling factor of the radius of earth (0.5 for half the radius, etc.) and $b_m$ is the scaling factor for mass.

If you're working in density terms, $M=\frac{4}{3}\pi r^2 \rho$ (or, again, in Earth-centric units, $M=r^2 \rho$), with $\rho$ being the density in each case.

Wikipedia will tell you that "The total mass of the hydrosphere is about 1.4 quintillion metric tons ($1.4×10^{18}$ long tons or $1.5×10^{18}$ short tons), which is about 0.023% of Earth's total mass" – which isn't enough to make a significant difference to gravity.

  • $\begingroup$ Good answer - it can be improved by adding simple examples for SI and multiples of earth, and also by providing the value for G (6.67408 × 10^-11). That'll make it easy enough to use without additional sources... $\endgroup$ – G0BLiN Jul 27 '17 at 14:05

The force of gravity exerted on an object on the surface of a planet is proportional to the mass of the planet and inversely proportional to the square of the distance from the object to the center of the planet. So, if you work with values relative to those of Earth, you can forget about difficult units and large numbers.

$$ g = {M \over r^2} $$

where $M$ = mass of the planet, expressed in Earth masses; and $r$ = radius of the planet, expressed in Earth radii. Of course $g$, the gravity, will also be expressed in Earth gravities.

For example, a Super-Earth that is 1.2 times the size of Earth and with 1.8 times its mass will have a surface gravity of

$$ g = {1.8 \over 1.2^2} = {1.8 \over 1.44} = 1.25 $$

The actual value will be 1.25 times Earth gravity, i. e.

$$ g = 1.25 \times 9.8\space m / s^2 = 12.25\space m / s^2$$

Note that mass and radius are not independent. If your planet is larger than Earth but has the same mass, then it must have a lower density, but a Super-Earth (a large rocky planet) cannot be a lot less dense than Earth.

Oceans on Earth make no significant difference on the mass of the planet. In any case, an ocean is not made of special stuff; water (and ice) only add to the overall mass of the planet.


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