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I understand that pascals have the Newtons part (which changes when acceleration changes) and the $m^2$ part (which is the same here and on Venus or something). I was wondering if you would take the earth-pascals and divide by $9.8\,\mathrm{m/s}^2$, then multiply by your planets gravity, THEN put it over $m^2$, or if that is the wrong way to do it.

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    $\begingroup$ A $\mathrm{Pa}$ is a standard unit, and not dependent on place, and so is $1\,\mathrm{N}$. $\endgroup$ – Sayan Mandal Aug 23 '18 at 16:48
  • $\begingroup$ @SayanMandal But wouldn't the pascals exerted on the ground change if an object of equal mass was set down on the surface or Earth vs. Mars? $\endgroup$ – S. Horangic Aug 23 '18 at 16:57
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    $\begingroup$ I think you are confusing between pressure (which is a physical quantity) and a pascal (which is a standard unit used to measure pressure). Similarly, a newton is a standard unit of force. So, while the force indeed changes (according to the prescription you mentioned), the units don't. $\endgroup$ – Sayan Mandal Aug 23 '18 at 17:00
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A Pascal is a unit derived from the SI system of units.

There is no dependence on local gravity or other external effects. As far as possible these units are based on multiples of precisely defined physical quantities which should be universal.

In the case of the Pascal it is one Newton per meter squared ( $1\,kg\,m^{-1}s^{-2}$ ). The meter and second are very precisely defined in modern terms. The kilogram has a definition based on a physical object of reference, but steps are under way to replace that definition with a more modern version which will remove the need for a reference object ( The International Kilogram Prototype ) which is difficult to reproduce.

None of these units are dependent on local gravity.

I was wondering if you would take the earth-pascals and divide by $9.8m/s^2$, then multiply by your planets gravity, then put it over $m^2$, or if that is the wrong way to do it.

When you divide the atmospheric pressure by the acceleration due to gravity you get the mass per unit area ( in SI units $kg\,m^{-2}$ ). When you multiple it by a different acceleration due to gravity you again get a pressure value.

However for a given planet the atmospheric surface pressure is not as simply related to gravity as this. There is no simple formula, but broadly speaking the thickness of the atmosphere (and it's mass per unit area) is going to be affected in complex ways by gravity (and other factors). You might reasonably expect a planet with a higher gravity than Earth's to retain a denser and thicker atmosphere (other things being equal) and hence have an higher surface pressure than your simple method would suggest.

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