Calculating a planet's mass & gravitational pull

[Asmala]

Originally posted by dogscoff:
For Earth I get a gravity value of 9840866.19 (starting with a diameter of 12756km and density of 5520kg/m^3).

Did you remember convert the diameter to meters? The answer (9840866.19) is right, but the unit would be μm/s^2.

I'm still a little fuzzy on the kg-1 s-2, although I now understand that it's just the unit by which gravity is measured. How would you pronounce it? What does that s stand for? Is it seconds?

Yes s is seconds. s-2 is equal to s^-2 which is equal to 1/s^2

[dogscoff]

Hmm... I hadn't done that. However if I divide the diameter by 1000 to get metres, my gravity figure comes out even bigger. Or is that right?

[Andrés]

Let's see what I remember from my physics classes.
Gravity on earht should be 9.8 m/s2 pronounced "meters per square second". Meaning the speed of an object in free fall will increase in 9,8 m/s every second.
So you seem to have the point shifted several spaces.

No, rotation should actually create a centrifugal force that will pull you away from the planet and tend to lower gravity close to the equator.

[Suicide Junkie]

You need to multiply by 1000 instead of divide, since there are more meters than kilometers.

[Slick]

DOH! That's pretty much how NASA missed Mars with the Last mission. Or maybe they hit Mars, but they certainly didn't get into orbit as desired. It was a unit conVersion error.

Slick.

[Fyron]

It helps if you set up "conVersion factors" (lame, yes; but still useful). An example:

code:

---

         / 1000 m \

12 km x  | ------ |  =  12000 m

         \  1 km  / 

---

(those are parenthesis)

[dogscoff]

Yeah, one of my planned features will be the ability to select different units for input/output. I'll just have to filter all calculations through a "conVersion" function.