@jbriggs444 I don't think that's right about gravity scaling linearly. Also, I think pi helps to rule out the linear relationships.
What about my equation (linked above) is wrong?
I took the surface gravity equation here:
http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html -- g = G * M /...
That definitely seems reasonable. The gravity at the surface also has the inverse square law from the center of mass to consider as well though, which is why it is maybe bigger than it first would seem...or my equation is wrong.
I was thinking that it wasn't quite small enough either, but could the cubic nature of the volume of a sphere lead to a slightly unintuitive answer?
@Dave I don't think it'd work that was as it wouldn't be a linear comparison.
I think I got it with wolfram's help:
https://www.wolframalpha.com/input/?i=solve+1%3D6.67*10^-11%2818.9%284%2F3r^3%29%29%2Fr^2+for+r
Gets 5.949x10^8 which seems high. I think one of the equations is solving for centimeters while the other is using kilometers
Converting the cm to km does...
Thank you.
So if I understand correctly, I'd need two equations then. One to figure out the mass (with the known density of U238) given a particular radius. And another to figure out the radius given a particular mass in order to equal 1g. I'm guessing I could use some method to graph the two...
Physics is not my forte, yet.
If I understand correctly do I just solve for R (where g=1)?from http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html:
You have learned that the surface gravity (g)of a body depends on themass (M)and theradius (r)of the given body.
The formula which relates...