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albear · Apr 15, 2007, 07:17 AM

The earth has density rho(p) and radius R. the gravitational field strength at the surface is g. what is the gfs at the surface of a planet of density 2p and radius 2R.
A g
B 2g
C 4g
D 16g

what is the equation I would use to solve this

ebaines · Apr 17, 2007, 04:11 PM

Originally Posted by albear

The earth has density rho(p) and radius R. the gravitational field strength at the surface is g. what is the gfs at the surface of a planet of density 2p and radius 2R.
A g
B 2g
C 4g
D 16g

what is the equation i would use to solve this

The general expression for gravitational force of attraction between two bodies is GMm/R^2, where G is the gravitational constant, M and m are the masses of the two bodies, and R is the distance betweentheir centers of gravity. Applied to determine the weight of a body at the surface of the earth, G, M (mass of the earth) and R (radius of the earth) are constant. Hence we express the weight of the body at the surface of the earth as g*m, so g is equal to GM/R^2.

Does this help?

albear · Apr 18, 2007, 08:11 AM

thank you for answering, yes it does help but I should have been more specific when asking what equation to use, I was referring to how do I link density (p) to Gm so that I can then try to solve the question. Do you know how I can relate them ?

asterisk_man · Apr 18, 2007, 08:25 AM

mass=density * volume
therefore, mass is directly proportional to density

albear · Apr 18, 2007, 08:33 AM

OK so g=G(p*v)/R^2 right so therefore if p and R were both *2 then the 2s would cancel out giving g as the answer right?

asterisk_man · Apr 18, 2007, 08:34 AM

I think the difference in r that would be in the v will have an effect also

albear · Apr 18, 2007, 08:36 AM

? I'm sorry I don't quite understand that

asterisk_man · Apr 18, 2007, 08:38 AM

$GFS_{earth}=G\frac {M_{earth}} {{r_{earth}}^2} = g\\ M_{earth}=\rho_{earth} \frac 4 3 \pi {r_{earth}}^3 \\ GFS_{other}=G\frac {M_{other}} {{r_{other}}^2} \\ M_{other}=2\rho_{earth} \frac 4 3 \pi {\left(2r_{earth}\right)}^3 = 2\rho_{earth} \frac 4 3 \pi 8{\left(r_{earth}\right)}^3 = 16 \rho_{earth} \frac 4 3 \pi {\left(r_{earth}\right)}^3 = 16 M_{earth}\\ GFS_{other}=G\frac {16M_{earth}} {{\left(2r_{earth}\right)}^2} = G\frac {16M_{earth}} {4{\left(r_{earth}\right)}^2} = 4G\frac {M_{earth}} {{\left(r_{earth}\right)}^2}=4GFS_{earth}=4g\\ $

I think this is right.
Can you see how the extra radius increased the mass substantially while simultaneously decreasing the field strength?

albear · Apr 18, 2007, 09:00 AM

4/3piR^3 is the volume of a sphere but yes I can see the reason why now it's the inverse square law thanks (on the second to last line of working there is a mistake in
2p(4/3)pi8R^3=16p(4/3)pi8R3) so g=Gm/R2 but m=g*V and V=(4/3)piR3. Thank you for your help

asterisk_man · Apr 18, 2007, 09:38 AM

Thanks for catching the mistake! I've fixed it now. Let me know if you need any more help with this.