|
|
|
|
Ultra Member
|
|
Apr 15, 2007, 07:17 AM
|
|
density, radius and g
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
|
|
|
Expert
|
|
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?
|
|
|
Ultra Member
|
|
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 ?
|
|
|
Full Member
|
|
Apr 18, 2007, 08:25 AM
|
|
mass=density * volume
therefore, mass is directly proportional to density
|
|
|
Ultra Member
|
|
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?
|
|
|
Full Member
|
|
Apr 18, 2007, 08:34 AM
|
|
I think the difference in r that would be in the v will have an effect also
|
|
|
Ultra Member
|
|
Apr 18, 2007, 08:36 AM
|
|
? I'm sorry I don't quite understand that
|
|
|
Full Member
|
|
Apr 18, 2007, 08:38 AM
|
|
I think this is right.
Can you see how the extra radius increased the mass substantially while simultaneously decreasing the field strength?
|
|
|
Ultra Member
|
|
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
|
|
|
Full Member
|
|
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.
|
|
Question Tools |
Search this Question |
|
|
Check out some similar questions!
Setting up Mesh network to cover 5 km radius
[ 2 Answers ]
Hi all,
I wish to setup a mesh Wi-fi network, to cover a area of about 5 sq km. I have about 150 clients who will be connecting to this mesh network.
Any solution for this?
Radius for Arched Doorway
[ 8 Answers ]
I am trying to properly lay out an arched doorway and can not seem to get this... what I do know is as follows.. the two side walls are 8'9" from the floor to the bottom point of the arch...from a center point between these two vertical lines taking a vertical measurement from the horizontal...
Density question pelvis
[ 12 Answers ]
can anyone help me figure out what this means? This is an x-ray result.
4mm density overlying the right pelvis. While I doubt this is due to a PHLEBOLITH, this cannot be entirely excluded. What is a phlebolith. Thanks
Converting Density
[ 2 Answers ]
Hello,
I am trying to help someone out with their homework and I've managed to help her with all kinds of metric conversions.
But now I am stuck.
The problem is convert 5.3 kg/L to g/cm^3
That's 5.3 kilograms per liter to grams per centimeter cubed.
View more questions
Search
|