Gravitational potential energy of a sphere of non-uniform density

In summary: DI think I understand this now, I was just lacking in understanding what the equations were actually telling me :)In summary, the total gravitational potential energy stored in a sphere with a 1/r2 density distribution can be found by first integrating to determine the density constant in terms of the known values (total mass and total radius), then using this constant to find an expression for the mass of the sphere at any given radius, and finally summing the contributions from each shell using the expression for gravitational potential energy, dU(r)=-((GM(r)dM)/r). This results in the equation U=G(16/15)π2ρ02R5, where ρ0=(M/(4πR)).
  • #1
mell0r
11
0

Homework Statement



Find the total gravitational potential energy stored in a sphere with a 1/r2 density distribution if the total mass is 6.7 solar mass and the radius is 1.3 solar radius. Express you answer in units of 1041 Joules.


The Attempt at a Solution



To derive the equation, i attempted to follow the same procedure in deriving the gravitational potential energy of a constant sphere, but with also considering the altering value of density throughout.

I started off by figuring out an equation for the mass in a thin shell of mass dM and width dr at radius r:

dM=4πr2ρ0dr

Then integrated to obtain an equation of the total mass of material already assembled:

M(r)=(4/3)πr3ρ0

After this, i plugged these equations into dU(r)=-((GM(r)dM)/r) and integrated again for an equation for gravitational potential energy:

U=G(16/15)π2ρ02R5

The problem is that i can't for the life of me right now figure out how to re-arange this into an equation of the form ((GM2)/R), as i can't simply plug in ρ=((M)/(4/3)πr3) like i did for the sphere of constant density...

Any help would be greatly appreciated - this has been irritating me for a while now!

Thanks.
 
Physics news on Phys.org
  • #2
mell0r said:

Homework Statement



Find the total gravitational potential energy stored in a sphere with a 1/r2 density distribution if the total mass is 6.7 solar mass and the radius is 1.3 solar radius. Express you answer in units of 1041 Joules.


The Attempt at a Solution



To derive the equation, i attempted to follow the same procedure in deriving the gravitational potential energy of a constant sphere, but with also considering the altering value of density throughout.

I started off by figuring out an equation for the mass in a thin shell of mass dM and width dr at radius r:

dM=4πr2ρ0dr
How does the above incorporate the fact that density varies as 1/r2 ?

Presumably you should let the density be given by some function of r like ##\rho(r) = \frac{\rho_0}{r^2}##
Then integrated to obtain an equation of the total mass of material already assembled:

M(r)=(4/3)πr3ρ0

After this, i plugged these equations into dU(r)=-((GM(r)dM)/r) and integrated again for an equation for gravitational potential energy:

U=G(16/15)π2ρ02R5

The problem is that i can't for the life of me right now figure out how to re-arange this into an equation of the form ((GM2)/R), as i can't simply plug in ρ=((M)/(4/3)πr3) like i did for the sphere of constant density...

Any help would be greatly appreciated - this has been irritating me for a while now!

Thanks.
 
  • #3
mell0r said:

dM=4πr2ρ0dr

Then integrated to obtain an equation of the total mass of material already assembled:

M(r)=(4/3)πr3ρ0


Potential energy is the work done in bringing a unit mass from infinity to the center of sphere.
U=integral[from infinty to zero](F.dr.cos(x)) here x=0=>cosx=1.
now f is not constant. it is different outside the sphere and also inside(non uniform density).
Break the integral for outside the sphere and inside. Like for outside do integral[infinity to R]((G{M,integrated mass of the sphere)/r^2).dr).
 
  • #4
gneill said:
How does the above incorporate the fact that density varies as 1/r2 ?

Presumably you should let the density be given by some function of r like ##\rho(r) = \frac{\rho_0}{r^2}##

Ah, yes, sorry - i meant to include that.

Even so, my mind is completely blank now.. How can i relate that function to my derivation from my ρ0 variable? :confused:
 
  • #5
If you integrate dM to find the expression for the mass (which is a given value), then you can solve that for your ##\rho_0## (probably best to leave it in symbolic form for now).

Then you can redo the integration for the mass to find the mass of the sphere as a function of r. That is, M(r). Then you can go back to summing the potential energies of the shells.
 
  • #6
So overall i'd be performing 3 integration's? What would be the difference between integrating for the mass and the mass as a function? This math seems completely new to me, and i would have Google searched to figure it out, but I'm not even sure what to search..

Sorry for being annoying haha! Thanks for your time so far though :smile:
 
  • #7
The first integration is to determine the value (or expression for) your density constant in terms of the known values (total mass, total radius).

The second integration uses this constant expression to replace your previously unknown constant ##\rho_0## when determining an expression for the mass of a sphere of arbitrary radius. That is, M(r).

The third integration uses the above M(r) and dM(r) to sum the shell-wise contributions to the potential energy.
 
  • #8
so I integrated dM=4πr2ρ0dr over 0 and R, i get M=[itex]\frac{4}{3}[/itex]πR3ρ0 and thus worked out ρ0 from re-aranging that equation. Is that correct so far?

I'm finding this so confusing, yet I can do the same derivation for a uniform density perfectly fine..
 
  • #9
mell0r said:
so I integrated dM=4πr2ρ0dr over 0 and R, i get M=[itex]\frac{4}{3}[/itex]πR3ρ0 and thus worked out ρ0 from re-aranging that equation. Is that correct so far?

I'm finding this so confusing, yet I can do the same derivation for a uniform density perfectly fine..
You need to integrate, [itex]\displaystyle\ dM=4\pi r^2\rho(r) dr=4\pi r^2\frac{\rho_0}{r^2} \,dr\,\ [/itex] because [itex]\displaystyle\ \rho(r)=\frac{\rho_0}{r^2}\ .[/itex] Integrate this from 0 to R and set it equal to M to evaluate ρ0.
 
  • #10
SammyS said:
You need to integrate, [itex]\displaystyle\ dM=4\pi r^2\rho(r) dr=4\pi r^2\frac{\rho_0}{r^2} \,dr\,\ [/itex] because [itex]\displaystyle\ \rho(r)=\frac{\rho_0}{r^2}\ .[/itex] Integrate this from 0 to R and set it equal to M to evaluate ρ0.

Thank you! so the integral would be M=4πRρ0, and therefore ρ0=[itex]\frac{M}{4πR}[/itex]??
 
  • #11
mell0r said:
Thank you! so the integral would [STRIKE]be[/STRIKE] give M=4πRρ0, and therefore ρ0=[itex]\frac{M}{4πR}[/itex]??

Right !
 
  • #12
Wow, there you go. I Don't think I've ever been as happy to see an equation before as i just have haha! Thank you SammyS, gneill and smatik for the fantastic help and for coping with me :tongue:
 

1. What is gravitational potential energy?

Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. It is the energy required to move the object from its current position to a reference point, usually at infinity, where the gravitational potential energy is defined as zero.

2. How is gravitational potential energy related to non-uniform density?

The gravitational potential energy of an object is directly proportional to its mass and the strength of the gravitational field it is in. In the case of a sphere of non-uniform density, the mass is not evenly distributed, so the gravitational potential energy will vary depending on the distance from the center of the sphere.

3. How is the gravitational potential energy of a sphere of non-uniform density calculated?

The gravitational potential energy of a sphere of non-uniform density can be calculated by integrating the potential energy at each point on the surface of the sphere, taking into account the varying density and distance from the center. This can be represented by the equation: U = -G ∫(ρdV/r), where G is the gravitational constant, ρ is the density at a given point, dV is the volume element, and r is the distance from the center of the sphere.

4. How does the gravitational potential energy of a sphere of non-uniform density compare to that of a uniform density sphere?

The gravitational potential energy of a sphere of non-uniform density will be higher than that of a uniform density sphere with the same mass, as the varying density causes a greater variation in potential energy at different points on the surface.

5. Can the gravitational potential energy of a sphere of non-uniform density be negative?

Yes, the gravitational potential energy of a sphere of non-uniform density can be negative if the reference point is chosen to be closer to the center of the sphere. In this case, the potential energy at points closer to the surface will be negative, indicating that work would need to be done to move an object from that point to the reference point.

Similar threads

  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
354
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
846
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
945
  • Introductory Physics Homework Help
Replies
6
Views
667
Back
Top