Gravitation - gravitational attraction due to a nearby mountain range

  • #1

Homework Statement



The gravitational attraction due to a nearby mountain range might be expected to cause a plumb
bob to hang at an angle slightly different from the vertical. If a mountain range could be
represented by an infinite half-cylinder of radius a and density [itex]ρ[/itex] lying on a flat plane (see
attached image), show that a plumb bob at a distance [itex]r_0[/itex] from the cylinder axis would be deflected
by an angle [itex] θ ≈ \frac{πa^2Gρ}{r_0g}[/itex]. You can assume that [itex]θ << 1[/itex] and [itex]r_0 >> a[/itex]. In actual measurements
of this effect, the observed deflection is much smaller. Next assume that the mountain range
can be represented by a cylinder of radius a and density ρ which is floating in a fluid of density
[itex]2ρ[/itex] as illustrated. Show that the plumb-bob deflection due to the mountain range is zero in this
model. Since the latter result is in much better agreement with observations, it is postulated that
mountains, and also continents, are in isostatic equilibrium with the underlying mantle rock.

Homework Equations



[itex] ψ_g=\iint \vec{g_c}\cdot\vec{dA} = -4πρG [/itex] and [itex] ∑τ=0[/itex]

The Attempt at a Solution



Part A) First I need to calculate the gravitational field [itex] \vec{g_c} [/itex]. Because the mountain is approximated as an infinite half cylinder the only contribution comes from the thin slice of directly in front of the plum-bob.

[itex] ψ_g = \iint\|\vec{g_c}\|\|\vec{dA}||cosθ = -4πρG →\|\vec{g_c}\| \iint \|\vec{dA}\| = -4πρG → \|\vec{g_c}\|π(r_0) = -4πρG → \|\vec{g_c}\| = \frac{-4ρG}{r_0} \Rightarrow \|\vec{F_c}\| = \frac{-4ρGm(πa^2)}{2r_0}[/itex]

Now the other force acting on the bob is the Earth's gravitaional force given near surface of the earth is: [itex]\|\vec{F_g}\|=-mg[/itex]

Now consider the torque given by both forces acting on the plum-bob: [itex]∑τ=0 → -Lmgsin(θ)-L[\frac{-2ρGπa^2}{r_0}]sin(Θ) = 0 [/itex] where [itex]θ+Θ=90°[/itex] so [itex]θ<<1 \Rightarrow Θ≈90 [/itex] then the above equation simplifies to: [itex] mgθ≈\frac{2ρGπa^2m}{r_0} → θ≈\frac{2πa^2Gρ}{r_0 g}[/itex] and since the angle is small to begin with the factor of 2 can be neglected so I get the predicted result

Part B) I'm not really sure where to go with the next part so any hints would be greatly appreciated. What I do know: The earth's gravitation field is much stronger than the field due to the infinite half cylinder's (mountain's) and there is a bouyant force acting on the plum bob opposite to the earth's gravitational field. This lead me to believe there was a balancing viscous force due to the fluid, but I'm unsure of this.
 
Last edited:

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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What water? The continents are thought to float on the mantle, which is a fluid composed of molten rock.
 
  • #3
I'm sorry I meant to say mantle not water
 
  • #4
Hi Wavefunction. Have you considered that the mass densities on either side of the pendulum are the same? At least in the condition r>>a, the force due to the mountain would equal zero because all that would matter would be the mass densities and not the specific mass distributions.

Edit: Also, my first post here.
Is this what it's like to contribute to the internet?
 
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  • #5
Hi Wavefunction. Have you considered that the mass densities on either side of the pendulum are the same? At least in the condition r>>a, the force due to the mountain would equal zero because all that would matter would be the mass densities and not the specific mass distributions.

Edit: Also, my first post here.
Is this what it's like to contribute to the internet?
Could you explain how the mass density matters more than the specific mass distribution in this case? Are you basically saying that that I take my answer from a and then take the limit as [itex] r_0 →∞[/itex] (Apply [itex]r_0\ll a[/itex]) because that would give me my answer.
 

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