Schiehallion Experiment - Mathematical Procedure

Mathematical Procedure

Consider the force diagram to the right, in which the deflection has been greatly exaggerated. The analysis has been simplified by considering the attraction on only one side of the mountain. A plumb-bob of mass m is situated a distance d from P, the centre of mass of a mountain of mass MM and density ρM. It is deflected through a small angle θ due to its attraction F towards P and its weight W directed towards the Earth. The vector sum of W and F results in a tension T in the pendulum string. The Earth has a mass ME, radius rE and a density ρE.

The two gravitational forces on the plumb-bob are given by Newton's law of gravitation:

$F = frac {G m M_M} {d^2} ,quad W = frac {G m M_E} {r_E^2}$

Where G is Newton's gravitational constant. G and m can be eliminated by taking the ratio of F to W:

$frac {F} {W} = frac {(G m M_M) / d^2} {(G m M_E) / r_E^2} = frac {M_M}{M_E} {left( frac {r_E}{d} right)}^2 = frac {rho_M} {rho_E} frac {V_M} {V_E} {left( frac {r_E}{d} right)}^2$

Where VM and VE are the volumes of the mountain and the Earth. Under static equilibrium, the horizontal and vertical components of the string tension T can be related to the gravitational forces and the deflection angle θ:

$W = T cos theta ,quad F = T sin theta$

Substituting for T:

$tan theta = frac {F} {W} = frac {rho_M}{rho_E} frac {V_M}{V_E} {left( frac {r_E}{d} right)}^2$

Since VE, VM, d and rE are all known, and θ and d have been measured, then a value for the ratio ρE : ρM can be obtained:

$frac {rho_E}{rho_M} = frac {V_M}{V_E} {left( frac {r_E}{d} right)}^2 frac {1}{tan theta}$

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