During the eighteenth century an ingenious attempt to find the mass of the Earth

Question

During the eighteenth century an ingenious attempt to find the mass of the Earthwas made by the British Astronomer Royal, Nevil Maskelyne. He observed the extent to which a plumblinewas pulled out of true by the gravitational attraction of a mountain. The figure below illustrates the principleof the method.

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The change of direction of the plumbline was measured between the two sides of the mountain by sighting
on stars. After allowance had been made for the change in direction of the local vertical because of the curvature of the Earth, the residual angular divergence ? could be obtained from relating FM and FE, where
FM is the horizontal force on the plumb-bob due to the mountain and FE = GMEm/R2
E is the force on
the bob (of mass m) due to the Earth. The value was found to be about 10 seconds of arc for measurements
on opposite sides of the base of a mountain about 2000 m high and 4000 m wide.
Assume that the mountain can be approximated by a cone of rock of density 2.5 g/cm3
and volume V = ?r2h/3. Further assume that its mass can be approximated to be concentrated at the center of the base of
the mountain when computing the gravitational force of the mountain on the plumb-bob.
Given ? = 10" (I'm not sure what that notation means), find a result for the mass of the Earth and compare it to the accepted value. Using the
accepted value for the mass of the Earth, how much would you expect the bob to be deflected?

Explanation / Answer

Since F (gravitational) = G M m / R^2
Fm / Fe = (Re / Rm)^2 * Mm / Me    where m refers to mountain and e refers to earth
(Note that the m in the formula is the mass of the bob and cancels
Mm = pi * r^2 * h / 3 * 2500 kg / m^3) = 2.1 * 10E13 kg    for the mountain
Fm / Fe = (6.4 * 10E6 / 2000)^2 * 2.1 * 10E13 / 6 * 10E24 = 3.6 * 10E-5
(Using 5.98 * 10E24 for Me and 6.4 * 10E6 for Re)
tan alpha = Fm / Fe    where Fm is the force due to mountain and Fe due to earth
For small angles tan alpha = alpha where alpha is in radians
Now 1 second (1") = 1 / 3600 deg = 2.8 * 10E-4 deg
1 sec = 2.8 * 10E-4 deg / (57 deg / rad) = 4.8 * 10E-6 rad
alpha = 3.6 * 10E-5 rad / (4.8 * 10E-6 rad /sec) = 7.5"    for the expected deflection (each side)
Now you can use this to calculate Me in the above formula

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