Question 4.1: Calculate the tide-generating force on the Earth due to the ......

Calculate the tide-generating force on the Earth due to the Earth-Moon and Earth-Sun systems, given that the mass of the Earth, (m),  is $= 5.98 × 10^{24}$ kg, the mass of the moon, (M), = $7.35 × 10^{22}$ kg, the major semi-axis of the lunar orbit around the Earth, (r), is $3.84 × 10^{8}$ m, and the mean radius of the Earth, (s), is $6.37 × 10^{6}$ m.

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We may estimate the magnitude of the effect of the tide-generating forces by assuming that the water on the Earth’s surface is in equilibrium. (This is equivalent to assuming that the natural periods of tidal oscillation are small in comparison to the rotation period of the Earth.) In this case the surface of the water takes a shape on which the gravitational potential has a single value. Let the height through which the sea surface is raised be η(θ). Then, under the assumption that η(θ) is small, the change in the Earth’s gravitational potential due to a rise in the sea surface of η(θ) is approximately gη(θ). This change must be balanced by the potential due to the Moon to maintain equilibrium. Thus,

$g\eta(\theta)=\frac{GM s^2}{ r^3}\left\lgroup\frac{3 \cos ^2(\theta)-1}{2}\right\rgroup$ .              (4.8)

Thus, from Equation 4.1,

$F=\frac{GmM}{\left|r_{m}-r_{M} \right| ^{2} }$                (4.1)

g = Gm/s². Substituting this into Equation 4.8 gives

Equation 4.9

$\eta(\theta)=\frac{M s^4}{ mr^3}\left\lgroup\frac{3 \cos ^2(\theta)-1}{2}\right\rgroup$            (4.9)

gives the maximum tidal elevation for the Moon as 36 cm and for the Sun as 16 cm. Equation 4.9 describes a prolate spheroid and indicates a lowering of the sea surface away from the equator.

Question: 4.2

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