Chapter 6, Problem 75GP

The gravitational force at different places on Earth due to the Sun and the Moon depends on each point’s distance from the Sun or Moon, and this variation is what causes the tides. Use the values inside the front cover of this book for the Earth–Moon distance R_EM, the Earth–Sun distance R_ES, the Moon’s mass M_M, the Sun’s mass, M_S, and the Earth’s radius R_E. (a) First consider two small pieces of the Earth, each of mass m, one on the side of the Earth nearest the Moon, the other on the side farthest from the Moon. Show that the ratio of the Moon’s gravitational forces on these two masses is

${\left(\frac{F_{\text{near}}}{F_{\text{far}}}\right)}_{\text{M}}=\mathrm{1.0687.}$

(b) Next consider two small pieces of the Earth, each of mass m, one on the nearest point of Earth to the Sun, the other at the farthest point from the Sun. Show that the ratio of the Sun’s gravitational forces on these two masses is

${\left(\frac{F_{\text{near}}}{F_{\text{far}}}\right)}_{\text{S}}=\mathrm{1.000171.}$

(c) Show that the ratio of the Sun’s average gravitational force on the Earth compared to that of the Moon’s is

${\left(\frac{F_{\text{S}}}{F_{\text{M}}}\right)}_{\text{avg}}=178.$

Note that the Moon’s smaller force varies much more across the Earth’s diameter than the Sun’s larger force. (d) Estimate the resulting “force difference” (the cause of the tides)

$\Delta F=F_{\text{near}}-F_{\text{far}}=F_{\text{far}}\left(\frac{F_{\text{near}}}{F_{\text{far}}}-1\right)\approx F_{\text{avg}}\left(\frac{F_{\text{near}}}{F_{\text{far}}}-1\right)$

for the Moon and for the Sun. Show that the ratio of the tide-causing force differences due to the Moon compared to the Sun is

$\frac{\Delta F_{\text{M}}}{\Delta F_{\text{S}}}\approx \mathrm{2.3.}$

Thus the Moon’s influence on tide production is over two times as great as the Sun’s.