**QUESTION**

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8. Practitioners of some schools of yoga are warned not to perform yoga during the full or the new Moon, citing the tidal effect of the Moon at those times on other "watery bodies" such as the oceans. Let us investigate this idea. a. Verify the dramatic tidal effect of the Moon on the oceans by using Eq. 4.127 to calculate the ratio of the tidal "lifting force" $F_{\text {tide }}$ and the Earth's gravitational attraction $F_{\text {grav }}$, for a point mass on the surface of the Earth and on the Earth-Moon line. Use this ratio to estimate the change in water height, in $\mathrm{cm}$, between high tide and low tide, due to the Moon alone. Repeat the calculation for the tidal effect due to the Sun alone. Hint: The surface of the oceans traces an equipotential surface, $g R=$ constant, where $R$ is the distance from the Earth's center to the ocean surface at every point, and $g$ is the effective gravitational acceleration at every point on the surface. Translate the tidal-to-gravitational force ratio into a relative change in $g$ between a point at high tide (which experiences the full tidal force) and a point at low tide (which experiences no tidal force), and thus derive the relative change in $R$. Answers: $77 \mathrm{~cm}$ due to Moon, $33 \mathrm{~cm}$ due to Sun.

Note: While the Sun and Moon are the drivers of ocean tides, a reliable calculation of tides at a particular Earth location must take into account additional factors, including the varying distances between Earth, Moon, and Sun (due to their elliptical orbits), their inclined orbital planes, the latitude, coastline shape, beach profile, ocean depth, water viscosity and salinity, and prevailing ocean currents. b. Calculate by how much (in milligrams) you are lighter when the full or new Moon is overhead or below, compared to when it is rising or setting, assuming your body weight is $50 \mathrm{~kg}$. c. Calculate in dynes and in gram-force (i.e., in dynes divided by the gravitational acceleration $g=980 \mathrm{~cm} \mathrm{~s}^{-2}$ ) the tidal stretch exerted on your body by the Moon plus the Sun when you are standing up with the full or new Moon overhead. Assume your body weight is $50 \mathrm{~kg}$ and your height is $180 \mathrm{~cm}$.

The ratio between the forces is \[ \frac{F_{\text {tide }}}{F_{\text {grav }}}=\frac{2 M_{2}}{M_{1}}\left(\frac{\Delta r}{r}\right)^{3} . \]Please do not copy from other answers or I will definitely downvote. Thanks!