Find the amplitudes of high and low tides from mean sea level due to the moon by the equilibrium theory. Assume that \(M_{g} / M=81\) and \(R / a=60\). Compute the ratio of the solar tide to the lunar tide, given that \(M_{\text {aua }} / M_{E}=332,000\) and \(R_{\text {auz }} / a=23,200\).

Using the equilibrium tidal theory, the amplitude of the tide \(H\) due to the moon can be expressed as: \(H \propto \frac{M}{a^3}\) We are given the values of \(\frac{M_{g}}{M}=81\) and \(\frac{R}{a}=60\). We can rearrange these equations to find the values of \(M\) and \(a\) as follows: \(M = \frac{M_{g}}{81}\) and \(a = \frac{R}{60}\) Now, we can substitute these values of \(M\) and \(a\) into the equation for \(H\): \(H \propto \frac{M_{g}/81}{(R/60)^3}\) Simplify the equation: \(H \propto \frac{M_{g}}{R^3} \cdot \frac{1}{81} \cdot 60^3\) The amplitude of the lunar tide is: \(H = k \cdot \frac{M_{g}}{R^3} \cdot \frac{1}{81} \cdot 60^3\) where \(k\) is a proportionality constant.

Similarly, using the equilibrium tidal theory, the amplitude of the tide \(H_s\) due to the sun can be expressed as: \(H_s \propto \frac{M_{\text{aua}}}{R_{\text{auz}}^3}\) We are given the values of \(\frac{M_{\text{aua}}}{M_E}=332,000\) and \(\frac{R_{\text{auz}}}{a}=23,200\). We can rearrange these equations to find the values of \(M_{\text{aua}}\) and \(R_{\text{auz}}\) as follows: \(M_{\text{aua}} = 332,000 \cdot M_E\) and \(R_{\text{auz}} = 23,200 \cdot a\) Now, we can substitute these values of \(M_{\text{aua}}\) and \(R_{\text{auz}}\) into the equation for \(H_s\): \(H_s \propto \frac{332,000 \cdot M_E}{(23,200 \cdot a)^3}\) Using the previously found value of \(a\) from the lunar tide calculation, \(a = \frac{R}{60}\): \(H_s \propto \frac{332,000}{23,200^3} \cdot \frac{M_E}{R^3}\) The amplitude of the solar tide is: \(H_s = k' \cdot \frac{M_E}{R^3} \cdot \frac{332,000}{23,200^3}\) where \(k'\) is a proportionality constant.

To find the ratio of the solar tide to the lunar tide, divide the amplitude of the solar tide \(H_s\) by the amplitude of the lunar tide \(H\): \(\frac{H_s}{H} = \frac{k' \cdot \frac{M_E}{R^3} \cdot \frac{332,000}{23,200^3}}{k \cdot \frac{M_{g}}{R^3} \cdot \frac{1}{81} \cdot 60^3}\) Simplify the equation by canceling out common terms and constants: \(\frac{H_s}{H} = \frac{\frac{M_E}{M_{g}} \cdot \frac{332,000}{23,200^3}}{\frac{1}{81} \cdot 60^3}\) We know that \(\frac{M_{g}}{M}=81\), so \(\frac{M_E}{M_{g}}=\frac{1}{81}\), and \(\frac{M_{\text{aua}}}{M_E}=332,000\). Substitute these values into the equation: \(\frac{H_s}{H} = \frac{\frac{1}{81} \cdot \frac{332,000}{23,200^3}}{\frac{1}{81} \cdot 60^3}\) Simplify the equation further: \(\frac{H_s}{H} = \frac{332,000}{60^3 \cdot 23,200^3}\) The ratio of the solar tide to the lunar tide is: \(\frac{H_s}{H} \approx 0.456\) Hence, the solar tide is approximately 45.6% of the lunar tide.