Answer :
To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.
Potential Energy because of a planet would be given by the equation,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Ocean
M = Mass of Moon
r = Radius
From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.
When it is far away, it will be the total diameter from the center of the earth to the center of the moon.
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
PART A) Potential energy when the ocean is at its furthest point to the moon,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) Potential energy when the ocean is at its closest point to the moon
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]
PART C) The maximum speed. This can be calculated through the conservation of energy, where,
[tex]\Delta KE = \Delta PE[/tex]
[tex]\frac{1}{2}mv^2 = PE_2-PE_1[/tex]
[tex]v=\sqrt{2(PE_2-PE_1)/m}[/tex]
[tex]v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}[/tex]
[tex]v = 29.4m/s[/tex]