Imagine yourself weightless inside a spherical space station, orbiting a planet. Yhere are no windows. you have four marbles. Find out where the planet is that you are orbiting. (stolen from some old SciFi book). hint: think of orbiting marbles at slightly different altitudes. W. ----- Original Message ----- From: "Eugene Salamin" <gene_salamin@yahoo.com> To: <math-fun@mailman.xmission.com> Sent: Sunday, May 11, 2003 3:47 AM Subject: [math-fun] Re: Tides
Because the Earth is free falling in the Moon's gravitational field, Earth does not experience a force due to the gravitational acceleration at its center. But the Moon's force is not constant over the Earth's surface, and this is responsible for the tides.
The gravitational potential at Earth due to the Moon, expressed in coordinates centered at Earth's center, and expanded to second order, is
Vm = (G Mm / L^3)(y^2/2 - x^2), L = distance to Moon, x points toward Moon,
in which the constant and linear terms have been removed, the constant because it is unimportant, the linear because we're in free fall. The gravitational potential due to Earth at height h above its surface is
Ve = - G Me / (Re + h),
which becomes, if the value at h = 0 is removed,
Ve = (G Me / Re^2) h.
The ocean surface is an equipotential of V = Ve + Vm. In the absence of the Moon, the geoid h = 0 corresponds to V = 0. In the presence of the Moon the V = 0 geoid is perturbed to
h = (Mm / Me) (Re^2 / L^3) (x^2 - y^2/2)
= (Mm / Me) (Re^4 / L^3) (cos(theta)^2 - (1/2)sin(theta)^2).
The bulge at theta = 0, on the side facing the Moon, equals the bulge at theta = pi, on the side opposite the Moon, while at theta = pi/2, the ocean drops by one half the amount.
Re = 6.40e6 m L = 3.85e8 m Me = 5.97e24 kg Mm = 7.35e22 kg
The predicted height of the Moon's contribution to the tidal bulge is 0.36 m or 1.2 ft., and the high-to-low tide difference is 1.8 ft. The Sun's effect has been ignored, as well as the interaction between the rotating tidal bulge and the bathymetric terrain.
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