Ocean tides are caused by the difference between the force the moon exerts on the ocean on one side of the Earth and the ocean on the other side. Calculate the difference between the force on 1 kg of water located on the side of the Earth closest to the moon and that on 1 kg of water on the side farthest from the moon. (Possibly) relevant data:

Earth-Moon distance R_EM = 3.84 x 10⁸ m

Radius of the Earth = 6.37 x 10⁶ m

Mass of the Earth M_E = 5.98 x 10²⁴ kg

Mass of the Moon M_m = 7.26 x 10²² kg

G = 6.67 x 10^-11 Nm²/kg²

This will just require applying Newton's law of universal gravitation twice, to the two different bits of water. For the mass closest to the moon, we have

F = GmM_m/(R_EM - R_E)²

since this kilogram of water is nearer the moon by the radius of the Earth, and where m is the mass of the water, and M_m is the mass of the moon, and where we are only calculating the magnitide of the force. Similarly, the force on the other kilogram of water is

F = GmM_m/(R_EM + R_E)²

The difference in the forces is then

F = GmM_m[ 1/(R_EM - R_E)²   -   1/(R_EM + R_E)²]

Substituting in numbers gives

= 2.18 x 10^-6 N

The force difference is tiny - nevertheless, it is sufficiently large to cause the tides. The mass of the Earth was not needed in this problem - it was a red herring...

Note: This is essentially the same as problem 12-42 from the suggested extra problems, and it is also essentially identical to the calculation of the tidal forces on a barbell at the surface of a neutron star ( problem 12-56 from the homework).

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