Question

In class we described the tidal forces that are responsible for raising and lowering thewater level near the shore of the ocean.(a) Suppose the moon is directly overhead. What is the magnitude of the tidal forceon a person of massm, as a fraction of the person’s weight? We are looking fora numerical answer.(b) Explain why the tides are high twice a day. You make make any simplifyingassumptions about the oceans that you find necessary. Why are there tides in theocean but not in, say, swimming pools?

Answer 1

The newton's law of universal gravitation used to describe how a particle attracts every other particle in the universe.

The equation is given by,

$F= G\frac{m_1 m_2}{r_2}$

Where,

F= Gravitational force

$m_1, m_2 =$ masses of the objects

r= is the distance between the center of their masses

G= Gravitational constant

For our problem we have defined that,

$M_m = 7.24*10^{22}kg$ (mass of the moon)

$d= 3.84*10^8 km$ (distance Earth-moon)

G= 6.671*10^{-11}Nm^2/kg^2

M=Mass of a person

We have then,

$F_m = M\frac{6.67*10^-11*7.34*10^22}{3.84*10^8}$

$F_m = M*3.320*10^{-5} N$

In the other hand we have the force on m-mass due to earth

$F_E= MG = M*9.8N$

Ratio is given by

$\frac{F_m}{F_E} = {M*3.320*10^{-5}}{9.8} = 3.387*10^{-6}$

B) Suppose there is a group of young people surfing in the moonlight. They are directly under the moon. At this time the moon exerts its gravitational effect of the earth that causes the tide to rise. Around 6 hours later, when the earth has moved a quarter of the moon, the force on that point decreases, so the tide drops. However, after another 6 hours, people return and experience the same process. In this case the moon is not above them, but on the other side. This is because the moon having an orbit on the earth, generates an external force, similar to the previous one, but the earth reacts in the opposite way. It is like going in a car and turning it, all people will tend to get out of it, because a centrifugal force is experienced.