We have seen that normal tides are due to the gravitational force exerted by the

Question

We have seen that normal tides are due to the gravitational force exerted by the Moon on the Earth's oceans. When the Moon, Sun, and Earth are aligned as shown in the figure below, the magnitude of the tide increases due to the gravitational force exerted by the Sun on the oceans (at the times of the "new" Moon and the "full" Moon during the course of a month). Calculate the ratio of the gravitational force of the Sun to that of the Moon on the oceans. This "extra" force from the Sun does make a difference! (For simplicity, use the distance from the center of the Earth to the Moon and Sun in your calculation.)

=

Explanation / Answer

force between two bodies is F = Gm1m2/r^2

where

F = gravitational force between the earth and the moon,

G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2,

m = mass of the moon = 7.36e 22 ) kg

M = mass of the earth = 5.9742 e 24 and

r = distance between the earth and the moon = 384,402 km

Fearth -moon = Fem = 6.67e -11 * (7.36e 22* 5.97e 24 10^(24) / (3.84 e 8 )^2

Fem = 1.985 e 20 N

the force distance between the centre of the earth and the centre of the sun,

d = 1.5*10^ 11m

F = (6.67*10^-11 * 6.0*10^24 * 2.0*10^30)/ (1.5*10^11)

so

Fsun earth = F = 3.557*10^22 N

Fsun/Fmoon = 3.557 e 22/(1.9865 e 20)

Fsun/Fmoon = 179.05

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