7. Isaac Newton was the first person to correctly explain the twice-daily occurence of high and low tides. This happens because gravity is a force that depends on distance. Thus, the part of the Earth closest to the Moon (position 1 in Fig. 1) feels a stronger pull than the diametrically opposite part (position 2) which is the farthest from the Moon. The magnitude of the gravitational force between any two masses is given by where a is a constant that contains the two masses and d is the distance between the object feeling the force and the centre of the object exerting the force. Let R be distance from the centre of the Moon to the centre of the Earth and Rg be the radius of the Earth (see Fig. 1). Compute the tidal force using Fide 0 2. where 1,2 are the positions described above (red dots in the figure). To do this, let ar represent RE/R which is much smaller than 1 since R << R. Then using the Maclaurin series for d2 and d₂2 in Eq. (1) obtain the first non-zero term for the Fride as a series in a multiplied by constants. Finally, substitute x = RE/R into your expression to find that the tidal force depends on the inverse of the cube of R. RE 1 F1-F20 • (-a) R Figure 1: The Earth-Moon system (not to scale). Rg is the radius of the Earth and R is the Earth-Moon distance measured from the centres.

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7. Isaac Newton was the first person to correctly explain the twice-daily occurence
of high and low tides. This happens because gravity is a force that depends on
distance. Thus, the part of the Earth closest to the Moon (position 1 in Fig. 1)
feels a stronger pull than the diametrically opposite part (position 2) which is the
farthest from the Moon. The magnitude of the gravitational force between any
two masses is given by
where a is a constant that contains the two masses and d is the distance between
the object feeling the force and the centre of the object exerting the force. Let R
be distance from the centre of the Moon to the centre of the Earth and RE be
the radius of the Earth (see Fig. 1). Compute the tidal force using
α
d²
2.
RE
Ftide=Fg,1Fg,2 = a
where 1,2 are the positions described above (red dots in the figure). To do this,
let a represent RE/R which is much smaller than 1 since RE < R. Then using
the Maclaurin series for d2 and d₂2 in Eq. (1) obtain the first non-zero term for
the Ftide as a series in a multiplied by constants. Finally, substitute x =
RE/R
into your expression to find that the tidal force depends on the inverse of the cube
of R.
1
1
(-2)
R
Figure 1: The Earth-Moon system (not to scale). RE is the radius of the Earth and R
is the Earth-Moon distance measured from the centres.
Transcribed Image Text:7. Isaac Newton was the first person to correctly explain the twice-daily occurence of high and low tides. This happens because gravity is a force that depends on distance. Thus, the part of the Earth closest to the Moon (position 1 in Fig. 1) feels a stronger pull than the diametrically opposite part (position 2) which is the farthest from the Moon. The magnitude of the gravitational force between any two masses is given by where a is a constant that contains the two masses and d is the distance between the object feeling the force and the centre of the object exerting the force. Let R be distance from the centre of the Moon to the centre of the Earth and RE be the radius of the Earth (see Fig. 1). Compute the tidal force using α d² 2. RE Ftide=Fg,1Fg,2 = a where 1,2 are the positions described above (red dots in the figure). To do this, let a represent RE/R which is much smaller than 1 since RE < R. Then using the Maclaurin series for d2 and d₂2 in Eq. (1) obtain the first non-zero term for the Ftide as a series in a multiplied by constants. Finally, substitute x = RE/R into your expression to find that the tidal force depends on the inverse of the cube of R. 1 1 (-2) R Figure 1: The Earth-Moon system (not to scale). RE is the radius of the Earth and R is the Earth-Moon distance measured from the centres.
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