13.6 Tidal forces (Page 3/7)

University physics volume 1 Page 3 / 7

Solution

Substituting the mass of the Moon and mean distance from Earth to the Moon, we have

F_{12} = G \frac{m_{1} m_{2}}{r^{2}} = (6.67 \times 10^{−11} N \cdot m^{2} {/kg}^{2}) \frac{(1.0 kg) (7.35 \times 10^{22} kg)}{{(3.84 \times 10^{8} \pm 6.37 \times 10^{6} m)}^{2}} .

In the denominator, we use the minus sign for the near side and the plus sign for the far side. The results are

F_{near} = 3.44 \times 10^{−5} N and F_{far} = 3.22 \times 10^{−5} N.

The Moon’s gravitational force is nearly 7% higher at the near side of Earth than at the far side, but both forces are much less than that of Earth itself on the 1.0-kg mass. Nevertheless, this small difference creates the tides. We now repeat the problem, but substitute the mass of the Sun and the mean distance between the Earth and Sun. The results are

F_{near} = 5.89975 \times 10^{−3} N and F_{far} = 5.89874 \times 10^{−3} N.

We have to keep six significant digits since we wish to compare the difference between them to the difference for the Moon. (Although we can’t justify the absolute value to this accuracy, since all values in the calculation are the same except the distances, the accuracy in the difference is still valid to three digits.) The difference between the near and far forces on a 1.0-kg mass due to the Moon is

F_{near} = 3.44 \times 10^{−5} N - 3.22 \times 10^{−5} N = 0.22 \times 10^{−5} N,

whereas the difference for the Sun is

F_{near} - F_{far} = 5.89975 \times 10^{−3} N - 5.89874 \times 10^{−3} N = 0.101 \times 10^{−5} N.

Note that a more proper approach is to write the difference in the two forces with the difference between the near and far distances explicitly expressed. With just a bit of algebra we can show that

F_{tidal} = \frac{G M m}{r_{1}^{2}} - \frac{G M m}{r_{2}^{2}} = G M m (\frac{(r_{2}^{} - r_{1}^{}) (r_{2}^{} + r_{1}^{})}{r_{1}^{2} r_{2}^{2}}),

where $r_{1}^{}$ and $r_{2}^{}$ are the same to three significant digits, but their difference $(r_{2} - r_{1})$ , equal to the diameter of Earth, is also known to three significant digits. The results of the calculation are the same. This approach would be necessary if the number of significant digits needed exceeds that available on your calculator or computer.

Significance

Note that the forces exerted by the Sun are nearly 200 times greater than the forces exerted by the Moon. But the difference in those forces for the Sun is half that for the Moon. This is the nature of tidal forces. The Moon has a greater tidal effect because the fractional change in distance from the near side to the far side is so much greater for the Moon than it is for the Sun.

Check Your Understanding Earth exerts a tidal force on the Moon. Is it greater than, the same as, or less than that of the Moon on Earth? Be careful in your response, as tidal forces arise from the difference in gravitational forces between one side and the other. Look at the calculations we performed for the tidal force on Earth and consider the values that would change significantly for the Moon. The diameter of the Moon is one-fourth that of Earth. Tidal forces on the Moon are not easy to detect, since there is no liquid on the surface.

Consider the last equation above. The values of $r_{1}$ and $r_{2}$ remain nearly the same, but the diameter of the Moon, $(r_{2}^{} - r_{1}^{})$ , is one-fourth that of Earth. So the tidal forces on the Moon are about one-fourth as great as on Earth.

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Other tidal effects

Tidal forces exist between any two bodies. The effect stretches the bodies along the line between their centers. Although the tidal effect on Earth’s seas is observable on a daily basis, long-term consequences cannot be observed so easily. One consequence is the dissipation of rotational energy due to friction during flexure of the bodies themselves. Earth’s rotation rate is slowing down as the tidal forces transfer rotational energy into heat. The other effect, related to this dissipation and conservation of angular momentum, is called “locking” or tidal synchronization. It has already happened to most moons in our solar system, including Earth’s Moon. The Moon keeps one face toward Earth—its rotation rate has locked into the orbital rate about Earth. The same process is happening to Earth, and eventually it will keep one face toward the Moon. If that does happen, we would no longer see tides, as the tidal bulge would remain in the same place on Earth, and half the planet would never see the Moon. However, this locking will take many billions of years, perhaps not before our Sun expires.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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