Example 13.14 Comparing Tidal Forces Compare the Moon's gravitational force on a 1.0-kg mass located on the near side and another on the far side of Earth. Repeat for the Sun and then compare the results to confirm that the Moon's tidal forces are about twice that of the Sun. Strategy We use Newton's law of gravitation given by Equation 13.1. We need the masses of the Moon and the Sun and their distances from Earth, as well as the radius of Earth. We use the astronomical data from Appendix D. Solution Substituting the mass of the Moon and mean distance from Earth to the Moon, we have = (6.67 x 10-1! N - m²/kg?)_ (1.0 kg)(7.35 x 1022 kg) (3.84x 10% + 6.37 × 106 m)²" F12 = G" 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 x 10-5 N and Frar= 3.22 x 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 Fnear = 5.89975 x 10–3 N and Ffar= 5.89874x 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 Fnear = 3.44 x 10-S N – 3.22x 10-5 N= 0.22 × 10–5 N, whereas the difference for the Sun is Fnear – F far = 5.89975 × 10–3 N – 5.89874× 10–3 N= 0.101 × 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
Example 13.14 Comparing Tidal Forces Compare the Moon's gravitational force on a 1.0-kg mass located on the near side and another on the far side of Earth. Repeat for the Sun and then compare the results to confirm that the Moon's tidal forces are about twice that of the Sun. Strategy We use Newton's law of gravitation given by Equation 13.1. We need the masses of the Moon and the Sun and their distances from Earth, as well as the radius of Earth. We use the astronomical data from Appendix D. Solution Substituting the mass of the Moon and mean distance from Earth to the Moon, we have = (6.67 x 10-1! N - m²/kg?)_ (1.0 kg)(7.35 x 1022 kg) (3.84x 10% + 6.37 × 106 m)²" F12 = G" 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 x 10-5 N and Frar= 3.22 x 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 Fnear = 5.89975 x 10–3 N and Ffar= 5.89874x 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 Fnear = 3.44 x 10-S N – 3.22x 10-5 N= 0.22 × 10–5 N, whereas the difference for the Sun is Fnear – F far = 5.89975 × 10–3 N – 5.89874× 10–3 N= 0.101 × 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
Related questions
Question
(a) What is the difference between the forces on a 1.0-kg mass on the near side of Io and far side due to Jupiter? Io has a mean radius of 1821 km and a mean orbital radius about Jupiter of 421,700 km. (b) Compare this difference to that calculated for the difference for Earth due to the Moon calculated in Example 13.14. Tidal forces are the cause of Io’s volcanic activity.
Transcribed Image Text:Example 13.14
Comparing Tidal Forces
Compare the Moon's gravitational force on a 1.0-kg mass located on the near side and another on the far side of
Earth. Repeat for the Sun and then compare the results to confirm that the Moon's tidal forces are about twice that
of the Sun.
Strategy
We use Newton's law of gravitation given by Equation 13.1. We need the masses of the Moon and the Sun and
their distances from Earth, as well as the radius of Earth. We use the astronomical data from Appendix D.
Solution
Substituting the mass of the Moon and mean distance from Earth to the Moon, we have
= (6.67 x 10-1! N - m²/kg?)_ (1.0 kg)(7.35 x 1022 kg)
(3.84x 10% + 6.37 × 106 m)²"
F12 = G"
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 x 10-5 N and Frar= 3.22 x 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
Fnear = 5.89975 x 10–3 N and Ffar= 5.89874x 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
Fnear = 3.44 x 10-S N – 3.22x 10-5 N= 0.22 × 10–5 N,
whereas the difference for the Sun is
Fnear – F far = 5.89975 × 10–3 N – 5.89874× 10–3 N= 0.101 × 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
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