Question
(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon's acceleration to the Sun's and comment on why the tides are predominantly due to the Moon in spite of this number.
1. $3.33 \times 10^{-5}\textrm{ m/s}^2$
2. $5.93\times 10^{-3}\textrm{ m/s}^2$
3. The acceleration due to gravity on the Earth's surface due to the sun is 178 times that due to the moon. The tides are cased by the difference in gravitational force between the near and far sides of the Earth. The difference for the moon is $2.2\times 10^{-6}\textrm{ m/s}^2$ whereas for the sun the difference is $1.0\textrm{-6}\textrm{ m/s}^2$. The difference due to the sun is about half that due to the moon, which is still significant enough to manifest as spring tides every two weeks when the sun-Earth-moon line up.
Solution Video

# OpenStax College Physics Solution, Chapter 6, Problem 34 (Problems & Exercises) (5:14)    