- $3.33 \times 10^{-5}\textrm{ m/s}^2$
- $5.93\times 10^{-3}\textrm{ m/s}^2$
- 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 $1.0\times 10^{-6} \textrm{ m/s}^2$. 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, Chapter 6, Problem 34 (Problems & Exercises)

### Calculator Screenshots

*g*and the mass of the Moon and then so on and so on... plug in numbers. So we are subtracting the Earth's radius and then adding the Earth's radius to this Earth-Moon distance and we end up with 2.2 times 10 to the minus 6 meters per second squared is the difference in acceleration due to gravity on the near versus far side due to the Moon. And then for the Sun, we have the Sun's mass here and we have the Earth-Sun distance minus the Earth radius and then Earth-Sun distance plus the Earth radius and we get 1.0 times 10 to the minus 6 meters per second squared. So this number is about half of that but it's nevertheless still big and there are in fact solar tides. So the Sun does cause some of the tides but you don't really notice it very much because the Moon has a bigger effect. But where you do notice the Sun's effect is when the Moon and the Sun line up. So these are called spring tides and they occur about every 2 weeks at full and new moons. And when you have a full moon, you have the Moon here and you have Earth here and then the Sun's over here and they all line up. And this being the dark side of the Earth, you see this bright Moon here or it also happens during new moons in which case the Moon is is here and you are on the dark side of the Earth and you don't see any Moon at all because the Moon's over here on the bright side of the Earth. Okay so that's the story!

## Comments

Hello,

I get that the answer for this problem (a) is incorrect. Any idea why? I have tried changing the significant figures.

Hello earljohns,

Thank you for your question. I'm not sure what you're noticing is wrong with the answer in part (a)?

Where is all the info in green coming from?

Hi mikayla, thank you for the question. The info in green comes from the OpenStax textbook appendix.

Hi Shaun,

I ran through this problem and it came up wrong, even after changing G to 6.674 x 10^-11. I believe the problem may be that we should consider the distance between earth and moon centers and then subtract the radius of earth since we're looking for gravity on the surface of earth rather than at its center. Interested to hear your thoughts. Thanks.

Hi bjappell, thank you for the great comment. The assumption from the perspective you're suggesting is that the question is asking for the acceleration due to the moon on the Earth's surface *where the surface is closest to the moon*. The question isn't clear about which position on the Earth's surface we're meant to consider. One could likewise suggest *adding* the radius of the Earth to find the acceleration due to the moon's gravity on the *far* side of the Earth. Using the center of the Earth is a kind of average, which I think is the best one can do without more specificity from the question. There will be a ring of positions on the surface with the answer calculated here, where the ring is the perimeter of an Earth cross-section somewhere between the near point and far point from the moon.

When you mention "came up wrong", what are you comparing to? Did your calculation not equal the answer in the video here, or are you comparing to a different resource?

All the best,

Shaun