# The Effect of the Moon

1. May 1, 2008

### KMjuniormint5

1. The problem statement, all variables and given/known data
Moon effect. Some people believe that the Moon controls their activities. The Moon moves from being directly on the opposite side of Earth from you to be being directly overhead. Assume that the Earth-Moon (center-to-center) distance is 3.82 x 10^8 m and Earth's radius is 6.37 x 10^6.
(a) By what percent does the Moon's gravitational pull on you increase?

2. Relevant equations
m*g*$$\wedge$$h

(G*M*m)/(R^2) <--effect of two forces on each other

3. The attempt at a solution

G*M*m/(3.7563*10^8)^2 would be for when the moon is directly overhead
G*M*m/(3.8837*10^8)^2 would be for when the moon is directly opposite side of Earth

$$G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}$$
M = 7.3477×10^22 kg . . .mass of moon

Am I approaching this question correctly or am I way off?

2. May 1, 2008

### lzkelley

that look just right.

3. May 1, 2008

### KMjuniormint5

ok then what? would I use my mass for 'm'? and set the two equation equal to each other??

4. May 2, 2008

### alphysicist

Hi KMjuniormint5,

I don't think you can set them equal to each other, because they're not equal; you want to find out what percentage increase there is in going from the smaller number to the larger. How can you find the percent increase?

5. May 2, 2008

### KMjuniormint5

Ok i found out the first part . . .i was just had the signs mixed up. . .(on going theme with me) so i solved part a.) by using:
Side note: G*M*m gets cancelled out so all you have to worry about is the r (distance)

[(1/375630000^2)-(1/38837000^2)]/(1/388370000^2)

and then multiply by 100 to get the percent

Last edited: May 2, 2008
6. May 2, 2008

### KMjuniormint5

The second part of the question asks:
(b) By what percent does your weight (as measured on a scale) decrease?
wouldnt we just use the same formula just backwards? so. . .

[(1/38837000^2)-(1/375630000^2)]/(1/375630000^2)

and multiply by 100 to get the percent

7. May 2, 2008

### kamerling

I don't think so. one of the M's is the mass of the earth, the other M is the mass of the moon.

To get the weight difference measured on a scale you have to calculate the tidal force, which is even smaller.

8. May 2, 2008

### alphysicist

This looks like the right idea to me (except I think you're missing a zero in the second number).

In the first part of the problem you were asked about the change in just the Moon's gravitational force. Now you are asked about the change in the apparent weight, which is a scale reading. This reading will include the effects of the earth's gravity.

9. May 2, 2008

### KMjuniormint5

how am i suppose to calculate the tidal force? use the answer in part a and 9.8*answer. . . take that answer and subtract it from 9.8 . . . to get how much the moon decreases the "gravity" on me. . .and then use the same formula to calculate percentage?

10. May 2, 2008

### alphysicist

Assume the moon is directly overhead. If you're standing on a scale, what would be it's reading? (You have three forces acting on you, and no acceleration.)

What would be the reading if it's directly opposite that?

Those would give you the two scale readings, and you want to know how much of a percent decrease is involved in going from the larger to the smaller.

11. May 2, 2008

### JimChampion

I don't know what your knowledge of calculus is like, but (a) can be solved (to a first level of approximation) using calculus...

(1) If F is the gravitational attraction between you and Moon, and R is separation between you and Moon...

$$\frac{dF}{F} = 2\:\frac{dR}{R}$$

If you can work out why this is true then this is the solution for you!

(2) You know the change in separation and the mean separation, so you can work out the fractional change in force like this...

$$\frac{\delta F}{F} = 2\:\frac{\delta R}{R} = 2\:\frac{12740}{382000} = 0.0667$$

i.e. a change of approximately 7% (to a first-order approximation)

Nice thing about this method is you don't need to know $$G$$, $$M_{moon}$$, $$M_{you}$$, you just need to know the data you were given in the original question and the fact that $$F$$ is inversely proportional to $$R^2$$.

[I've left the detailed out - is this the right thing to do in this forum? Otherwise it seems like you've giving it away!]

Last edited: May 2, 2008
12. May 2, 2008

### D H

Staff Emeritus
The tidal "forces" are a kind of pseudo-force. Pseudo-forces result when one performs calculations in a non-inertial reference frame. For example, a centrifugal force and Coriolis effect arise if you want to use Newton's second law (F=ma) in a rotating reference frame. Similar issues arise in an accelerating reference frame. This is exactly what is happening in this problem. You are implicitly working in an Earth-centered reference frame, and this is a non-inertial frame. The Earth is accelerating toward the Moon.

Suppose you are standing on the point on the surface of the Earth that is on the line segment connecting the Earth's center of mass and the Moon's center of mass. You're acceleration toward the Moon due to the Moon's gravitational field is

$$a_{\text{person}} = \frac {GM_{\text{moon}}} {(R_{\text{moon}}-r_{\text{earth}})^2}$$

where $R_{\text{moon}}$ is the distance between the Earth and the Moon and $r_{\text{earth}}$ is the radius of the Earth.

The Earth itself is accelerating toward the Moon:

$$a_{\text{earth}} = \frac {GM_{\text{moon}}} {{R_{\text{moon}}}^2}$$

You are standing on a scale that in turn is on the surface of the Earth (i.e., you are stationary in an Earth-centered frame). Therefore, the impact of the Moon on the scale's reading is the difference between these two accelerations. Use a similar argument to find the impact of the Moon when you are standing on the point on the Earth that is opposite the Moon. What you the need to calculate is the difference between these two scale readings.