# Moon Effect (Gravitation)

1. Jan 18, 2014

### whoareyou

1. The problem statement, all variables and given/known data

Hello, I need somebody to double check and confirm my answer. I have absolutely no idea what the solution manual is doing.

Moon effect. Some people believe that the Moon controls
their activities. If the Moon moves from being directly on the
opposite side of Earth from you to being directly overhead, by
what percent does (a) the Moon’s gravitational pull on you
increase and (b) your weight (as measured on a scale) decrease?
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 m.

2. Relevant equations

Newton's Gravitation Equation(s)

3. The attempt at a solution

I got the first part of the question using a percent increase formula. I got 6.9% which is what the solution manual got. However for part b), I have no idea what they are doing. I'll attach their work below. What I did was used the percent increase formula again (absolute value since its a decrease) got the answer to be 0.000685%. This is by using free body diagrams (ie. the case where the moon is on the opposite side of the earth, both the earth's gravity and the gravity from the moon act in the same direction and in the case where the moon is on the same side, it would be the earth's gravity minus the moon's gravity.)

Solution manual:

2. Jan 18, 2014

### Andrew Mason

The problem is that the earth underneath you is subject to the same effect. Everything in or on the earth (other than right at the centre of mass) is going to experience a tidal effect, the most extreme cases being on the surface facing the moon and on the surface opposite the moon.

So what you want to do is determine at each of the points in question how your acceleration toward the moon differs from the acceleration of the centre of the earth toward the moon.

AM

Last edited: Jan 18, 2014
3. Jan 19, 2014

### ehild

They applied approximations. As the Earth-Moon distance is much greater than the radius of Earth, RE<<RME, x=RE/RME<<1. RME was factored out from the denominators and they became RME2(1±x)^2 .

For |x|<<1

$$\frac{1}{(1\pm x)^2}\approx 1 \mp 2x$$

Here x=0.017, 1/(1+x)2=0.967and 1-2x=0.966, and 1/(1-x)2=1.035 and 1+2x=1.034.

When you calculate the difference of quantities very close to each other, the approximation can give more accurate result than the direct calculation which is loaded with rounding errors possible greater than the inaccuracy of the approximation.

ehild

Last edited: Jan 19, 2014
4. Jan 19, 2014

### whoareyou

OK I think I figured out where I went wrong. This is the equation that I was using before:

$\frac{M_E r_E^{-2} -M_m(R_{EM} - R_E)^{-2}}{M_E r_E^{-2} + M_m(R_{EM} + R_E)^{-2}} - 1$

which gives the answer as 0.000685%. However, this is where I assumed that when the moon is directly below you, then the moon pulls you "down" and when the moon is above you, the moon pulls you "up." However, if I change it so that in either situation, the moon is pulling you "up," then the equation becomes:

$\frac{M_E r_E^{-2} -M_m(R_{EM} - R_E)^{-2}}{M_E r_E^{-2} - M_m(R_{EM} + R_E)^{-2}} - 1$

and then I get the answer that the solution manual has. So I guess the question now becomes, is the moon always pulling us "up" (as the solution manual seems to assume as well)?

5. Jan 19, 2014

### Andrew Mason

The moon is effectively pulling us "up". When it is overhead, it is literally pulling us away from the centre of the earth (ie. its pull per unit mass is greater on us than it is on the centre of the earth). When we are on the opposite side of the earth, it is pulling the centre of the earth away from us (ie. its pull per unit mass is greater on the centre of the earth than it is on us).

If you let aE-M = FE-M/ME, am-M = Fm-M/mE, and am-E = Fm-E/mE then (without taking into account the rotation of the earth on its axis), the acceleration of a mass on the surface on a line with a line between the earth-moon centres is:

$$\vec{a_{m-E}} = -\frac{GM}{R_{ME}^2}\hat R_E - ({a_{m-M}} - {a_{E-M}})\hat R_{m-M}$$

When the direction of $\hat R_E$ changes, the sign of $$({a_{m-M}} - {a_{E-M}})$$ changes also.

AM

6. Jan 19, 2014

### ehild

Question b) asks the relative change of your weight, but the solution gives the change of Moon's pull with respect to your weight. When on the opposite side of Earth , the pull of Moon adds to your weight, when overhead, it is subtracted from your weight. Your method is correct. The Manual is wrong.

ehild

7. Jan 19, 2014

### whoareyou

That's what I originally thought as well. But Andrew Mason's explanation above does make sense to me and that the Moon is always effectively pulling us "up." Do you agree or disagree?

8. Jan 19, 2014

### ehild

I disagree. See picture. The Moon pulls you away from the Earth centre when overhead, and pulls towards the Earth centre, when it is on the opposite side of earth.

ehild

#### Attached Files:

• ###### moonearth.JPG
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9. Jan 19, 2014

### whoareyou

But the moon is also pulling on the Earth as well. When the moon is overhead, you experience a greater gravitational force from the moon than the Earth so its obvious in this situation that the moon pulls you "up." When the moon is directly on the opposite side of the earth as you, then the earth experiences a greater gravitational force from the moon than you do so you should actually weigh less. So the Moon is in both situations effectively pulling us "up" right?

10. Jan 19, 2014

### ehild

The force of Moon experienced by you does not depend on the force the Moon exerts on Earth. The force of Moon is always less than the force of Earth, as its mass is much less and its distance is much more than the distance from you to the Earth centre. The Moon will not "pull away" the Earth from you. It pulls you away from Earth, or pulls you towards the Earth centre.
You experience two forces: The gravitational pull of Moon, and the gravitational pull of Earth. Once they are of the same direction, the other time they are opposite. (There are also fictitious forces due to the fact the the surface of Earth is not an inertial frame of reference. But let's ignore them.)

ehild

11. Jan 19, 2014

### Andrew Mason

The problem with that analysis (which fooled me once too) is that it does not work. In fact, high tides occur on both sides of the earth at the same time. It is difficult to see unless one analyses the real forces from an inertial reference frame. See this page for a good explanation of this.

AM

12. Jan 19, 2014

### ehild

Well, I did not want to mix inertial forces in, although the weight of a person depends also on the inertial forces, and it is different in the two cases (Moon above or below), as the Moon and Earth both orbit around their common CM. But the force acting on one body does not effect the forces acting on other bodies. The force the Moon exerts on something is not changed by the force it exerts on the Earth.

ehild

13. Jan 19, 2014

### Andrew Mason

You are quite correct that the force that the moon exerts on something is not changed by the force it (the moon) exerts on the earth. But the difference between the acceleration of a mass on the surface and the acceleration of the centre of the earth relative to an inertial point is changed. The earth is using some of its gravitational force to make up that difference and this will affect the weight.

There is no need to involve inertial (D'Alembert) forces. One just needs to do the analysis from an inertial frame. The way I would put it is: the weight (which is the mechanical tension between the mass on the surface and the earth) depends on the real forces acting on a body on the surface of the earth (gravitational forces between the mass on the surface and the earth and between that mass and the moon,), and its acceleration as measured in an inertial reference frame.

Suppose the mass of the moon was such that the barycentre was just at the earth surface. IN that case, a mass on the surface of the earth would pass through the earth-moon barycentre when it crossed the line between the centres of mass of the earth and moon. At that point its acceleration would be 0. So the sum of the gravitational forces of the earth and moon plus the normal force would be 0 (i.e. ignoring the effect of the rotation of the earth). So at that point, the moon reduces the mass' weight by the moon's gravitational force, which is a bit more than mREω2.

The same mass on the opposite side of the earth would be accelerating toward the barycentre at the rate: ac = 2REω2. That acceleration is greater than the gravitational force per unit mass of the moon at that point (which is a bit less than REω2, the force per unit mass of the moon at the earth centre of mass). The difference has to be supplied by earth gravity. So this results in a reduction in the normal force per unit mass of a tad more than REω2. So the moon's gravity reduces the weight on both sides of the earth by about the same amount.

AM

Last edited: Jan 19, 2014
14. Jan 20, 2014

### ehild

See the situation from an inertial frame of reference. The man stands on a bathroom scale. It experiences the gravity of Moon, the gravity of Earth and the normal force. Its distance from the CM of Earth cannot change. In an inertial frame of reference, the Earth accelerates toward the Moon, and the man accelerates together with it. The scale reads the normal force of magnitude N. The acceleration of the Earth centre towards the Moon is

$$a_E=G \frac{m_M}{D^2}$$.

The radius of Earth is R, the centre-to centre distance between Moon and Earth is D. The direction towards Moon is taken positive.

1) The man is at the opposite side of Earth as the Moon is. Both the Earth and the Moon attracts him towards the Earth centre. The normal force points away from the centre of Earth and Moon.

$$ma =ma_E=mG \left( \frac{M_E}{R^2}+\frac{M_M}{(D+R)^2} \right )-N1$$
$$N1=mG \left( \frac{M_E}{R^2}+\frac{M_M}{(D+R)^2}\right ) -ma_E$$

$$N1=mG \left( \frac{M_E}{R^2}+\frac{M_M}{(D+R)^2}-\frac{M_M}{D^2} \right )$$

As 1/(D+R)2-1/D2<0, the normal force is less than the pull of Earth.

2) The man and Moon are at the same side of Earth. The acceleration of the man and Earth towards the Moon is equal again. The normal force points in the direction of Moon. The net force on the man is

$$ma =ma_E=mG \left( -\frac{M_E}{R^2}+\frac{M_M}{(D-R)^2} \right )+N2$$
$$N2=mG \left( \frac{M_E}{R^2}-\frac{M_M}{(D-R)^2} +\frac{M_M}{D^2} \right )$$
As -1/(D-R)2+1/D2<0, the normal force is less than the pull of Earth again. So you were right

The difference of the normal forces (scale readings) is

$$N1-N2= m M_M G \left( \frac{1}{(D+R)^2}+\frac{1}{(D-R)^2}-2\frac{1}{D^2} \right)$$

With the notation x=R/D, it can be written

$$N1-N2= \frac{m M_M G } {D^2} \left( \frac{1}{(1+x)^2}+\frac{1}{(1-x)^2}-2\right)$$
The term in the parentheses is $$\frac{1-2x+x^2+1+2x+x^2-2(1-2x^2+x^4)}{1-2x^2+x^4}=\frac{6x^2+2x^4}{1-2x^2+x^4}\approx6x^2$$

The scale reading is less when the Moon is overhead, but not as much as given in the Manual.

I hope my derivation is correct.

ehild

Last edited: Jan 20, 2014
15. Jan 20, 2014

### Andrew Mason

It looks like you are setting the acceleration of the mass as if it were rotating about the moon. But in fact it is rotating about the earth-moon barycentre. So you get that extra -2 term in the brackets and a + sign between the other terms instead of a -. It is difficult to visualize with the centre of rotation inside the earth. That is why I simplified it to put the barycentre on the surface of the earth (so a mass at that point it is not accelerating at all).

I think the manual is correct that the difference in the normal force between the two positions is the decrease in the moon's force over the diameter of the earth:

$$m\left(\frac{GM_m}{(D-R)^2} - \frac{GM_m}{(D+R)^2}\right)$$

AM

16. Jan 20, 2014

### ehild

Could you please show how your derivation leads to that result.

ehild

17. Jan 20, 2014

### Andrew Mason

Let's start by supposing that the earth provided no normal force. We just want to determine what the force would be on a mass m on the surface. This is just to make it simple.

We will define two displacement vectors from the barycentre to m, and to the centre of the Earth as sm and sE respectively.

Let the radial displacement vector from the earth centre to the mass be r, the radial displacement vector from the centre of the moon to m be d and the radial displacement vector from the centre of the moon to the centre of the earth be R.

Since the only forces on m are the gravity of the moon and the earth:

(1)$$\ddot{s_m} = -\frac{GM_E}{r^2}\hat{r} - \frac{GM_M}{d^2}\hat{d}$$

Since we are interested in the acceleration of m relative to the centre of the earth we have to take into account the acceleration of the earth relative to the barycentre:

(2)$$\ddot{s_E} = -\frac{GM_M}{R^2}\hat R$$

So the acceleration of m relative to the centre of the earth is the difference between these two accelerations which is just (1) - (2):

$$\ddot{s_m} - \ddot{s_E} = -\frac{GM_E}{r^2}\hat{r} - \left(\frac{GM_M}{d^2}\hat{d} -\frac{GM_M}{R^2}\hat{R}\right)$$

The left side is just $\ddot{r}$, the acceleration of m relative to the centre of the earth.

When the m is in position 1 (moon directly overhead), d = R-r and the vectors all align but R and d are in the opposite direction to r, so this becomes:

(1)$$\ddot{r}_1 = -\frac{GM_E}{r^2}\hat{r} - \left(\frac{GM_M}{R^2}\hat{r} - \frac{GM_M}{(R-r)^2}\hat{r}\right)$$

When m is in position 2 (on the opposite side of the earth to moon), d = R + r and all vectors are in the same direction, so the expression is:

(2)$$\ddot{r}_2 = -\frac{GM_E}{r^2}\hat{r} - \left(\frac{GM_M}{(R+r)^2}\hat{r} - \frac{GM_M}{R^2}\hat{r}\right)$$

The difference between the magnitude of those two accelerations is:

$$|\ddot{r}_1| - |\ddot{r}_2| = \frac{GM_M}{(R-r)^2} - \frac{GM_M}{(R+r)^2}$$

If we make the earth solid, the normal force/unit mass is exactly equal and opposite to $\ddot{r}$.

AM

Last edited: Jan 21, 2014
18. Jan 21, 2014

### ehild

I think you meant $\ddot r$ vector so it should have been written $\ddot {\vec r }$.

The normal force points radially outward, and as you said, it is opposite and equal to the radial acceleration. $$\vec N= N \hat e_r$$. Let be m=1 kg. Also, N has to be positive. I continue your derivation, as you made some mistakes I am afraid.

$$N_1\hat{r}=- \ddot{\vec r}_1 = \frac{GM_E}{r^2}\hat{r} + \left(\frac{GM_M}{R^2}\hat{r} - \frac{GM_M}{(R-r)^2}\hat{r}\right)$$, that is

$$N_1=\frac{GM_E}{r^2} + \left(\frac{GM_M}{R^2} - \frac{GM_M}{(R-r)^2}\right)>0$$

$$N_2\hat{r}=-\ddot{\vec r}_2 = \frac{GM_E}{r^2}\hat{r} + \left(\frac{GM_M}{(R+r)^2}\hat{r} - \frac{GM_M}{R^2}\hat{r}\right)$$

$$N_2=\frac{GM_E}{r^2} + \left(\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}\right)>0$$

$$N_2-N_1=\frac{GM_E}{r^2} + \left(\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}\right)-\left( \frac{GM_E}{r^2} + \left(\frac{GM_M}{R^2} - \frac{GM_M}{(R-r)^2}\right) \right)$$

$$= \left(\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}\right)- \left(\frac{GM_M}{R^2} - \frac{GM_M}{(R-r)^2}\right)$$

$$=\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}- \frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}$$

$$N_2-N_1=\frac{GM_M}{(R+r)^2} - 2\frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}$$

ehild

19. Jan 21, 2014

### Andrew Mason

Thanks. Yes, $\ddot r$ is a vector. (I was trying to bold the r but couldn't do it within Latex).

You are right. Very good. I was careless in my subtracting there.

The difference is as you have calculated, which is approximately 6GMMr2/R4 or about the book answer x 1.5(r/R) (i.e. quite a bit less).

The tidal effect is proportional to r/R3 but the difference in tidal effect from one side to the other is proportional to r2/R4.

AM

Last edited: Jan 21, 2014
20. Jan 21, 2014

### ehild

There is some difference between the tide and the man standing on the scales. The surface of the ocean can move away from the centre of the Earth. The man stays on the surface.

ehild

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