How does the position of the Moon affect gravitational pull and weight on Earth?

[whoareyou]

Homework Statement

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.

Homework Equations

Newton's Gravitation Equation(s)

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:

 

[Andrew Mason]

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.

Homework Equations

Newton's Gravitation Equation(s)

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.)

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

 

[ehild]

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.)

They applied approximations. As the Earth-Moon distance is much greater than the radius of Earth, R_E<<R_ME, x=R_E/R_ME<<1. R_ME was factored out from the denominators and they became R_ME²(1±x)^2 .For |x|<<1

[tex]\frac{1}{(1\pm x)^2}\approx 1 \mp 2x[/tex]

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

 

[whoareyou]

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

[itex]\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[/itex]

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:

[itex]\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[/itex]

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)?

 

[Andrew Mason]

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

[itex]\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[/itex]

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:

[itex]\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[/itex]

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)?

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 a_E-M = F_E-M/M_E, a_m-M = F_m-M/m_E, and a_m-E = F_m-E/m_E 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:

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

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

AM

 

[ehild]

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

[itex]\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[/itex]

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."

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

 

[whoareyou]

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

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?

 

[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

 

[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?

 

[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

 

[Andrew Mason]

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

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

 

[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

 

[Andrew Mason]

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.

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 mR_Eω².

The same mass on the opposite side of the Earth would be accelerating toward the barycentre at the rate: a_c = 2R_Eω². That acceleration is greater than the gravitational force per unit mass of the moon at that point (which is a bit less than R_Eω², 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 R_Eω². So the moon's gravity reduces the weight on both sides of the Earth by about the same amount.

AM

 

[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

[tex]a_E=G \frac{m_M}{D^2}[/tex].

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.

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

[tex]N1=mG \left( \frac{M_E}{R^2}+\frac{M_M}{(D+R)^2}-\frac{M_M}{D^2} \right ) [/tex]

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

[tex]ma =ma_E=mG \left( -\frac{M_E}{R^2}+\frac{M_M}{(D-R)^2} \right )+N2[/tex]
[tex]N2=mG \left( \frac{M_E}{R^2}-\frac{M_M}{(D-R)^2} +\frac{M_M}{D^2} \right )[/tex]
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

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

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

[tex]N1-N2= \frac{m M_M G } {D^2} \left( \frac{1}{(1+x)^2}+\frac{1}{(1-x)^2}-2\right)
[/tex]
The term in the parentheses is [tex] \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
[/tex]

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

 

[Andrew Mason]

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

[tex]N1-N2= \frac{m M_M G } {D^2} \left( \frac{1}{(1+x)^2}+\frac{1}{(1-x)^2}-2\right)
[/tex]
The term in the parentheses is [tex] \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
[/tex]

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

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

 

[ehild]

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

ehild

 

[Andrew Mason]

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

ehild

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 s_m and s_E 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)[tex]\ddot{s_m} = -\frac{GM_E}{r^2}\hat{r} - \frac{GM_M}{d^2}\hat{d} [/tex]

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)[tex]\ddot{s_E} = -\frac{GM_M}{R^2}\hat R[/tex]

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

[tex]\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)[/tex]

The left side is just [itex]\ddot{r}[/itex], 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)[tex]\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)[/tex]

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)[tex]\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)[/tex]

The difference between the magnitude of those two accelerations is:

[tex]|\ddot{r}_1| - |\ddot{r}_2| = \frac{GM_M}{(R-r)^2} - \frac{GM_M}{(R+r)^2}[/tex]If we make the Earth solid, the normal force/unit mass is exactly equal and opposite to [itex]\ddot{r}[/itex].

AM

 

[ehild]

(1)[tex]\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)[/tex]

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)[tex]\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)[/tex]

The difference between the magnitude of those two accelerations is:

[tex]|\ddot{r}_1| - |\ddot{r}_2| = \frac{GM_M}{(R-r)^2} - \frac{GM_M}{(R+r)^2}[/tex]


If we make the Earth solid, the normal force/unit mass is exactly equal and opposite to [itex]\ddot{r}[/itex].

AM

I think you meant [itex]\ddot r [/itex] vector so it should have been written [itex]\ddot {\vec r }[/itex].

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

[tex]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)[/tex], that is

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

[tex]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)[/tex]

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

[tex]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) [/tex]

[tex]= \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) [/tex]

[tex]=\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}- \frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}[/tex]

[tex]N_2-N_1=\frac{GM_M}{(R+r)^2} - 2\frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}[/tex]


ehild

 

[Andrew Mason]

I think you meant [itex]\ddot r [/itex] vector so it should have been written [itex]\ddot {\vec r }[/itex].

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

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

[tex]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)[/tex], that is

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

[tex]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)[/tex]

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

[tex]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) [/tex]

[tex]= \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) [/tex]

[tex]=\frac{GM_M}{(R+r)^2} - \frac{GM_M}{R^2}- \frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}[/tex]

[tex]N_2-N_1=\frac{GM_M}{(R+r)^2} - 2\frac{GM_M}{R^2} +\frac{GM_M}{(R-r)^2}[/tex]


ehild

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

The difference is as you have calculated, which is approximately 6GM_Mr²/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

 

[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

 

[Andrew Mason]

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

The land surface moves away from the centre of the Earth too. The reason we don't notice it is because land does not flow when there is a change in the height of the surface. At the equator, the surface rises and lowers by about half a metre twice a day.

The ocean surface, however, forms an equipotential surface and water flows laterally as a result of the change in the shape of that equipotential surface. Shore line features can enhance these flows. So the ocean tides are more noticeable because they cause lateral tidal flows.

AM

 

[D H]

The moon is effectively pulling us "up".

That's only true in two extreme cases. The moon is indeed effectively pulling us "up" when the Moon is directly overhead or directly underfoot, but only then. When the Moon is on the horizon, the Moon is effectively pulling us "down", with about half the magnitude of the upward force at the zenith (Moon directly overhead) or nadir (Moon directly underfoot). At intermediate angles, the vertical component of the tidal force varies from upward when the Moon is directly overhead to downward when the Moon is at the horizon and then upward again when the Moon is directly underfoot. This means there are places where the vertical component of the tidal force is zero.

There is also a horizontal component to the tidal force. This is zero when the Moon is directly overhead, directly underfoot, or on the horizon. The horizontal component of the tidal force is non-zero at intermediate angles.

The vertical component of the tidal force is the driver behind the Earth tides, but it's the horizontal component that drives the ocean tides.

 

[Andrew Mason]

That's only true in two extreme cases. The moon is indeed effectively pulling us "up" when the Moon is directly overhead or directly underfoot, but only then.

Right. These are the two cases referred to in the OP.

When the Moon is on the horizon, the Moon is effectively pulling us "down", with about half the magnitude of the upward force at the zenith (Moon directly overhead) or nadir (Moon directly underfoot). At intermediate angles, the vertical component of the tidal force varies from upward when the Moon is directly overhead to downward when the Moon is at the horizon and then upward again when the Moon is directly underfoot. This means there are places where the vertical component of the tidal force is zero.

There is also a horizontal component to the tidal force. This is zero when the Moon is directly overhead, directly underfoot, or on the horizon. The horizontal component of the tidal force is non-zero at intermediate angles.

The vertical component of the tidal force is the driver behind the Earth tides, but it's the horizontal component that drives the ocean tides.

The actual tidal force/unit mass is given by the term in brackets in this equation (from my post #17):

[tex]\ddot{\vec{r}}= -\frac{GM_E}{r^2}\hat{r} - \left(\frac{GM_M}{d^2}\hat{d} -\frac{GM_M}{R^2}\hat{R}\right)[/tex]

The directions of the term in brackets are shown in this diagram. The horizontal component is zero when the tidal force is directed toward the Earth centre. This occurs in the two extreme cases referred to and at points on the surface that are the same distance from the moon as the centre of the earth.

AM

 

[whoareyou]

OK, so I want to follow along the proof you posted in post #17. What do the symbols with two dots above them mean? Is that the second derivative of "r" and "s" wrt to time (ie. acceleration)?

 

[Andrew Mason]

OK, so I want to follow along the proof you posted in post #17. What do the symbols with two dots above them mean? Is that the second derivative of "r" and "s" wrt to time (ie. acceleration)?

Yes:

[tex]\ddot{\vec{r}} = \frac{d^2\vec{r}}{dt^2}[/tex]

AM

 

[whoareyou]

Ok I so understand ehild's proof so far. But I though we were ignoring rotation? Why was your alternative proof necessary?

Edit: I plugged in all the known values keeping the mass of the object at 1kg and both proofs work out to be about the same percent differences (using (F2/F1 - 1) * 100%) (ie. about 5.[...] x 10^(-9) %)

 

[Andrew Mason]

Ok I so understand ehild's proof so far. But I though we were ignoring rotation? Why was your alternative proof necessary?

We are ignoring the rotation of the Earth on its axis because that complicates the analysis as it introduces acceleration about a point that is non-inertial. The rotation of the Earth and moon about their common centre of mass (barycentre) is not a problem. The barycentre is an inertial point. But, actually any inertial point will do. You don't really have to assume that the moon and Earth are orbiting each other in order to see tidal effects. It is just that the effects do not last very long if there is no rotation!

Ehild's analysis provided the tidal forces at the extrema (directly overhead and on opposite side of earth). He had it right to begin but I thought he was wrong. So I provided a more general analysis for the tidal force at any point on the surface but I then made a mistake in stating the difference between the two tidal forces. In the end I agreed with him: The book is wrong by about a factor of 40 in its answer.

AM

 

[ehild]

Ok I so understand ehild's proof so far. But I though we were ignoring rotation? Why was your alternative proof necessary?

Edit: I plugged in all the known values keeping the mass of the object at 1kg and both proofs work out to be about the same percent differences (using (F2/F1 - 1) * 100%) (ie. about 5.[...] x 10^(-9) %)

AM and me got the same result, only the notations are different. My and your first attempt was not correct, as we ignored the acceleration of the Earth towards the Moon and did not include the normal force. The bathroom scale reads the normal force, and you need the difference of the scale readings. I think the detailed derivation in Post#17 is correct. AM thought I used a wrong expression for the acceleration of the Earth. The acceleration of the Earth towards the Moon is the same as its acceleration towards the common barycentre. You can use any inertial frame of reference in the derivation. The barycentre is not needed.

The 24-hour rotation of the Earth causes the same effect to the weight either the man is at the same side of Earth as the Moon, or at the opposite side, so it will cancel when taking the difference of the weights.

ehild

 

[whoareyou]

Ok and in both proofs, have we assumed that the point where the person is standing is an inertial frame? Or are we saying since we are ignoring rotation, any point on the surface of the Earth is inertial? Or are we saying that the common centre of mass of the system is assumed to be at the point where the person is standing which makes that point inertial?

 

[ehild]

Ok and in both proofs, have we assumed that the point where the person is standing is an inertial frame? Or are we saying since we are ignoring rotation, any point on the surface of the Earth is inertial? Or are we saying that the common centre of mass of the system is assumed to be at the point where the person is standing which makes that point inertial?

No. The Earth accelerates towards the Moon (or the common barycentre). Those points accelerate together with the Earth. Neither the surface of Earth nor the centre of Earth are "inertial".

ehild

 

[whoareyou]

So then all points on Earth are still non-inertial? So I'm guessing then our inertial frame is just somewhere in space?

 

[ehild]

It can be somewhere in space or at barycentre between the Earth and Moon).

ehild

 

[whoareyou]

Hold on, when we defined a_E in ehild's proof, it was due to the gravity of the moon. Doesn't the person also contribute to the Earth's acceleration?

 

[ehild]

The mass of the man is very small with respect to the mass of the Earth. You can ignore its effect on Earth.

ehild