How much does Earth's gravity affect the gravity that we feel on the Lunar Surface?

  • #1
artriant
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TL;DR Summary
How much Earth affects the gravity that we feel on the Lunar Surface?
Hello everyone, happy holidays!

Y/day i googled that question (see title), and i found no clear answer, and I was really suprised,
So I had to investigate my self, this is a super basic question,

Let me know if i got this right:

Earth R: 6,371 km
Moon R: 1,737.1 km
d1: 384,400 km (center to center)
d2: 382,663 km (E core to near side)
d3: 386,137 km (E core to far side)
Assuming Earths gravity is 1g, M and E are perfect spheres, orbits are circular, distribution of mass is perfectly uniform, no other forces complicating our system like for example the pull of the sun.
---

A man at the center of the near side of the moon, is standing 60.06 times further away from the Earths CORE compared to a person standing at the surface of Earth. Applying inverse square law we can find that: Earths gravityis 3,488.4 times weaker up there.
Moons surface gravity averaging ~0.166g, the Earths pull is -0.0002772g at that point and has a completely oposite orientation (1.66% in comparison),

Simmilar calculations for the far side of the moon (at the point where Earths and Moons gravitational vectors align): 386,137/3671=60.60|sqr(60.60)=3,673.39|That is: +0.000272g or 1.638% in comparison to Moon Average surface gravity

To put that in perspective a cargo box weighting 1000kg will have an average weigth 166Kg on the Lunar surface,
But at the center of NEAR side it will weight (~1.66% less) 163.24Kg, and (+1.638%)168.72Kg @ the center of opposite side,

Thats a whole 5.48 Kg difference for that container, or 548 grams difference for a man that originally weights 100Kg on Earth

(3.35% icrease in weigth) just because you changed location on the Moon, isn't that crazy considering we just took in account the change due to Earths gravity gradient?
 

Answers and Replies

  • #3
artriant
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Sorry but i don't get your point,

Let me be more specific: I think i used this exact equation (or sort of) :

G, M, and m are constants (M=Earths mass, m = mass of astronaut and G is constant by definition)
So i assume Earths gravity can be calculated based on distance alone and here is how:

1 Earth radius= 1g
2 Earth radiuses 1/2^2 = 4 times weaker
3 Earth Radiuses 1/3^2 Etc = 9 times weaker

Are you saying that if i use masses, it will change results? Sorry I am not a physics student and i might think in unconvensional ways. I just need to know if I am close to the real answer.

Im actually suprised that Earths 'contribution' is not mentioned like anywhere.
 
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  • #4
Keith_McClary
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what did i miss?
The Moon is revolving around the Earth at a speed such that its centrifugal force balances gravity. The people are also moving at comparable speeds, and subject to forces.
 
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  • #5
artriant
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Thank you Keith 🙏 , that makes perfect sense!

I completely ignored centrifugal, maybe I was too focused on the Moon it self.

From what i understand, Earths pull completely canceling out at lunar orbit distance, due to the motion of the Moon. A man orbiting at the center of the Moon (instead of the Moon it self) would experience complete weigtlessness.

However, I wonder:

Is that also true for the man at the center of the near side? I mean his significantly closer. Does his speed matches an actuall orbit, or his simply constrained to the tidaly locked motion of the Moon?

In general we know that the closer we get to Earth the faster orbits become (their rotational rate as well), but the man standing at the center of the near side (or anywhere on the Moon actually) is remaining constrained to the Moon. Hmm, it gets too complicated for me, i will investigate another day, thanks again!
 
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  • #6
Charles Link
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You can use Kepler's ## R^3/T^2 ## law to see that if the object under study is closer to the earth, ## T ## will be shorter, and there will be a slight tendency for the object to pull ahead of the moon in its orbit. For an object farther away, it will have a larger ## T ##, and tend to lag behind. These effects are normally small, but it is this type of effect that makes rings happen around planets such as Saturn. The (large) object's own gravity is incapable of holding it together as a single object, and it comes apart if ## R ## is too small. For more info, see the Roche limit https://en.wikipedia.org/wiki/Roche_limit
 
  • #7
PeroK
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The Moon is revolving around the Earth at a speed such that its centrifugal force balances gravity. The people are also moving at comparable speeds, and subject to forces.
The Earth's gravity is a centripetal force on the Moon. There is no centrifugal force counteracting gravity.
 
  • #8
PeroK
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Thank you Keith 🙏 , that makes perfect sense!

I completely ignored centrifugal, maybe I was too focused on the Moon it self.

From what i understand, Earths pull completely canceling out at lunar orbit distance, due to the motion of the Moon. A man orbiting at the center of the Moon (instead of the Moon it self) would experience complete weigtlessness.
It may make sense, but it's not right! There is no force cancelling out the Earth's gravity. As above, the Earth's gravity provides a centripetal force on the Moon. There is no counteracting centrifugal force.

If we stick to Newtonian gravity, then there are four significant forces on a man on the Moon: the gravitational forces from the Sun, Earth and Moon; and a normal force from the surface of the Moon. These explain the man's motion, which is a combination of an orbit of the Sun and an orbit of the Earth, while remaining on the Moon's surface.
 
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  • #9
davenn
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Thank you Keith 🙏 , that makes perfect sense!

which is a pity since it was a bad answer
 
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  • #10
DrStupid
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Earths gravityis 3,488.4 times weaker up there.

But Earth's gravity is not what you are looking for. You already know how to do it correctly:

we just took in account the change due to Earths gravity gradient?

The the gravity that we feel on the Lunar surface is not altered by Earth's gravitational force

##F = - \frac{{G \cdot M}}{{\left| r \right|^3 }} \cdot r##

but by it's gradient. The first approximation for the resulting tidal force is

##\Delta F \approx F' \cdot \Delta r = - \frac{{G \cdot M}}{{\left| r \right|^3 }} \cdot \left( {\Delta r - 3 \cdot r \cdot \frac{{r \cdot \Delta r}}{{r^2 }}} \right)##

In the radial direction it results in an decreased weight

##\Delta F_r \approx 2 \cdot \frac{{G \cdot M}}{{\left| r \right|^3 }} \cdot \Delta r##

and in tangential direction in an increased weight

##\Delta F_t \approx - \frac{{G \cdot M}}{{\left| r \right|^3 }} \cdot \Delta r##

The total difference between minimum and maximum is just 0.003 % of the average weight on the Moon (according to my calculation).
 
  • #11
Richard R Richard
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It seems that what you want to find out is the variation of the apparent weight of an object,spinning very fine, depending on its position on the surface of the Moon. If you make other assumptions like
  • The moon rotates on its axis with the same angular speed as that which rotates around the Earth with parallel axes.
  • You neglect the effect of rotation with respect to the sun and only observe the factors dependent on the Earth and the Moon

What you are looking for is how much the normal force of the Moon's surface varies as a function of latitude ## \beta ## and longitude ## \alpha ##
[tex] N = m_oa_{app}\cong \dfrac {Gm_oM_M}{R_M ^ 2} - \dfrac {Gm_oM_E}{(R_{orbit} + R_M \cos \alpha \cos \beta) ^ 2} + m_o \omega_E ^ 2(R_{orbit} + R_M \cos \alpha \cos \beta) - m_o \omega_E ^ 2 (R_M \cos \beta) [/tex]

Where longitude varies between ## \alpha \in [0, \pi] ## and latitude ## \beta \in [- \dfrac {\pi} {2}, \dfrac {\pi} {2}] ##
and ## \omega_E = \dfrac {2 \pi} {27.32d \, 24h / d \, 3600s / h} ##

Resulting in the acceleration of the object independent of mass,, what you propose is to compare the results when ## \alpha = 0 ## and when ## \alpha = \pi ## for ## \beta = 0 ##
 
  • #12
DrStupid
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  • The moon rotates on its axis with the same angular speed as that which rotates around the Earth with parallel axes.

The rotation results in another 0.001 % in addition to the tidal forces. That means that the effects discussed in this thread are negligible compared to the 0.6 % measured variations of the surface gravity (e.g. due to mascons):

711352main_Zuber-3-pia16587-43_946-710.jpg
 
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  • #14
PeroK
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OK, "pseudoforce".

Should I also mention that the "gravitational force" is a pseudoforce? :rolleyes:
The general relativistic description of planetary motion is very different. In Newtonian mechanics, which is where the thread started, gravity is a force.

You only invoke centrifugal force in the frame of orbiting object, to explain why there is no motion in that frame. I.e. why an object in circular motion is non-accelerating relative to itself. It doesn't explain orbital motion.
 
  • #15
artriant
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which is a pity since it was a bad answer

@PeroK & @davenn
I dissagree, I think that @Keith_McClary is the first guy that suspects I am describing the problem from rotating frame, so introducing centrifugal force was smart. Its not a bad answer its a different POV. You can all blame me for not defining a frame.

@Charles Link thank you for your reply, Keplers third law seems very useful, i can find the theoretical T for imaginary far and near side orbits, super roughly:

Tn=650.4hours
Tf=659.3hours

And compare it with the actual T=654.9hours,
+ I can even calc the inertial centrifugal forces at near and far side centers, and so on *(again on rotating frame)

Roches limit is very interesting stuff,
--

@Richard R Richard Yes that's most likely what I am looking for, but it will take me a while to digest your math ,use of coordinates blew my brain, you guys way too smart :)

@DrStupid 0.003% is most likely the answer to my question, indeed a small figure compared to 0.6%. I did some rough calcs on rotating frame and i found a simmilar results ~0.0025%
 
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  • #16
PeroK
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im describing the problem from rotating frame

So, this analysis is from what reference frame? Looks very like the Earth's reference frame to me!

Let me know if i got this right:

Earth R: 6,371 km
Moon R: 1,737.1 km
d1: 384,400 km (center to center)
d2: 382,663 km (E core to near side)
d3: 386,137 km (E core to far side)
Assuming Earths gravity is 1g, M and E are perfect spheres, orbits are circular, distribution of mass is perfectly uniform, no other forces complicating our system like for example the pull of the sun.
---

A man at the center of the near side of the moon, is standing 60.06 times further away from the Earths CORE compared to a person standing at the surface of Earth. Applying inverse square law we can find that: Earths gravityis 3,488.4 times weaker up there.
Moons surface gravity averaging ~0.166g, the Earths pull is -0.0002772g at that point and has a completely oposite orientation (1.66% in comparison),

Simmilar calculations for the far side of the moon (at the point where Earths and Moons gravitational vectors align): 386,137/3671=60.60|sqr(60.60)=3,673.39|That is: +0.000272g or 1.638% in comparison to Moon Average surface gravity

To put that in perspective a cargo box weighting 1000kg will have an average weigth 166Kg on the Lunar surface,
But at the center of NEAR side it will weight (~1.66% less) 163.24Kg, and (+1.638%)168.72Kg @ the center of opposite side,

Thats a whole 5.48 Kg difference for that container, or 548 grams difference for a man that originally weights 100Kg on Earth

(3.35% icrease in weigth) just because you changed location on the Moon, isn't that crazy considering we just took in account the change due to Earths gravity gradient?
 
  • #17
DrStupid
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Looks very like the Earth's reference frame to me!

I don't see it. This reather looks like an inertial frame:

M and E are perfect spheres, orbits are circular

Earth doesn't orbit in it's own rest frame (neither in its co-rotating frame nor in the non-rotating rest frame of its center of mass). It doesn't orbit in the co-rotating frame of the Earth-Moon system either. There are rotating frames where Earth and Moon are moving on circular paths but it wouldn't make sense to use them.
 
  • #18
artriant
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@DrStupid I only mentioned circlular orbits, so distances relative to Earth and Moon remain constant,

@PeroK Yes i confirm that i originally imagined the Earth and the Moon in inertial (rotating reference frame) , with our "camera" looking from the top like that (circles represent the equators repectively):

New Project.png

But the frame doesn't really matter, as long as we arive in the same conclusions:

The first conclusion is that my original post is wrong because i calc apparent weigth only based on gravity, therefore i find some very exagerated variations between the near and the far side of the Moon (that do not exist in the real world.

The second conclusion is that despite the above, there is space for smaller variations: It seems that the distance that we have from Earths core play some very subtle role in the apparent weigth in various locations on the Lunar surface (~0.003% between min and max ),

That would be just 0.5 miligrams difference in weigth (for a subject that originally weights 100kg on Earth), a negligible amount until someone states otherwise, for example a future Architect that will win a contract just because of that! xd
 
  • #19
Richard R Richard
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A view, my attempt to perform accurate apparent weight calculations would start by placing an inertial reference frame on the ground for example

calculating the force on a mass m due to the mass of the Moon

$$ F_M = \dfrac {GmM_M} {R_M ^ 2} $$

calculating the force on a mass m due to the mass of the Earth on the visible side of the moon

$$ F_{Ec} = \dfrac {GmM_T} {(D_{EM} -R_M) ^ 2} $$

calculating the force on a mass m due to the mass of the Earth on the far side of the moon

$$ F_{Ef} = \dfrac {GmM_T} {(D_{EM} + R_M) ^ 2} $$

The effect on centripetal acceleration due to the rotation of the moon around the Earth on the visible side of the moon

$$ F_{cEc} = m \omega_E ^ 2 (D_{EM} -R_M) $$

The effect on centripetal acceleration due to the rotation of the moon around the Earth on the visible side of the moon

$$ F_{cEf} = m \omega_E ^ 2 (D_{EM} + R_M) $$

Spinning fine

The effect on the centripetal acceleration due to the rotation of the moon on its axis
$$ F_{cMc} = m \omega_E ^ 2 (R_M) $$

On the visible side the apparent weight N

$$ N_c = F_M-F_{Ec} + F_{cEc} - F_{cMc} $$

On the far side the apparent weight N

$$ N_f = F_M-F_{Ef} + F_{cEf} + F_{cMc} $$

The contribution of the centripetal force of the Moon is so small, that there are other more important factors to analyze, for example

  • The center of mass of the moon is not exactly located in the geometric center of the ideal sphere that we could observe, the center of mass is slightly closer to the Earth, on the axis that joins the geometric centers of Earth and Moon.
  • Due to this the center of gravity is even more displaced from the center of the Moon.A consequence of this is that in order to rotate on its axis at a different angular speed than that of rotation with respect to the earth, it must generate a very large torque, so over time it has begun to rotate synchronously with the earth, always showing us the same face.
  • In addition, there are other effects such as those already named "mascons" (mass concentrations or regions of higher density), which must be taken into account to accurately calculate the orbits of satellites such as the LRO.

These effects make the precision of Newtonian mathematics calculations not as accurate as we wish without a deep study of lunar geology and gravity. And if these factors are taken into account, the effect of the eccentricity of the Moon's Orbit, however small, is also calculated in the same order.
 
  • #20
Charles Link
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I did a couple of calculations computing how much lighter the object would appear on the moon, due to it being closer to Earth by the radius of the moon, and also on the far side of the moon. The results, a weight reduction in both cases, were virtually identical. The factor of reduction is ## \approx .00002 ## from the weight of the object as observed on the moon, without the moon orbiting the earth, with ## g_{moon}=g_{earth}/6 ##.

The calculations involved calculating ## ma=mv^2/r ## and comparing it to the gravitational force from the earth. These two were just slightly different, and resulted in the weight reduction that would be observed by a scale placed under the object on the moon.

My calculations are in near agreement with post 10, (@DrStupid ), but I get ## 3GMm \Delta r /r^3 ##, with a 3 rather than a 2. (## \Delta r ## is the radius of the moon).
 
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  • #21
PeroK
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On the visible side the apparent weight N

$$ N_c = F_M-F_{Ec} + F_{cEc} - F_{cMc} $$

On the far side the apparent weight N

$$ N_f = F_M-F_{Ef} + F_{cEf} + F_{cMc} $$
Are you sure about the signs here? The ##F_{cMc}## should be the same (minus) in both cases, surely. And, the ##F_{cEf}## should be negative, I imagine?
 
  • #22
PeroK
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My calculations are in near agreement with post 10, (@DrStupid ), but I get ## 3GMm \Delta r /r^3 ##, with a 3 rather than a 2. (## \Delta r ## is the radius of the moon).
Following the approach of @Richard R Richard above, I got the difference to be $$\frac{6GMm (\Delta r)^2}{r^4}$$
 
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  • #23
Charles Link
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Following the approach of @Richard R Richard above, I got the difference to be $$\frac{6GMm (\Delta r)^2}{r^4}$$
I agree with your (@PeroK ) result. My calculation, doing it just to first order in ## \Delta r ##, compared the results for orbiting the Earth vs. without any earth. To first order in ## \Delta r ##, the difference was zero between near side and far side in the orbiting case=they were both lighter by the same amount. Upon working it to second order, we are in agreement.

Note also the size of this effect: ## \Delta r/r \approx 1/240 ##. Even the first order result is very small.
 
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  • #24
DrStupid
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Upon working it to second order, we are in agreement.

Yes, the Taylor series for the 1-dimensional case is

##F_g \left( {r + \Delta r} \right) = \frac{{G \cdot M \cdot m}}{{\left( {r + \Delta r} \right)^2 }} = \frac{{G \cdot M \cdot m}}{{r^2 }} \cdot \sum\limits_{i = 0}^\infty {\left( {i + 1} \right) \cdot \left( { - \frac{{\Delta r}}{r}} \right)^i }##

and the second order term results in the difference of

##6\frac{{G \cdot M \cdot m}}{{r^4 }} \cdot \Delta r^2##

between both sides. I also stopped at the first order term (that's what I meant with "first approximation"), not only because the higher order terms are very small but also because they become quite ugly in case of more than one dimension.
 
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  • #25
Charles Link
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I question the correctness of having a ##F_{cMc} ## in @Richard R Richard 's calculation (post 19). In the rest frame of the earth, there is no additional rotation of the object about the axis of the center of the moon, when the same side of the moon always faces the earth. The object on the surface of the moon simply moves in a circular orbit around the earth, with no additional correction.
 
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  • #26
DrStupid
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I question the correctness of having a ##F_{cMc} ## in @Richard R Richard 's calculation (post 19).

That means it's not just me. I also got a notion that the rotation has been considered twice. Maybe I'll start my own derivation.
 
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  • #27
Richard R Richard
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Are you sure about the signs here? The ##F_{cMc}## should be the same (minus) in both cases, surely. And, the ##F_{cEf}## should be negative, I imagine?
The sign depends on whether or not it favors the acceleration of the mass in the reference frame ... if we propose an external one at rest with the earth, (which does not rotate) with it or with the moon, we will then see that the sign has the direction of the resulting acceleration ## F_ {cMc} ## must always be in the direction of the center of the Moon, since it is a centripetal resultant towards the Moon, of course I can be wrong with the interpretation I gave it, because you think it would take the opposite sign?.
The ## F_ {cEf} ## is also a centripetal resultant when you raise N2 so it has the opposite sign to the force that maintains the rotation if both terms are on the same side of equality.

For example $$ \dfrac {GmM} {r^2} = m \dfrac {v^2} {r} = m \omega^2r $$
O well
$$ \dfrac {GmM} {r^2} -m \omega^2r = 0 $$
For a circular path, the sign is negative if they are on the same member

I question the correctness of having a ##F_{cMc} ## in @Richard R Richard 's calculation (post 19). In the rest frame of the earth, there is no additional rotation of the object about the axis of the center of the moon, when the same side of the moon always faces the earth. The object on the surface of the moon simply moves in a circular orbit around the earth, with no additional correction.

Why is there no rotation?
I see the red dot first on the left and then on the right of the moon, or vice versa, seen from a static earth, or first above and then below, or vice versa.
rotacion.png
 
  • #28
PeroK
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On the visible side the apparent weight N

$$ N_c = F_M-F_{Ec} + F_{cEc} - F_{cMc} $$

On the far side the apparent weight N

$$ N_f = F_M-F_{Ef} + F_{cEf} + F_{cMc} $$

Hmm. If we take the Earth out of the equation, then (if I interpret your equations correctly) we have:

On the visible side the apparent weight N

$$ N_c = F_M - F_{cMc} $$

On the far side the apparent weight N

$$ N_f = F_M + F_{cMc} $$

And the effective gravity on Moon depends on which side you're standing?

And, if we take the centripetal forces out of the equation, we have:

$$ N_c = F_M-F_{Ec}$$

On the far side the apparent weight N

$$ N_f = F_M-F_{Ef}$$

And the Earth's gravity is reducing the normal force, even though on the far side I imagine that Earth's gravity is increasing the normal force?

I'm just really confused by what your equations mean.
 
  • #29
DrStupid
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This is my derivation including the rotation:

In addition to the tidal forces mentioned in #10 there are also corresponding differences of centrifugal forces

##\Delta F_c = F'_c \cdot \Delta r = m \cdot \left[ {\omega ^2 \cdot \Delta r - \omega \cdot \left( {\omega \cdot \Delta r} \right)} \right]##

in the co-rotating frame of the Earth-Moon system. Depending on the location of the moon in the rotating frame there might be a term ##\omega ^2 \cdot r## in the equation for the centrifugal force ##\Delta F_c## but not in the gradient ##{F'}##.

Neglecting the displacement of the center of Earth with respect to the center of mass results in the approximation

##\omega ^2 \approx \frac{{G \cdot M}}{{\left| r \right|^3 }}##

for the angular velocity. With Earth and Moon on the X-axis and the definition of longitude and latitude from #11 the total variation of the appearent weight is given by

##
\Delta F = \Delta F_g + \Delta F_c \approx \frac{{G \cdot M \cdot m}}{{\left| r \right|^3 }} \cdot r_M \cdot \left( {\begin{array}{*{20}c}
{3 \cdot \cos \alpha \cdot \cos \beta } \\
0 \\
{ - \sin \beta } \\
\end{array}} \right)
##

The corresponding normal component is

##\Delta F_N = \frac{{\Delta F \cdot \Delta r}}{{\Delta r^2 }} \cdot \Delta r \approx \frac{{G \cdot M \cdot m}}{{\left| r \right|^3 }} \cdot \left( {3 \cdot \cos ^2 \alpha \cdot \cos ^2 \beta - \sin ^2 \beta } \right) \cdot \Delta r##

That is a reduction by 0.00226 % on the near and far side and an increase by 0.00075 % on the poles, resulting in a maximum difference of 0.003 %.
 
  • #30
PeroK
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Why is there no rotation?
I see the red dot first on the left and then on the right of the moon, or vice versa, seen from a static earth, or first above and then below, or vice versa.

As @Charles Link pointed out:

The motion of a point on the Moon's surface is a circle centred on the Earth/Moon barycentre. The (real) acceleration is, therefore, simply the centripetal acceleration associated with this orbit. There is no additional acceleration associated with the motion relative to the centre of the Moon.
 
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  • #31
Charles Link
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If you look at the red dots and continue them through the whole circle, you will see they are doing circular motion about the center of the earth. Yes, the moon has a rotation whose period is the same as its period of revolution, but that rotation simplifies this problem, by making the motion of the object simply as a circle around the earth.

The object is lighter than the gravity by the moon, because the period of its orbit is determined by the time ## T ## that the center of the moon orbits. ## F_{gravity \, earth}=GMm/(r-\Delta r)^2## , while centripetal force needed is ## F_c=[(2 \pi (r- \Delta r)/T]^2/(r- \Delta r) ##, with center of moon in orbit obeys ## GMm/r^2=(2 \pi r/T)^2/r ##. The difference between the Earth's gravity and the centripetal force needed is how much lighter the scale reads from the gravity of the moon pulling on the object.
 
  • #32
Charles Link
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If you solve for ## F_c ## eliminating ## T ##, you get ## F_c=GMm(r-\Delta r)/r^3 ##. This gives
## F_g-F_c=GMm \frac{r^3-(r-\Delta r)^3}{r^3 (r-\Delta r)^2} ##.
With a little algebra you get ## F_g-F_c \approx 3GMm \Delta r/r^3=3 (GMm/r_e^2) r_e^2 \Delta r /r^3 ##.
In more detail, the above expression gives ## F_g-F_c=(GMm/r^2) (3(\frac{\Delta r}{r})+3 (\frac{\Delta r}{r})^2+...) ##
## GMm/r_e^2 ## is the weight of the object on earth.

You can do a similar thing when the object is on the far side of the moon. Then the reduction in weight is ## F_c-F_g ##. To first order this gives an identical result to the above. The second order result for the difference ## |F_{c1}-F_{g1}|-|F_{c2}-F_{g2}| =6GMm (\Delta r)^2 /r^4 ##. ( Note this difference is zero in first order). This result was first obtained by @PeroK post 22. (The second order term for the far side expression has the opposite sign, thereby the factor of 6).
 
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  • #33
Richard R Richard
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[QUOTE = "PeroK, post: 6435672, member: 493650"]

$$ N_f = F_M + F_ {cMc} $$



And the effective gravity on the Moon depends on which side you are standing on?



[/QUOTE]



You're right, maybe I mix ideas, actually on the opposite side in the same frame of reference I must have written something like

[tex] -N_f = -F_M + F_ {cMc} [/tex]



That is, when we include the Earth



[tex] N_c = F_M-F_ {Ec} F_ {cEc} -F_ {cMc} [/tex]


[tex] -N_f = -F_M - F_ {Ef} + F_ {cEf} + F_ {cMc} [/tex]

Which are the same equations, but the sign only indicates the vector sense in that frame of reference.



Actually, the difference between the measured values between one side and the other is due to the difference between

$$ F_ {Ec} - F_ {cEc} \neq F_ {Ef} - F_ {cEf} $$



Edit I still haven't read the additional messages I've been replying to
 
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  • #34
Charles Link
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It may be worth mentioning for the arithmetic of the results of post 32, etc., it is kind of simple, if you use for the radius of the Earth ## r_e=4000 ## miles, the Earth to moon distance ##r=240,000 ## miles, and the radius of the moon ## \Delta r=1000 ## miles. Then ## r_e/r=1/60 ##, and ## \Delta r/r=1/240 ##.
 
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