Gravitational acceleration from moon

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Discussion Overview

The discussion centers on calculating the gravitational acceleration of an object on Earth's surface due to the Moon's influence, particularly when the Moon is directly overhead. Participants explore different formulas and approaches to understand this gravitational effect, including comparisons to tidal influences.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculated the force of attraction from the Moon to be 1.99 x 10^(20) N and questioned whether to use the mass of the Earth or the Moon to find gravitational acceleration.
  • Another participant suggested using the formula for tidal influence, a = GMm/(R-r)^(2) - GMm/(R^(2)), leading to a calculated gravitational acceleration of 0.000001131 m/s^(2).
  • A later reply proposed a different formula for net acceleration, a = GMearth/R2 - GMmoon/D2, clarifying the variables involved.
  • One participant expressed confusion regarding the difference between the formula for gravitational acceleration and the formula used for calculating tidal effects.
  • Another participant raised concerns about the relevance of mass in the tidal formula, questioning the accuracy of treating all ocean water as a single massive object.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the appropriate formula to use for calculating gravitational acceleration due to the Moon. There are multiple competing views regarding the relevance of different formulas and the treatment of mass in these calculations.

Contextual Notes

There are unresolved questions regarding the assumptions behind the formulas used, particularly in relation to the mass of the object and the definitions of variables in the context of tidal influences.

Irishwolf
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Hi ,
How could I calculate the acceleration of an object on the Earth's surface , say at the equator , from the influence of the moon , when the moon is directly overhead?

I have calculated the force of attraction from the moon to be 1.99 x 10^(20) N.

Now to find the gravitaional acceleration do I use ==> a=F/M
Where big M = mass earth?
I do I use the mass (m) of the moon?
Or is this not the way to calculate it?
-------------------------------------------------------
Another attempt of solving this was i used the formula for influence of moon on tides:
a= GMm/(R-r)^(2) - GMm/(R^(2))
and grav acc due to moon is 0.000001131 m/s^(2)
But someone said I should solve it with the above equation?

Help please
 
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Irishwolf said:
Hi ,
How could I calculate the acceleration of an object on the Earth's surface , say at the equator , from the influence of the moon , when the moon is directly overhead?

I have calculated the force of attraction from the moon to be 1.99 x 10^(20) N.

Now to find the gravitaional acceleration do I use ==> a=F/M
Where big M = mass earth?
I do I use the mass (m) of the moon?
Or is this not the way to calculate it?
-------------------------------------------------------
Another attempt of solving this was i used the formula for influence of moon on tides:
a= GMm/(R-r)^(2) - GMm/(R^(2))
and grav acc due to moon is 0.000001131 m/s^(2)
But someone said I should solve it with the above equation?

Help please

I'm not sure I understand your question. If your question is what is the net acceleration of an object on Earth considering the gravitational effect of the moon, you may look at it with this formula:

a = GMearth/R2 - GMmoon/D2

Mearth = mass of earth
Mmoon = mass of moon
R = Earth radius
D = distance from center of moon to the object.
 
Hi yes that was my question thanks,
But why is that a different formula than the one used to calculate the moons influence on tides?
 
Not sure about tide formula. What does that do?
Note, mass of the object is irrelevant. For tide, is there a
well defined mass? You can treat all ocean water as one massive object,
but then acceleration of what water? Our simple formula then may not be
very accurate.
 

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