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How to calculate the sun and moon's gravitional pull in the Earth ?


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Posted (edited)

According to measurements we know that the influence of the tide is about 100 µGal.

This is for example illustrated by the example (image) below (this shows relative and absolute measurement during 24 hours)

 

But how can such measurements be compared to calculations ?

 

For example , the influence of Jupiter should be less that 1µGal, but when calculated based on the Newtonian equation (MG/r^2), I get 10 times so large values

Also when I calculate the influence of the Moon and the Sun, I get very high values.

 

For example the Moon....

Based on MG/r^2

M = Mass (of the moon) 7,35E22

G = Gravitational constant 6,67E-11

r = radius 384000000 meter^2

 

384000000*6,67€-11/6,32E11^2 = 0,00003324m/s (3324µGal)

This value is much higher as the 100 µGal we measure

 

In the same way the influence of the Sun can also be calculated to a much higher value.

 

Where did I went wrong ?

Why can I not use the Newtonian equation normally used to calculate acceleration due to gravity ?

Which equation must be used ?

 

231.jpg

Edited by Bjarne
Posted

Tides are the result of a difference in gravity between two points, not the gravity at that point. i.e. it's not g, it's dg/dr that matters, making it an inverse-cube relation, not inverse-square.

Posted

To paraphrase: Earth already follows the pull by Sun and Moon. What you feel with the accelerometer is only how much this pull differs where you are from the mean pull averaged over Earth's volume - which is nearly the pull at Earth's center.

Posted

Tides are the result of a difference in gravity between two points, not the gravity at that point. i.e. it's not g, it's dg/dr that matters, making it an inverse-cube relation, not inverse-square.

 

Thanks' for both answers

Can you show me an example how to calculate it (correct) ?

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