Bjarne Posted September 10, 2013 Share Posted September 10, 2013 (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 ? Edited September 10, 2013 by Bjarne Link to comment Share on other sites More sharing options...
swansont Posted September 10, 2013 Share Posted September 10, 2013 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. 2 Link to comment Share on other sites More sharing options...
Enthalpy Posted September 10, 2013 Share Posted September 10, 2013 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. 1 Link to comment Share on other sites More sharing options...
Bjarne Posted September 10, 2013 Author Share Posted September 10, 2013 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) ? Link to comment Share on other sites More sharing options...
swansont Posted September 10, 2013 Share Posted September 10, 2013 You need to calculate the difference in acceleration between the center and the surface. http://mb-soft.com/public/tides.html Link to comment Share on other sites More sharing options...
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