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Gravitational perturbations = tidal forces..?

by Deadstar
Tags: forces, gravitational, perturbations, tidal
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Jun14-11, 08:28 AM
P: 106
Hey folks I'm learning about graviational perturbations on satellites due to planets and, mainly for now, the Moon. In the book I'm reading on this, they say that gravitational perturbations due to celestial bodies are not casued by the full gravitational attraction of them but only by the corresponding tidal terms (i.e. by the difference between forces on the Earth and that on the satellite).

Can anyone elaborate on this? Say if we wanted to know the perturbations due to the Moon on a satellite, the book says to use the formula...
[tex]2 \frac{GM}{r_m^3}a_{sat}[/tex]
where M is mass on the Moon, [itex]r_m[/itex] the distance from the Moon to Earth and [itex]a_{sat}[/itex] the distance from the centre of the Earth to the satellite. I would have just gone and used
with d being the distance from Moon to satellite but why is this wrong?

The satellites they use this formula on are all 'close' to the Earth (i.e. with the Moons orbital radius). Would the formula they used still be valid for a satellite beyond the Moons orbital radius..?
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Jun14-11, 10:42 AM
Sci Advisor
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The thing to remember is that that moon pulls on the Earth too. If the Earth and satellite were an equal distance from the the Moon, then their accleration towards the Moon would be equal and there would be no influence acting to pull them apart.

As it is, there is a small difference between the distances so the Earth and Satellite have different accelerations due ot he Moon's gravity and this tends to pull the satellite and Earth apart. Note that his happens even if the Satellite is further from the Moon than the Earth is, as when the satellite is on the opposite side of the Earth from the Moon.

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