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    Gravitational field strength (g)

    If a test mass, m1, experiences a force F at some point in space, then the field
    strength, g, at that point is given by g = (F/m1)
    (F is calculate using:
    F =G m1 m2 )
    r2


    Note: The reason a small test mass is used is because a big mass might change the
    field that you are trying to measure.

    But I don't get the point of the note because when you equate, the masses m1, cancel out anyway, right?

    Sorry about the formatting with the formula of F...
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    If the gravitational field strength is g = F / m1 and F = G*m1*m2/r², so it follows that g = G*m1*m2/r² * 1/m1 = G*m2/r². Yeah, you are right the mass m1 is cancelled out.
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    Its me again!

    Why do you think that g = F/m1? after the link above the gravitational field strength is g = F/m...
    http://www.cyberphysics.co.uk/Q&A/KS.../equations.png
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    (Original post by Kallisto)
    Its me again!

    Why do you think that g = F/m1? after the link above the gravitational field strength is g = F/m...
    http://www.cyberphysics.co.uk/Q&A/KS.../equations.png
    They have stated the formulae separately. If we were using them in the same sum, and m2 were to be the bigger mass, then thought the force between the two objects would be the same, the smaller mass i.e. m1 would accelerate towards m2. In order to calculate the acceleration and, in other words, the gravity of m2, we would use the equation, F/m1 = (G*m1*m2/r^2)/m1.

    Is the logic wrong?
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    Oh God, silly me!

    I just realised that both the masses would accelerate and if the mass is much smaller in ratio to the other mass than the big mass' acceleration will be negligible and can be ignored but if it is big too then the big mass' acceleration will be substantial too and cannot be ignored.

    Yes, I got the point!
    Thanks a lot for asking just the right question! The questions often help me way more than direct answers.
    PRSOM
 
 
 
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