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    So gravitational field strength is defined as the force exerted per unit mass at a point in the field. So g=F/M so then in my textbook it then goes onto say that it's independent of the test mass used, but how can this be? from the equation surley a leager test mass means a smaller g field so it's dependent?
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    bree crazy gravitons with infinite range cause they have no mass but are affected by mass you get me ?
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    (Original post by Emissionspectra)
    So gravitational field strength is defined as the force exerted per unit mass at a point in the field. So g=F/M so then in my textbook it then goes onto say that it's independent of the test mass used, but how can this be? from the equation surley a leager test mass means a smaller g field so it's dependent?
    doesn't the M in g=F/M represent the Mass of the thing creating the gravitational force though, eg M would be the earth then a larger test mass wont affect it
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    (Original post by Benniboi1)
    doesn't the M in g=F/M represent the Mass of the thing creating the gravitational force though, eg M would be the earth then a larger test mass wont affect it
    thinking about it i think its becuase Force has mass in its units so the m's cancel out.
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    Grav force on a mass is prop to the mass.

    F=GMm/r^2

    g = F/m
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    It's independent because the "F" in your equation is the gravitational force on the mass (which is proportional to the smaller mass in the field) and so the two masses cancel.


    g=\frac{F_{gravitational}}{m}, where m is the mass in the gravitational field

    F_{gravitational}=\frac{GMm}{r^{  2}}, where M is the mass causing the gravitational field and m is the mass in the gravitational field.

    Combining these two, we get g=\frac{\frac{GMm}{r^{2}}}{m}, and so both the lowercase ms cancel out, giving g=\frac{GM}{r^{2}}. From this we can see that g is proportional only to the mass causing the field and the square of the distance from that mass; it is not proportional to the mass experiencing the field.
 
 
 
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