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    g=9.81m/s^2, but what does the 9.81 mean & what is it due to (the force of gravity?).
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    Hi there.

    9.81 is simply how fast an object will accelerate each second under the force of gravity (assuming no other forces)


    For example, a rock starts at 0 m/s when dropped. After a second it will be travelling at 9.81 m/s, after 2 seconds it will be travelling at 19.62 m/s (2 x 9,81)

    Note: It doesn't matter how heavy the object is. Everything falls at the same rate (assuming no other forces)
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    (Original post by Paranoid_Glitch)
    g=9.81m/s^2, but what does the 9.81 mean & what is it due to (the force of gravity?).
    Yes, it's the approximation we for the gravitational field strength close to the earth where we can model it as a uniform field. Further away, we have that: \displaystyle F = -\frac{Gm_1 m_2}{r^2}.

    Edit: oh, and moved to Physics.
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    g is the force at which donald trump grows real hair.
    it is an SI unit used worldwide and was pioneered by Napoleon and the Illuminati conspiring to destroy the creation of rare pepes everywhere
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    (Original post by Zacken)
    Yes, it's the approximation we for the gravitational field strength close to the earth where we can model it as a uniform field. Further away, we have that: \displaystyle F = -\frac{Gm_1 m_2}{r^2}.

    Edit: oh, and moved to Physics.
    And to follow on from this, from \displaystyle F = -\frac{Gm_1 m_2}{r^2} we can also see that the acceleration is independent of the mass of the object, since F=M2 A, so the M2 will cancel such that the acceleration doesn't depend on the mass of the object. You can also verify the value of 9.81 by plugging in the earths radius as R, and for small differences in height (such as those found in projectile questions) you can assume that g is 9.81 (since the change in height is tiny compared to the radius of the earth, so the difference in the field strength maybe affects say the 9th decimal place and therefore we can assume it is a uniform field
 
 
 
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