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Can anyone give a mathematical demonstration of this? I assume it has to do with the fact that Earth’s mass is much bigger than the mass of any object on Earth, but I think it would be interesting to see a precise mathematical derivation, from the universal formula for gravitation to the Earth-specific $m*g$ formula.

Qmechanic
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Steve
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2 Answers2

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The acceleration due to gravity at the surface of the earth ($9.8 m/s^2$) is a direct result of the formula for gravitational attraction.

The universal law of gravitation states that:

$F=GMm/r^2 $

You can plug in values for $G, M$, and $r$ to find the acceleration due to gravity at the surface of the earth, which is a handy quantity denoted $g$.

$g = GM/r^2 = 9.8 m/s^2$

Plugging this into the original equation yields the force of the earth's gravity on an object on the surface:

$F = mg$

Nuclear Hoagie
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This is actually by definition. The acceleration due to gravity ($g$) at the surface of a planet is defined as $$g = \frac{GM_p}{R_p^2}$$ where $R_p$ and $M_p$ are parameters of the planet. If you apply Newton's law to a particle close to the planet (say Earth), the equation becomes: $$\sum F_y = \frac{GmM_E}{R^2_E} = m(\frac{GM_E}{R_E^2}) =mg$$ This equation also explains the origins of the value $9.80 \frac{m}{s^2}$ for $g$ on the surface of the Earth.

Hope this helps.

Cross
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