# 2.2.4: Free Fall, Satellite Motion and the Mechanism of Tides


The formula for the force of gravity acting on a body, commonly referred to as "the body's weight", is:

$F_{\rm g}=m\cdot g$

with $$g$$ = 9.81 m/s $$^2$$, is valid only near the Earth's surface. At higher altitudes the value of $$g$$ changes. Its dependence on altitude, $$g(H)$$, where $$H$$ is the height over the Earth's surface (conventionally, measured from the sea level) can be readily found, if one invokes Newton's Law of Universal Gravitation, which states that two bodies of masses $$m_1$$ and $$m_2$$, $$R$$ apart, attract one another with a gravitational force $$F_{\rm g}$$, given by the equation:
$F_{\rm g} = G\frac{m_1 m_2}{R^2}$
where $$G$$ is the universal gravitational constant: $$G$$ = 6.674\)\times 10^{−11}\) N $$\cdot$$(m/kg)\)^2\). Then, the weight of an object of mass $$m$$ positioned near the Earth's surface -- i.e., the magnitude of the force attracting it towards the Earth's center is given by:
$F_{\rm g} = G\frac{m\cdot M_{\rm e}}{R_{\rm e}^{\; 2}}$
where $$M_{\rm e} = 5.972 \times 10$$^{24}\) kilograms is the Earth mass, and $$R_{\rm e} = 6,371 kilometers$$1 is the Earth radius. By equating the right sides of Eqs.(11) and (13), we obtain a formula for the value of $$g$$ near the earth surface2
$g = G\frac{M_{\rm e}}{R_{\rm e}^{\; 2}} = 9.8195 {\rm ~m/s}^2$
At altitude $$H$$ the distance of an object from the Earth center is $$R = R_{\rm e} +H$$. So, for the value of $$g$$ at this altitude, $$g(H)$$, we get:
$g(H) = G\frac{M_{\rm e}}{R^2} = G\frac{M_{\rm e}}{(R_{\rm e}+H)^2}$

2.2.4: Free Fall, Satellite Motion and the Mechanism of Tides is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.