metastable said:
^re: C = A * ~2.99... = A * ~3 -- So where did I do my math wrong?
To do the problem correctly, you have to use conservation of energy. The equation for finding the gravitational potential per unit mass anywhere inside a sphere of uniform mass is
$$ E = \frac{2}{3}\pi G p (r^2-3R^2) $$
where p is the density, r is the distance from the center, R is the radius of the mass and G the universal gravitational constant.
Given that for a spherical body,
$$ M = \frac{4 \pi R^3}{3} $$
we can make this
$$ E = \frac{GM}{2} (r^2-3R^2)$$
Given that we are looking for the difference between the surface where r=R and the Center where r=0, we are looking for the difference between
$$E= -\frac{GM}{R}$$
and
$$E= \frac{3}{2}\frac{GM}{R}$$
For the Earth, GM = 2.987e14 m
3/ s
2
and if we calculate out the difference we get 31255879.6 joules/kg
Since this is the energy that is converted to kinetic energy for the falling body, and KE = mv^2/2,
Then
$$ v = \sqrt { 2 (31255879.6)} $$
( we can ignore the "m" as we are looking for "energy per unit mass".
which equals 7906.4 m/s.
orbital velocity for an object just at the surface of the Earth is
$$ v= \sqrt {\frac{GM}{r}} $$
which equals 7906.4 m/s
This is no accident as
$$ -\frac{GM}{R} - \left (- \frac{3}{2}\frac{GM}{R} \right ) = \frac{GM}{2R} $$
Thus
$$ v = \sqrt{ 2 \frac{GM}{2R}} = \sqrt {\frac{GM}{r}}$$
