v=(2gr)^.5
g=Gm/r^2
30 miles = 4.828032e5 meters
m = 5.9742e24 kilograms
r = 6.378e6 + 4.828032e5 = 6.8608032e6 meters
( 2 * (6.673e-11 * 5.9742e24 / 6.8608032e6^2) * 6.8608032e6) = 1.078e4
which is roughly 1 km less per second than what the surface requires.
remember orbit and escape...
maybe this will make it easy for you to understand.
when in Earth at a radius of 9.27e4, you tell me what the escape velocity is?
when in the moon at a radius of 2.14e4, you tell me what the escape velocity is?
when in the sun at a radius of 6.43e6, you tell me what the escape velocity...
really? really? did you even read my post?
look we have 1.1e4 km/s of escape velocity on Earth at surface. the surface is at a radius of 6.378e6. one is smaller than the other! well... the closer to the center of the planet with less radius then the more escape velocity you get. THERE IS A...
v=(gr2)^.5
i assume that g is directly related to r as g = Gm/r^2
in such I started to wonder when, if ever, will v = r as an INTEGER ONLY NOT DIMENSIONS?
v = ( 2 * g * r )^.5
v = ( 2 * G * ( m / r^2 ) * r )^.5
v = ( 2 * G * m / r )^.5
v * r^.5 = ( 2 * G * m )^.5
REMEMBER! v=r at some point...