Is there a volosity to speed acheived ratio fixed to altitude of orbit in figuring space travel launch.?

Question

How fast can we leave the earths pull is my bottom line.
Is additional power of thrust at take off proportionally measured and calculated?

Answer 1

Sure -

F = mr (omega)^2 = mg

m's cancel out - we get g = r (omega)^2 where g is acceleration due to gravity, r is radius to earth's center of mass, and omega is rotational velocity of the object around the earth.

You can also go one step further and solve for v = r (omega) where v is the horizontal velocity.

Then g = v^2 / r

Checking units, we get ft / sec^2 = (ft^2/sec^2)/ft

So v = square root (rg)

Makes sense to me - perhaps a little oversimplified, but close enough for a reasonable estimate.


ADDED -

You may be looking for the velocity required to leave earth orbit. At launch, escape velocity is about 7 miles / second.