Escape velocity and circular velocity

As already established the velocity for a circular orbit is given by the formula – VL2 = gr2!d

where r is the radius of the earth, and d the distance from the centre of the earth.

At the earth’s surface d = r

V2 = gr-

Can the escape velocity be calculated equally simply? Very nearly so. If we go back to the idea of a body being projected vertically from the earth’s surface with sufficient kinetic energy to enable it to do work against the force of grav­itation all the way to infinity, then if the escape velocity is denoted by the symbol ve the

1

kinetic energy of a mass m will be ^inve

This kinetic energy must be sufficient to provide the energy needed to lift m from the earth’s surface to infinity. At a particular height above the earth, the energy needed to lift m one more metre is equal to its weight which equals mg at the earth’s surface, and decreases all the way to infinity when it will be zero. Since the weight changes and since the change is not a simple ratio but inversely as the square of the distance, it needs the principles of calculus to estimate the total work done, but the answer is very simple; it is the same as the weight at the earth’s surface mg X the radius of the earth, i. e. mgr.

but V* = gr

v =v X V2

e c

i. e. Escape velocity = V2 X Velocity of circular orbit at that radius.

Thus there is a simple relationship between all escape velocities and all circular velocities at a given radius from a mass such as that of the earth, or moon, or sun – the escape velocity is 1.41 X the circular velocity, or 41 per cent more.