Given:

  mdvdt=−mgR2(x+R)2 , Where x is the objects distance above the surface, R is earth’s radius, m mass and g is acceleration due to gravity.

Calculation:

Let h be maximum height reached by the object and v0 be the initial velocity.

Since velocity,

  v=dxdt .

By chain rule we can write,

  mvdvdx=mdxdtdvdx=mdvdt .

Therefore from the given information,

  mvdvdx=−mgR2(x+R)2

  vdv=−gR2(x+R)2dx

Integrating both sides,

  ∫v00vdv=−∫0hgR2(x+R)2dx

  [v22]v00=−[gR2(x+R)2]0h

  −v022=gR2h+R−gR2R

  −v022=−gR2h(h+R)R

  v02=2gRhh+R

  v0=2gRhh+R

i.e, the initial velocity is,

  v0=2gRhh+R

Given:

  ve=limh→∞v0

Calculation:

The initial velocity is given by,

  v0=2gRhR+h

As h→∞ , we can write,

  ve=limh→∞v0=limh→∞2gRhR+h

Using the L-hospital rule for limit, Taking derivative with respect to h on both the numerator and denominator,

  ve=limh→∞2gR1

Therefore, the escape velocity is,

  ve=2gR.

Given:

  R=3960 mi and g=32 ft/s2 .

Calculation:

The escape velocity is, ve=2gR.

From the given information,

  g=32 ft/s2=0.006 mi/s2 .

Therefore,

  ve=2gR=2×0.006×3960ve=47.5 mi/sec

In foot,

  ve=250800 ft/sec .