As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. neglecting air resistance, this acceleration is determined from the formula a = -g0[r2> (r + y) 2 ], where g0 is the constant gravitational acceleration at sea level, r is the radius of the earth, and the positive direction is measured upward. if g0 = 9.81 m> s 2 and r = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. hint: this requires that v = 0 as y s .

initialvelocity   Vo = √[4 go R S / (R + S) ]

Explanation:

As in this problem the acceleration is not constant we must use the basic formula of the acceleration of a body

     a = dv / dt

We use the L’hospital chain rule

     a = dv/dy  dy / dt

     a = dv/dy  v

     v dv = a dy

We replace and integrate

     ∫ vdv = -go ∫ [R² / (R + Y)²  dy

     ½ V² = -go R² [-2 / (R + y)]

We evaluate the integral you enter its lower limit with the velocity equal to Vo for when it is at ground level (y = 0) and the end point for when it has zero velocity (Vf = 0) for a height S (y = S), let's calculate

     ½ (0- Vo²) = 2 go R² [1 / (R + S) - 1 / (R + 0)]

     Vo² = 4 or R² [-S / (R + S) R]

     Vo² = 4 go R S / (R + S)

     Vo = √[4 go R S / (R + S) ]

what do the notation's look like?

the answer is 46 mls because when you add the speed per second it increses the speed