## Question:

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 = –${ g }_{0 }[{ R }^{ 2 }/{ (R + y) }^{ 2 }]$ , where${ g }_{ 0 }$is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If ${ g }_{ 0 }$= 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 $y\rightarrow \infty$.

## Step-by-step

v dv = a dy \\ \int _{ v }^{ 0 }{ vdv } =-{ g }_{ 0 } { R }^{ 2 }\int _{ 0 }^{ \infty }{ \frac { dy }{ { (R+y) }^{ 2 } } } \\ \frac{ { v }^{ 2 } } { 2 } |^{ 0 }_{ v } = \frac { { g }_{ 0 } { R }^{ 2 } } { R + y } | ^{ \infty }_{ 0 } \\ \begin{aligned} v &= \sqrt{ { { 2 }_{ g } }_{ 0 } R } \\ &= \sqrt { 2(9.81)(6356){ (10) }^{ 3 } } \\ &= 11167\space m/s = 11.2\space km/s \end{aligned}