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# We need to find t t but finding an explicit solution

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We need to find T + t , but finding an explicit solution for above equation is out of our reach. However, for practical problems we still could find num- berical solutions. For instance, one could use Mathematics. On the other hand, from the last equation we can calculate the distance fallen up to any given time t. 3 Escape velocity For a mass at a position x above the surface of the earth, Newton’s law is given by d 2 x dt 2 = - GM ( R + x ) 2 = - gR 2 ( x + R ) 2 , (15) where g = GM R 2 . (16) By making use of the trick d 2 x dt 2 = dv dt = dv dx dx dt = v dv dx , (15) becomes v dv dx = - gR 2 ( x + R ) 2 . (17) Example 5 (hw 11) Find the escape velocity for a body projected upward with an initial velocity v 0 from a point x 0 = ξR above the surface of the earth, where R is the radius of the earth and ξ is a constant greater than unity. Neglect air resistance. Find the initial altitude from which the body must be launched in order to reduce the escape velocity to 85% of its value at the earth’s surface. Start from the separable equation (17), one has Z v v 0 vdv = - gR 2 Z x x 0 dx x 2 , 1 2 ( v 2 - v 2 0 ) = gR 2 ( 1 x - 1 x 0 ) . 6

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Recall that v 0 = v escape when v 0 as x → ∞ . Therefore, v 2 escape = 2 gR 2 x 0 , v escape = s 2 gR 2 x 0 = s 2 gR ξ . The escape velocity at the earth’s surface is equal to 2 gR , thus for this example we should set r 1 ξ = 85% , wich gives us ξ = 1 . 38408 . 2 7
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