Question: An object g mass \(m\) is thrown vertically upward from the surface of the earth with initial velocity \(V_{0}\). We will calculate the value of \(V_{0}\), called the escape velocity, with which the object can escape the pull of gravity and never return to earth. since the object is moving far from the surface of the earth, we must take in to account the variation gravity with altitude, If the acceleration due to gravity sea level is g, and \(R\) is the radius of the earth, the gravitational force, \(F\), on the object of mass \(m\) at an altitude h above the surface of the earth is given by \(F=\frac{m g R^{2}}{(R+h)^{2}}\)

a.) Suppose \(v\) is the velocity of the object (measured upward) at time t. Use Newton's Law of Motion to show that \(\frac{dv}{dt}=\frac{-g{{R}^{2}}}{{{(R+h)}^{2}}}\).

b.) Rewrite the equation with \(h\) instead \(g\) as the independent variable using the chain rule

\(\frac{d v}{d t}=\frac{d v}{d h} \cdot \frac{d h}{d t}\)

Hence show that \(v\frac{dv}{dh}=\frac{-g{{R}^{2}}}{{{(R+h)}^{2}}}\)

c.) Solve the differential equation in part (b).

d.) Find the escape velocity, the smallest value of \(v_{0}\) such that \(c\) is never zero.

Price: $2.99

Solution: The downloadable solution consists of 3 pages
Deliverable: Word Document