This is a problem we were given to practice differential equations and I have not the darndest clue of what to do. Can someone walk me through it so I can practice on my own.
According to Newton’s Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from Earth’s surface is
F = (mgR^2) / ((x+R)^2)
where x = x(t) is the object’s distance above the surface at time t, R is Earth’s radius, and g is the acceleration due to gravity. Also, by Newton’s Second Law, F = ma = m( dv/dt ) and so
m(dv/dt) = (−mgR^2) / ((x+R)^2)
a) Suppose a rocket is fired vertically upward with an initial velocity v0. Let h be the maximum height above the surface reached by the object. Show that
v0 = sqrt((2gRh)/(R+h))
(Hint: by the chain rule, m(dv/dt) = mv(dv/dx).)
b) Calculate Ve = lim(as h goes to infinity) V0. This limit is called the escape velocity for Earth.
c) Use R = 3960mi and g = 32ft/sec^2 to calculate Ve in feet per second and in miles per second.