A rocket starts vertically upward with speed `v_(0)`. Show that its speed `v` at height `h` is given by `v_(0)^(2)-v^(2)=(2hg)/(1+h/R)`
where `R` is the radius of the earth and `g` is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed `0.9` times the escape velocity.

Explanation

The gravitational potential energy of a mass `m` on earth's surface and at a height `h` is given by
`U(R)=-(GMm)/R`
and `U(R+h)=-(GMm)/(R+h)`
`:.U(R+h)-U(R)=-GMm(1/(R+h)-1/R)`
`=(GMmh)/((R+h)R)=(mhg)/(1+h/R)=` [`:'GM=gR^(2)`]
This increse in potential energy occurs at the cost of kinetic enegy which correspondingly decreases. If `v` is the velocity of the rocket at height `h`, then the decrease in kinetic is `1//2 mv_(0)^(2)-1//2mv^(2)`. Thus,
`1/2mv_(0)^(2)-1/2mv^(2)=(mhg)/(1+h/R)` or `v_(0)^(2)-v^(2)=(2gh)/(1+h/R)`
Let `h_("max")` be the maximum height reached by the rocket at which its velocity has ben reduced to zero. Thus, substituting `v=0` and `h=h_("max")` in the last expression, we have
`v_(0)^(2)=(2ghi_("max"))/(1+h_("max")//R)`
or `v_(0)^(2)(1+(h_("max"))/r)=2gh_("max")`
or `v_(0)^(2)=h_("max")[2g-(v_(0)^(2))/R]`
or `h_("max")=(v_(0)^(2))/(2g-(v_(0)^(2))/R)`
Now, it is given that
`v_(0)=0.9xx` escape velocity
`=0.9xxsqrt((2gR))`
`therefore h_("max")=((09xx0.9)2gR)/(2g-((09xx0.9)2gR)/R)`
`=(1.62gR)/(2g-1.62R)`
`=(1.62R)/0.38=4.26R`