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 that 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 increases in potential energy occurs at the cost of kinetic energy which correspondingly decreases If v is the velocity of the rocket at height h, then the decrease in kinetic energy is `(1)/(2) mv _(0)^(2) - (1)/(2)mv^(2)`
Thus `(1)(2) mv_(0)^(2) - (1)/(2) mv^(2) = (mgh)/(1+(h)/(R))orv_(0)^(2)-v^(2)=(2gh)/(1+(h)/(R))`
Let `h_(max)` be the maximum height reached by the rocket at which its velocity has been reduced to zero Thus substituting `v =0` and `h = h_(max)` in the last expression we have
`v_(0)^(2)=(2gh_(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) =0.9 xx` escape velocity `=0.9xxsqrt((2gR)):.h_(max)=((09xx0.9)2gR)/(2g-((09xx0.9)2gR)/(R))`
`= (1.62 gR)/(2g - 1.62R) = (1.62R)/(0.38) = 4.26R` .