Question

A rocket starts vertically with speed ${{v}_{o}}$. Show that its speed v at height h is given by ${{v}_{o}}^{2}-{{v}^{2}}=\dfrac{2hg}{1+\dfrac{h}{R}}$ where R is the radius of the Earth and g is the acceleration due to gravity at the Earth’s surface. Deduce an expression for maximum height reached by the rocket fired with speed of 0.9 times escape velocity.

Answer 1

Hint: It is given that the rocket is fired with initial velocity ${{v}_{o}}$. As the rocket moves away from the surface of the Earth, the gravitational potential energy increases. As a result of this the kinetic energy of the rocket will keep on decreasing to overcome the Earth’s gravitational potential. Hence by using the conservation of energy principle we will obtain the above expression for the velocity of the rocket at height h from the surface.
Formula used:
The gravitational potential energy at a height h from the surface of the earth is given by, $U=-\dfrac{GMm}{R+h}$ and the kinetic energy is given by when the rocket has velocity v is given by
$K.E=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
Let us say the rocket has mass m. Hence the gravitational potential energy as the rocket moves to a height h is given by, ${{U}_{h}}=-\dfrac{GMm}{R+h}J$ where g is the gravitational constant, M is the mass of the Earth and R is the radius of the Earth. Initially the rocket is at h=0. Hence the increase in the gravitational potential energy is equal to,
$\begin{align}
  & {{U}_{h}}-{{U}_{0}}=-\dfrac{GMm}{R+h}-\dfrac{-GMm}{R+(0)} \\
 & \Rightarrow {{U}_{h}}-{{U}_{0}}=-GMm\left[ \dfrac{1}{R+h}-\dfrac{1}{R} \right] \\
 & \Rightarrow {{U}_{h}}-{{U}_{0}}=GMm\left[ \dfrac{R-R+h}{\left( R+h \right)R} \right] \\
 & \Rightarrow {{U}_{h}}-{{U}_{0}}=GMm\left[ \dfrac{h}{\left( R+h \right)R} \right]\text{, }\because \text{GM=g}{{\text{R}}^{2}} \\
 & \Rightarrow {{U}_{h}}-{{U}_{0}}=\dfrac{\text{g}{{\text{R}}^{2}}mh}{\left( R+h \right)R} \\
 & \Rightarrow {{U}_{h}}-{{U}_{0}}=\dfrac{mgh}{\left( 1+\dfrac{h}{R} \right)}.....(1) \\
\end{align}$
This potential energy is overcomed by the kinetic energy of the rocket. Let us say the velocity of the rocket at time is v. Hence the change in kinetic energy with respect to the initial kinetic energy is used to overcome the gravitational potential energy is given by,
$\begin{align}
  & K.{{E}_{o}}-K.{{E}_{h}}=\dfrac{1}{2}m{{v}_{0}}^{2}-\dfrac{1}{2}m{{v}^{2}} \\
 & K.{{E}_{o}}-K.{{E}_{h}}=\dfrac{1}{2}m\left( {{v}_{0}}^{2}-{{v}^{2}} \right)....(2) \\
\end{align}$
Equating equation 1 and 2 we get,
$\begin{align}
  & \dfrac{1}{2}m\left( {{v}_{0}}^{2}-{{v}^{2}} \right)=\dfrac{mgh}{\left( 1+\dfrac{h}{R} \right)} \\
 & \Rightarrow {{v}_{0}}^{2}-{{v}^{2}}=\dfrac{2gh}{\left( 1+\dfrac{h}{R} \right)} \\
\end{align}$
When the rocket reaches its maximum height $\left( {{H}_{MAX}} \right)$ its velocity at that instant will be zero. It is asked what will be the height reached by the rocket if it was launched 0.9 time the escape velocity. The escape velocity is given by $\sqrt{2gR}$. Hence from the above equation we get,
$\begin{align}
  & {{v}_{0}}^{2}-{{v}^{2}}=\dfrac{2gh}{\left( 1+\dfrac{h}{R} \right)} \\
 & {{v}_{0}}^{2}=\dfrac{2g{{H}_{MAX}}}{\left( 1+\dfrac{{{H}_{MAX}}}{R} \right)} \\
 & \Rightarrow {{v}_{0}}^{2}\left( 1+\dfrac{{{H}_{MAX}}}{R} \right)=2g{{H}_{MAX}} \\
 & \Rightarrow {{v}_{0}}^{2}={{H}_{MAX}}(2g-\dfrac{{{v}_{0}}^{2}}{R}) \\
 & \Rightarrow {{H}_{MAX}}=\dfrac{{{v}_{0}}^{2}}{2g-\dfrac{{{v}_{0}}^{2}}{R}}\text{, }\because {{v}_{0}}=0.9\sqrt{2gR} \\
 & \Rightarrow {{H}_{MAX}}=\dfrac{{{\left( 0.9\sqrt{2gR} \right)}^{2}}}{2g-\dfrac{{{\left( 0.9\sqrt{2gR} \right)}^{2}}}{R}} \\
 & \Rightarrow {{H}_{MAX}}=\dfrac{0.81\left( 2gR \right)}{2g-\dfrac{0.81\left( 2gR \right)}{R}} \\
 & \Rightarrow {{H}_{MAX}}=\dfrac{1.62g{{R}^{2}}}{gR\left( 2-1.62 \right)} \\
 & \Rightarrow {{H}_{MAX}}=\dfrac{1.62R}{\left( 0.38 \right)} \\
 & \Rightarrow {{H}_{MAX}}=4.26R \\
 & \\
\end{align}$
Hence the maximum height reached by the rocket is 4.26R.

Note:
It is to be noted that the final velocity of the rocket is taken to be zero. This is basically at a stage where the rocket keeps on decelerating and ultimately reaches to zero velocity. If the rocket is not launched with the escape velocity, then the rocket will no longer exit the gravitational field of the Earth and eventually come down.