91Ó°ÊÓ

The mass of the rocket $m=150.0kg$.

The velocity of the rocket when engine shuts off,
$\begin{array}{l}v=3.70\frac{km}{s}\\ =3.70×{10}^3\frac{m}{s}\end{array}$

The height of the rocket when engine shuts off

$\begin{array}{l}{h}_0=200km\\ =2.0×{10}^{5}m\end{array}$

The radius of earth $R_0=6.37×{10}^{6}m.$

We can use the concept of conservation of energy to find the kinetic energy and maximum height of the rocket.

Formulae:

$$\begin{gathered} K_i+U_i=K_f+U_f \\ U=\frac{GMm}{R} \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

We have the height of the rocket when its engine shuts off, which is$R_0=R_E+h_0$.

And maximum height of the rocket is$R=R_E+h$.

So, the equation for conservation can be written as

$$\begin{gathered} K_i+U_i=K_f+U_f \\ \frac{1}{2}m{v}^2−\frac{GMm}{R_0}=K−\frac{GMm}{R} \end{gathered}$$

Substituting the known values, we get

$$\begin{gathered} \frac{1}{2}\left(150\right){(3.7×{10}^3)}^2−\frac{6.67×{10}^{−11}\left(150\right)(5.98×{10}^{24})}{6.57×{10}^{6}m}=K−\frac{6.67×{10}^{−11}\left(150\right)(5.98×{10}^{24})}{7.37×{10}^{6}m} \\ K=3.82×{10}^{7}J \end{gathered}$$

The kinetic energy of the rocket at height 1000 m above earth surface is$K=3.82×{10}^{7}J$

Again, using energy conservation for the maximum height, we get

$\frac{1}{2}m{v}^2−\frac{GMm}{R_0}=0−\frac{GMm}{R_f}$

Solving this equation for $R_{f,}$we get

$R_f=7.40×{10}^{6}m$

Now, we know that

$$\begin{gathered} h_m=R_f−R_E \\ {h}_m=1.03×{10}^{6}m \end{gathered}$$

Maximum height of the rocket, $h_m=1.03×{10}^{6}m$.