Question

A $\mathbf{150}\mathbf{.0}\mathbf{}\mathit{k}\mathit{g}$ rocket moving radially outward from Earth has a speed of$\mathbf{3}\mathbf{.70}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{s}$ when its engine shuts off $\mathbf{200}\mathbf{}\mathit{k}\mathit{m}$above Earth’s surface. (a) Assuming negligible air drag acts on the rocket, find the rocket’s kinetic energy when the rocket is above Earth’s surface. (b) What maximum height above the surface is reached by the rocket?

Short answer

1. The kinetic energy of the rocket at height 1000 m above earth surface is$K=3.82\times {10}^{7}J.$
2. Maximum height of the rocket $h_m=1.03\times {10}^{6}m$ .

Step-by-step solution

Listing the given quantities

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

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

The height of the rocket when engine shuts off

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

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

Understanding the concept of conservation of energy

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

Formulae:

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

(a)Calculations of kinetic energy of the rocket at height 1000m above earth surface

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

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

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

(b)Calculations of maximum height of the rocket

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

$R_f=7.40\times {10}^{6}m$

Now, we know that

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

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