Question

A rocket is launched straight up from the earth’s surface at a speed of $15,000m/s.$What is its speed when it is very far away from the earth?

Short answer

The speed of the rocket when it is very far away from the earth is$18.7km/s.$

Step-by-step solution

Given information

Initial speed of rocket =$15000m/s.$

Calculation

The speed of the rocket when it is very far away from the earth can be found using the law of conservation of energy.

$K{E}_1+P{E}_1=K{E}_2+P{E}_2.$

Where, $K{E}_{1}andP{E}_1$represents the energy of the rocket at the surface of Earth

$K{E}_2,P{E}_2$represents the energy of the rocket at a far away distance.

Continuation of calculation

Now we get,

$\frac{1}{2}\times m{v}_1^2-\frac{G{M}_{e}m}{r}=\frac{1}{2}\times m{v}_2^2-\frac{G{M}_{e}m}{r}.$

Where,

'G' = universal gravitational constant.

${\mathrm{M}}_{\mathrm{e}}=$mass of earth.

'm' = mass of an object.

'R' = radius of the earth.

Here, the initial potential energy is zero.

Continuation of calculation

Here, the mass of the rocket is unknown, to solve the speed we have to eliminate the mass of the rocket, by dividing the equation by mass of rocket 'm' .

$$\begin{gathered} \frac{1}{2}m{v}_1^2-\frac{G{M}_{e}m}{r}=\frac{1}{2}m{v}_2^2-\frac{G{M}_{e}m}{r} \\ \frac{1}{2}m{v}_1^2+0=\frac{1}{2}m{v}_2^2-\frac{G{M}_{e}m}{r} \\ {v}_1^2=v_2^2-2\frac{G{M}_e}{r} \\ {v}_2^2=v_1^2+2\frac{G{M}_e}{r} \\ {v}_2^2={(15000m/s)}^2+2\times \frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)\times (5.98\times {10}^{24}kg)}{(6.37\times {10}^{6}m)} \\ {v}_2=18.7\mathrm{km}/\mathrm{s}. \end{gathered}$$

Final answer

The speed of the rocket when it is very far away from the earth is$18.7km/s.$