# A rocket is launched straight up from the earth's surface at a speed of 1.80 \times 10^4 m/s....

## Question:

A rocket is launched straight up from the earth's surface at a speed of {eq}1.80 \times 10^4 {/eq} m/s. What is its speed when it is very far away from the earth?

## Gravitational Potential Energy:

Gravitation has effects on forces but it also has effects on energy as well. This is the so called gravitational potential energy. This energy depends on the mass and radius of the planet and is written as:

{eq}\displaystyle U = -\frac{GMm}{r} {/eq}

where:

- r is the distance between the center of the planet and the object
- {eq}\displaystyle G = 6.67\ \times\ 10^{-11}\ m^3/kgs^2 {/eq} is the universal gravitational constant

- m terms are the masses of the two objects

## Answer and Explanation:

Given:

- {eq}\displaystyle v_0 = 1.8\ \times\ 10^4\ m/s {/eq} is the initial speed of the rocket

We can use the conservation of energy to solve this problem. In general, the kinetic and potential energies can be related together as:

{eq}\displaystyle K_1 + U_1 = K_2 + U_2 {/eq}

Our kinetic energy here is a classical form while our potential is due to gravitation effects. So we can expand this equation as:

{eq}\displaystyle \frac{1}{2} mv_0^2 - \frac{GMm}{r} = \frac{1}{2} mv^2 - \frac{GMm}{r_\infty} {/eq}

since we want the speed of the probe at very far distances from the Earth, we set the distance to be infinity. And since the gravitational potential energy is inversely proportional to *r*, then the gravitational potential energy will thus become zero, leaving us with:

{eq}\displaystyle \frac{1}{2} mv_0^2 - \frac{GMm}{r} = \frac{1}{2} mv^2 {/eq}

and here we cancel all *m* terms:

{eq}\displaystyle \frac{1}{2} v_0^2 - \frac{GM}{r} = \frac{1}{2} v^2 {/eq}

we also isolate the speed *v*:

{eq}\displaystyle v^2 = v_0^2 - \frac{2GM}{r} {/eq}

and we take the square root:

{eq}\displaystyle v = \sqrt{v_0^2 - \frac{2GM}{r}} {/eq}

now here we substitute. We use the given speed, and the mass/radius of the Earth:

{eq}\displaystyle v = \sqrt{(1.8\ \times\ 10^4\ m/s)^2 - \frac{2(6.67\ \times\ 10^{-11}\ m^3/kgs^2)(5.972\ \times\ 10^{24}\ kg)}{6,371,000\ m}} {/eq}

we simplify:

{eq}\displaystyle v = \sqrt{3.24\ \times\ 10^8\ m^2/s^2 - 1.2505\ \times\ 10^8\ m^2/s^2} {/eq}

{eq}\displaystyle v = \sqrt{1.9895\ \times\ 10^8\ m^2/s^2} {/eq}

and by taking the square root, we thus obtain:

{eq}\displaystyle \boxed{v = 14,104.96\ m/s} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6