# Escape from planet Earth

## Problem

A cannon ball is shot vertically upwards with speed v. How fast would the ball need to go such that it would never fall back to Earth?

How fast would you need to fire the cannon ball to escape from the moon, Jupiter or the surface of the sun? Before you try these parts of the calculation, do you think that these escape velocities will be lower or higher than that of Earth in each case? To get a feel for the relative sizes, you can look at this scale picture*.

Suppose that I go to planet Tiny, which is the same mass as Earth but a lot smaller. I fire a cannon ball again at 50% of the speed of light. How small would planet Tiny need to be so that the cannon ball cannot escape to infinity?

Supposing now that I go to planet Heavy, which is the same size as Earth but a lot more dense, and fire my cannon ball up at 50% of the speed of light. How many times more dense than Earth would planet Heavy need to be so that the cannon ball cannot escape its gravity?

Compare these results to trying to escape from a typical Neutron star which would weigh around twice the mass of the sun and have a radius of only 10km (this is around the same density as trying to squeeze the entire earth into a ball of 100m radius).

NOTES AND BACKGROUND

This problem introduces the concept of Escape Velocity . Whilst escape velocity is a real concept, attempting to reach such speeds in the atmosphere would cause immediate disintegration of almost any object. It was noted in the 1700s by Laplace that for a very massive body the escape velocity would approach the speed of light. This concept was developed in the 1920s following Einstein's discovery of general relativity. The black hole was discovered as a concept: an object so dense that even light beams would not be able to escape its gravitational pull. Over the last couple of decades evidence for the real existence of black holes in the galaxy has grown.

Name | Diameter relative to earth | Mass relative to earth |

Jupiter | 11.209 | 317.8 |

Moon | 0.273 | 0.0123 |

The sun | 109 | 332946 |

In this problem you will need to know that the gravitational potential energy of an object of mass m at a distance r from the centre of an object of mass M is given by PE = GMm/r where $G = 6.674\times 10^{-11}m^3kg^{-1}s^{-2}$ is Newton's gravitational constant. $M= 5.9763\times 10^{24}kg$ is the mass of the earth and the radius of the earth is 6378km

* Image adapted from http://commons.wikimedia.org/wiki/Image:Portrait_de_famille_%281_px_%3D_1000_km%29.jpg

## Getting Started

At what distance from the earth would an object have zero gravitational potential energy?

How fast would the balls need to be thrown so that they only just reach this point?

## Student Solutions

Steven from City of Sunderland College sent us in a wonderful, complete solution to this problem, which we recommend reading in full to any budding problem solver. The main points are as follows:

The gravitational potential energy of a cannon ball of mass $m$ at a distance $r$ from the centre of a planet of mass $M$ is

$$

V = -\frac{GMm}{r}\;.

$$

The kinetic energy of a cannon ball launched at speed $v$ is

$$

KE = \frac{1}{2}mv^2\;.

$$

Suppose that a cannon ball just escapes the pull of a planet and makes it to infinity. At this point, both its potential and kinetic energies will be zero. Thus, the initial kinetic and potential energies must sum to zero. So, if launched from a planet of radius $R$ we must have

$$

\frac{GMm}{R} = \frac{1}{2}mv^2\;.

$$

This gives the escape velocity $v$ as

$$

v =\sqrt{\frac{2GM}{R}}\;.

$$

Putting in the numbers for Earth gives

$$

v=\sqrt{\frac{2\times 6.674\times 10^{-11}\times 5.9763\times 10^{24}}{6.378\times 10^6}}=11.2\textrm{ km s}^{-1}\;.

$$

For the moon, Jupiter and the sun the escape velocity changes by the relative change in the factor $\sqrt{\frac{M}{R}}$. For the moon, Jupiter and the sun these are $0.2122$, $5.32$ and $55.26$, giving rise to escape velocities of

Moon: $2.37\textrm{ km s}^{-1}$,

Jupiter: $59.5\textrm{ km s}^{-1}$,

Sun: $619\textrm{ km s}^{-1}$.

## Teachers' Resources

### Why do this problem?

This problem provides an interesting application of conservation of kinetic and potential energy, and introduces students to ideas of non-constant gravity. The ideas raised, including the extrapolation to black holes, will be very interesting to most students and may inspire the students to look further into the ideas raised outside of the classroom.### Possible approach

Students may initially attempt to conserve energy using the approximation that gravity is a constant 9.8. They should be encouraged to see why this approach would be invalid. The activity best works as a hands-on task. One of the hurdles is accurate computation and students should be encouraged to provide numerical checks.### Key questions

- What assumptions will you need to make in this calculation?
- How significantly do you think that this will affect the result?