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Is Intergalactic space travel possible?

The Andromeda Galaxy is around 2.5 million light years away – a distance so large that even with the speed of light at traveling as 300,000,000m/s it has taken 2.5 million years for that light to arrive.  The question is, would it ever be possible for a journey to the Andromeda Galaxy to be completed in a human lifetime?  With the speed of light a universal speed limit, it would be reasonable to argue that no journey greater than around 100 light years would be possible in the lifespan of a human – but this remarkably is not the case.  We’re going to show that a journey to Andromeda would indeed be possible within a human lifespan.   All that’s needed (!) is a rocket which is able to achieve constant acceleration and we can arrive with plenty of time to spare.

Time dilation

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To understand how this is possible, we need to understand that as the speed of the journey increases, then time dilation becomes an important factor.  The faster the rocket is traveling the greater the discrepancy between the internal body clock of the astronaut on the rocket and an observer on Earth.  Let’s see how that works in practice by using the above equation.

Here we have

t(T): The time elapsed from the perspective of an observer on Earth

T: The time elapsed from the perspective of an astronaut on the rocket

c: The speed of light approx 300,000,000 m/s

a: The constant acceleration we assume for our rocket.  For this example we will take a = 9.81 m/s2 which is the same as the gravity experienced on Earth. This would be the most natural for a human environment.  The acceleration is measured relative to an inert observer.

Sinh(x): This is the hyperbolic sine function which can be defined as:

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We should note that all our units are in meters, seconds and m/s2 therefore when the astronaut experiences 1 year passing on this rocket, we first need to convert this to seconds:  1 year = 60 x 60 x 24 x 365 = 31,536,000 seconds.  Therefore T = 31,536,000 and:

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which would give us the time experienced on Earth in seconds, therefore by dividing by (60 x 60 x 24 x 365) we can arrive at the time experienced on Earth in years:

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Using either Desmos or Wolfram Alpha this gives an answer  of 1.187.  This means that 1 year experienced on the rocket is experienced as 1.19 years on Earth.  Now we have our formula we can easily calculate other values.  Two years is:

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which gives an answer of 3.755 years.  So 2 years on the rocket is experienced as 3.76 years on Earth.  As we carry on with the calculations, and as we see the full effects of time dilation we get some startling results:

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After 10 years on the space craft, civilization on Earth has advanced (or not) 15,000 years.  After 20 years on the rocket, 445,000,000 years have passed on Earth and after 30 years 13,500,000,000,000 years, which around 1000 times greater than the age of the Universe post Big Bang.  So, as we can see, time is no longer a great concern.

Distance travelled

Next let’s look at how far we can reach from Earth.  This is given by the following equation:

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Here we have

x(T): The distance travelled from Earth

T, c and a as before.

Cosh(x): This is the hyperbolic cosine function which can be defined as:

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Again we note that we are measuring in meters and seconds.  Therefore to find the distance travelled in one year we convert 1 year to seconds as before:

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Next we note that this will give an answer in meters, so we can convert to light years by dividing by 9.461×1015

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Again using Wolfram Alpha or Desmos we find that after one year the spacecraft will be 0.563 light years from Earth.  After two years we have:

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which gives us 2.91 light years from Earth.  Calculating the next values gives us the following table:

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We can see that as our spacecraft approaches the speed of light, we will travel the expected number of light years as measured by an observer on Earth.

So we can see that we would easily reach the Andromeda Galaxy within 20 years on a spacecraft and could have spanned the size of the observable universe within 30 years.   Now, all we need is to build a spaceship capable of constant acceleration, and resolve how a human body could cope with such forces and we’re there!

How likely is this?

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Well, the technology needed to build a spacecraft capable of constant acceleration to get close to light speed is not yet available – but there are lots of interesting ideas about how these could be designed in theory.  One of the most popular ideas is to make a “solar sail” – which would collect light from the Sun (or any future nearby stars) to propel it along on its journey.  Another alternative would be a laser sail – which rather than relying on the Sun, would receive pin-point laser beams from the Earth.

Equally we are a long way from being able to send humans – much more likely is that the future of spaceflight will be carried out by machines.  Scientists have suggested that if the spacecraft payload was around 1 gram (say either a miniaturized robot or digital data depending on the mission’s aim), a solar sail or laser sail could be feasibly built which would be sufficient to achieve 25% the speed of light.

NASA have begun launching continuous acceleration spacecraft powered by the Sun.  In 2018 they launched the  Near-Earth Asteroid Scout.  This will unfurl a solar sail and be propelled to a speed of 28,600 m/s.  Whilst this is a long way from near-light speeds, it is a proof of concept and does show one potential way that interstellar travel could be achieved.

You can read more about the current scientific advances on solar sails here, and some more on the mathematics of space travel here.

IB Revision

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If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

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The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

Bullet Projectile Motion Experiment

This is a classic physics experiment which counter to our intuition.  We have  a situation where 1 ball is dropped from a point, and another ball is thrown horizontally from that same point.  The question is which ball will hit the ground first?

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(diagram from School for Champions site)

Looking at the diagram above you might argue that the ball that is dropped falls to the floor quicker as it has a shorter path.  Or, you might think that the ball thrown sideways would travel faster to the ground because of its initial horizontal velocity.  Both of these views are wrong however – as both balls will land at exactly the same time.  To understand why, let’s look at the 2 situations in turn.

The ball launched sideways

To show that both balls would hit the ground at the same time we need to split the motion into its x and y components.  We have

x = v t \cos \theta 
y = vt \sin \theta - \frac{1}{2} g t^2

Where the angle theta is the angle of launch, v is the initial velocity, g is the gravitational constant 9.8 m/s.  If we have a launch from the horizontal direction, then this angle is 0, which gives the simplified equations:

x = vt

y = 0.5gt2

if we relabel y as the vertical distance (d), then we have:

\ t =\ \sqrt {\frac{2d}{g}}

which is the time taken (ignoring air resistance etc) for an object launched horizontally to fall a distance d, where g is the gravitational constant 9.8 m/s.

So if we have a ball launched at a speed of 1 m/s from a height of 1m, it would hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

So we can use this value of t to see how far in the x direction it has travelled:

x = vt

x = 1(0.45)

x = 0.45m.

The ball dropped vertically

We still start with:

x = v t \cos \theta 
y = vt \sin \theta - \frac{1}{2} g t^2

But this time we have no initial velocity as so we simply get:

x = 0

y = 0.5gt2

or as before, if we relabel y as the vertical distance (d), then we have:

\ t =\ \sqrt {\frac{2d}{g}}

So with a ball dropped from a height of 1m, it would also hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

But this time the distance in the x direction will of course be 0.

Showing this graphically 

We can also show this graphically using the tracker software.  This allows you to track the motion of objects in videos.  So using the video above we can set the axis, and the height of the table

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and then the motion capture software actually plots the parabola of the ball’s motion.

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This first graph shows the change in the y direction with respect to time for the ball launched horizontally.  We have large steps because the video was in super slow motion, so there were frames of very little movement.  Nevertheless we can clearly see the general parabola, with equation:

y = -0.43x2 -1.2x + 107

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The second graph shows the change in y direction with respect to time for the ball dropped vertically down.  As before we have a clear parabola, with equation:

y = -0.43x2 -1.2x + 106

Which is a remarkably close fit.  So, there we go, we have shown that the vertical motion of our 2 objects are independent of their horizontal motion.

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

This is a classic puzzle which is discussed in some more detail by the excellent Wired article.  The puzzle is best represented by the picture below.  We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch.  However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired.  The question is therefore, at what angle should the hunter aim so that he still hits the monkey?

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(picture from the Wired article – originally from a UCLA physics textbook)

The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls.  You can see a video of this experiment at the top of the page.

You can use tracking software (such as the free software tracker ) to show this working graphically.  Tracker provides a video demo with the falling monkey experiment:

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As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.

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We can then generate a graph to represent this data.  The red line is the height of the bullet with respect to time.  The faint blue line (with yellow dots) is the height of the monkey with respect to time.  We can see clearly that the red line can be modeled as a quadratic.  The blue line should in theory also be a quadratic (see below):

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but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink.  Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.

We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t2+39t +0.1

So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t2+39t +0.1

which will be around 0.45 seconds after the gun is fired.

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(A more humane version, also from Wired – where we can throw the monkey a banana)

Newtonian Mathematics

The next question is can we prove this using some algebra?  Of course!  The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa).  So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls.  So they still collide.  Mathematically we have:

The vertical height of the bullet given by:

y = V0t – 0.5gt2

Where V0 is the initial vertical speed, t is the time, g is the gravitational force (9.8)

The vertical height of the monkey is given by:

y = h – 0.5gt2

where h is the initial vertical height of the monkey.

Therefore these will intersect when:

V0t – 0.5gt2 = h – 0.5gt2
V0t = h
V0/h = t

And for any given non-zero value of V0 we will have a t value – which represents the time of collision.

Well done – you have successfully shot the monkey!

If you like this you might also like:

Throwing cannonballs off the Leaning tower of Pisa – why weight doesn’t affect falling velocity

War Maths – how cannon operators used projectile motion to win wars

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

Galileo:  Throwing cannonballs off The Leaning Tower of Pisa 

This post is inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

Galileo Galilei was an Italian mathematician and astronomer who (reputedly) conducted experiments from the top of the Tower of Pisa.  He dropped various objects from in order to measure how long it took for them to reach the bottom, coming to the remarkable conclusion that the objects’ weight did not affect the speed at which it fell.  But does that really mean that a feather and a cannonball would fall at the same speed?  Well, yes – as long as they were dropped in a vacuum.  Let’s have a look at how we can prove that.

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Newton’s Laws:

For an object falling through the air we have:

psgV – pagV – FD = psVa

ps = The density of the falling object
pa = The density of the air it’s falling in
FD = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

To understand where this equation comes from we note that Newton second law (Force = mass x acceleration) is

F = ma

The LHS of our equation (psgV – pagV – FD) represents the forces acting on the object and the RHS (psVa) represents mass x acceleration.

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. pa will be many magnitudes smaller than than ps – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

psgV – FD ≈ psVa

Next, we can note that in a vacuum FD (the drag force) will be 0 – as there is no air resistance.  Therefore this can also be ignored to get:

psgV ≈ psVa

g ≈ a

 But we have a = dU/dt where U = velocity, therefore,

g ≈ a
g ≈ dU/dt
g dt ≈ dU

and integrating both sides will give:

gt ≈ U

 Therefore the velocity (U) of the falling object in a vacuum is only dependent on time and the gravitational force.  In other words it is independent of the object’s mass.  Amazing!

This might be difficult to believe – as it is quite unintuitive.  So if you’re not convinced you can watch the video below in which Brian Cox tests this out in the world’s largest vacuum chamber.

If you liked this post you might also like:

War maths – how modeling projectiles plays an essential part in waging wars.

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

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