Physics: Gravity vs Sound

Isaac Newton, upset with his failed attempt to discard the apple that woke him from his afternoon nap, decided that it was time to take stronger measures to destroy the apple once and for all.  Deciding that he would give the stupid apple a lesson in gravity, he took it to the top of a tall building, and dropped it, so that it would hit the ground at great speed and disintegrate into lots of tiny pieces of apple sauce.

However, to his horror, he realised at the exact moment that he dropped the apple that there was a poor unfortunate man standing directly underneat the apple, and shouted down at the man to move out of the way of the doomed fruit.

But would the man hear Isaac's shout in time?  Or would he suffer the same fate as our unfortunate father of modern physics?

The question is - which will reach the man first:  the apple, accelerating due to gravity, or the sound of Mr Newton's shout, travelling at the constant speed of sound?

Firstly, the apple.  The apple is accelerating at 9.81 ms^-2, getting faster all the time.  Yes, I'm ignoring terminal velocity and air resistance for now.

The formula to use for the apple is   s = ut + 1/2 a t².
s = distance travelled
u = initial velocity = 0
t = time since apple was dropped
a = acceleration due to gravity

since u = 0, (when the apple was dropped it had zero initial speed), we have a simplified formula:  s = 1/2 a t²
This tells us s (how far the apple has fallen) after so much time has passed (t).  We can rearrange it to tell us how long it will take the apple to fall from a building of height s.  This gives us:

√(2s/a) = t

The formula for the sound of Isaac Newton's shout is simpler:
v = s / t

s = distance
v = velocity = the speed of sound, 330 ms^-1
t = time since Isaac shouted

Rearranging gives us t = s / v

Now, suppose the building Isaac standing on was 300 m high, just short of the Chrysler Building's 319m.  No, the buildings in Isaac Newton's time weren't this tall, but let's just suppose he had a time machine and he made the journey.


For the apple: √ (2s/a) = t
Time taken is 7.82 seconds

And for the shout,

t = s / v = 300 m / 330 ms^-1 = 0.909 seconds


So, the shout will reach the ground (7.82 - 0.909 = 6.91) seconds before the apple does, giving the potential victim time to move out of the way.


From here, we can go on to work out how high a building would have to be for the apple and the sound to reach the ground at the same time. From any height above this, the apple would always land before the shout (because the apple will keep accelerating - we're still ignoring terminal velocity) and no amount of shouting would save a victim from being hit on the head by the apple.

In order to do the calculation, we simply set the time for the apple and the time for the shout to be equal.  This gives us:

t = √(2s/a) = s / v

√(2s/a) = s / v
2s/a = s² / v²

2 v² /a = s

We can now put in (substitute) the values for the speed of sound (330 ms^-1) and acceleration due to gravity (9.81 ms^-2), and calculate the height of the building (s).  


2 x 330² / 9.81 = 22,201 metres, whiich is very, very high indeed.  To put it another way:

It's just over 73,200 ft, which is 2.5 times higher than Mount Everest.
It's twice the height on airliner's cruising altitude.

Air resistance will be less of a problem to start with, but sound needs air to travel through, and the air is so thin at those heights that the sound won't travel as well or as fast.  I'm not even going to discuss the lack of air pressure; lack of breathable oxygen; the temperature (frozen apples and frostbite); terminal velocity and air resistance.

When, during an A-level class, my maths teacher posed this question, I don't think she was thinking of such things either.  That's the problem with maths without science - it can give you an answer that is meaningless and useless when you actually consider the real world!

Other Physics related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
Calculating (and explaining) escape velocity

Measuring g using a pendulum

Physics discussion: Escape Velocity

The story goes that Isaac Newton was sitting under an apple tree, when an apple fell on his head, and prompted him to wonder why it fell downwards, and not upwards or even sideways.  However, what history doesn’t tell us is that he probably got quite upset at having his afternoon nap interrupted by an apple, and, in his annoyance, threw the apple away as far as he could, declaring, “Stupid apples!”  He then wondered why the apple fell back to the earth, despite him throwing it away as hard as he could.

The same applies today (gravity hasn't changed much since then).  Consider throwing a tennis ball:  the harder you throw it, the higher it goes.  How about throwing it upwards, or even aiming for the moon (it’s not a million miles away, you know)? How fast does it have to be travelling, or how quickly do I have to throw it, so that it doesn’t come back down again?  We call this initial speed (how fast you have to throw it) the escape velocity.

Thinking in scientific terms, we can say that the apple (or the tennis ball) has escaped from the Earth’s gravitational pull, and will not fall back down to the earth.  It has maximum gravitational potential energy, and no kinetic energy (i.e. it stopped moving).  This happens at the edge of the gravitational field.

Since the kinetic energy at the start (i.e. from the throw) has all been converted into potential energy, we can say that the two are equal.

The potential energy is:

And the kinetic energy is:

where m₂ is the mass of the object being thrown, and m₁ is the mass of the Earth. 

 I’ll explain a bit more here about how this works, because at school I was taught that gravitational potential energy = mgh¸ where m is mass, g is acceleration due to gravity, and h is height – so that potential energy continues to increase with height.  So, when does PE = mgh stop being correct?  PE = mgh is not true when h becomes large, and g becomes very small.  The value of g changes with height; close to surface of the earth, mgh is an acceptable approximation, however at high altitude, g becomes

very much smaller.  It’s different at the top of a mountain than it is at sea level for instance.

So, the definition of potential energy is something else, it’s not mgh, it’s taken as something else.PE for all locations is equal to the formula given above.

Since we can equate these two energies, we have that:

Solving for this revised equation gives an expression for escape velocity, v, as:

Where m₂ is the mass of the Earth (in this case) and r is the Earth’s radius from centre to surface (i.e. from the centre of gravity to the point we’re launching from), since we have a bit of a head-start on gravity (we don't have to launch from the centre of the Earth).

Solving for all the numbers gives us an escape velocity of 11,181 metres per second, which is 34 times the speed of sound (Mach 34).  If you tried to throw an object at this speed, you'd probably either break your arm, or suffer friction burns from the air resistance as the air particles tried to move out of the way of your arm (and failed).  

It's also worth mentioning that I've not looked at air resistance, which at Mach 34 is considerable.  The sonic boom caused by the apple (or the tennis ball) would be extremely loud... in fact, I imagine the apple would turn into apple sauce, and the tennis ball would melt into a sticky, furry goo before it got anywhere near earth orbit.

I should explain at this point that escape velocity isn’t the speed that space rockets travel at when they take off.  This is really important.

An important point about escape velocity

Remember at the start that we were talking about throwing objects – where all the energy, and force is transferred to the object at the start of its flight.  With space rockets, the engines keep pushing the space rocket while it’s in flight, so they don’t have to travel as quickly, they just have to push upwards with a force that constantly exceeds their weight until they achieve an earth orbit.  This means that space shuttles, and space rockets, don't have to reach escape velocity.  Instead, they just have to keep pushing upwards with a force that is greater than their weight, until they reach an orbital height.

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?
Calculating (and explaining) escape velocity

Physics Experiment: Determine g with a pendulum

Having done some work on determining pi by mathematical methods, I'm now going to use it in conjunction with some experimental work to determine the value of g, which is acceleration due to gravity. Any reference book will tell you the value of g is approximately 9.81 ms-2, but I'm going to do an experiment to show what it is. It's not a difficult experiment, and it doesn't require any specialised scientific equipment. To give you an idea, I did this using a toddler fireguard for my vertical surface, a piece of sewing cotton for my pendulum, and in the absence of any respectable small mass, used a small pine cone tied to the end of it. I also used a standard stopwatch on a digital watch (it's accurate to 1/100th of a second, although I'm not).

The relationship between a pendulum and g is described in the following limerick:

If a pendulum's swinging quite free
Then it's always a marvel to me
That each tick plus each tock
Of the grandfather clock
Is 2 pi root L over g

In order to improve the accuracy of my results, I counted the time taken for ten complete 'swings' or periods, and quoted this. I also repeated each ten-swing measurement three times, so that I could take an average and identify any anomalies. And somehow, saying that, I feel like I'm writing up a GCSE science coursework piece!

Here are my results...

length (l, in metres)	10 swings (10T, seconds)	Run 2	Run 3
0.162	8.45	8.31	8.35
0.237	10.09	9.90	10.04
0.321	11.63	11.67	11.66
0.344	12.09	11.90	12.17
0.410	12.98	12.95	13.00
0.475	14.23	14.13	14.03

I calculated the average value of 10T, and hence T and then T², which I can use to determine g, with the following rearrangement:

An alternative, if I'd wanted to plot a graph of my data, is to determine g by finding the slope of the appropriate plot.  Using the following rearrangement, it's possible to plot T²  against l and have a slope of 4pi²/g

However, I'm going at it in number-crunching form, using the formula above.  My results for g are as follows:

length (l, in metres)	g in ms-2
0.162	9.136
0.237	9.338
0.321	9.332
0.344	9.3436
0.410	9.612
0.475	9.395

So, not perfect, but given the nature of the experiment - me with a fireguard and a pine cone - it's not too bad at all, and I feel quite pleased at having worked out something so massively significant with such basic equipment, and I feel it proves that science isn't just for big-budget departments!

Next time, determining the distance to the moon using the same principle as for geostationary satellites (except that this one is a bit bigger, a bit further away and not geostationary!).