# Big and small numbers in The Physical World

## Problem

Think about how best to approximate these things from the physical world around us. You will need to make some estimations and find information from friends or other sources, as would any scientist! Take care to represent all of your answers using a sensible number of decimal places and be sure to note all of your assumptions clearly.

- Light travels at $c=3\times 10^8$ metres per second. How fast in this in miles per hour? How many times faster is this than a sports car?
- The Milky Way is a spiral galaxy with diamater about 100,000 light years and thickness about 1000 light years. There are estimated to be between 100 billion and 400 billion stars in the galaxy. Estimate the average distance between these stars.
- Density of lead $11.34$g/cm$^3$. How big would a tonne of lead be?
- Estimate the mass of ore it takes to produce a roll of aluminum kitchen foil.
- How many AA batteries contain enough charge between them to run a laptop for an hour?
- Estimate how many atoms there are in a staple.
- Einstein's equation tells us that the enegy $E$ stored in matter equals $mc^2$, where $m$ is the mass and $c$ is the speed of light. How much energy is contained in the staple from question 6? How long could this energy run your laptop for?
- How much energy would it take to raise the air temperature of the room you are in by 1$^\circ$C? How much gas must be burned to produce this much energy? What is the cost of that much gas?

An obvious part of the skill with applying mathematics to physics is to know the fundamental formulae and constants relevent to a problem. By not providing these pieces of information directly, you need to engage at a deeper level with the problems. You might not necessarily know all of the required formulae, but working out which parts you can and cannot do is all part of the problem solving process!

## Getting Started

Note that most of the ideas used here are typically covered at school before the age of 16, although possibly in mathematics, physics or chemistry.

In estimation questions don't be afraid to have a go with a guess at some numbers in the problem and then to refine your estimate after checking it makes some sort of sense.

Although there is no 'right' answer to an estimation, there are good or bad estimates and sensible or over detailed calculations.

Think how you might make your estimation a good one, and think how it makes sense to ignore certain complexities in particular calculations.

## Student Solutions

1. Let's change miles per hour to meters per second: 1 mile = 1.61 km and 1 hour = 3600 seconds. Thus $c=3\times 10^8$m/s = $3\times 10^8$$\times 3600$/$1610$ mph =$6.7\times 10^8$mph. The fastest car in 2010 was the Bugatti Veyron with max speed 250 mph. The speed of light is $6.7\times 10^8$ / $250$ = $2.7\times 10^6$ times bigger than the speed of a sport car.

2. We are given that a diameter of the galaxy is $D = 100,000$ ly and a thickness about $d = 1000 ly$. We can approximate our galaxy as a cylinder then calculate a volume of the galaxy. $V = d\times \pi (D/2)^2$. An average number of starts in the galaxy is $N = 250$ billion stars. Now, we can find out a density of the stars: $N/V$ = $(250\times 10^9) : (1000\times \pi (100,000/2)^2)$ = $0.032 stars/ly^3$. Thus, one star occupies the volume which is equal $1/0.032$ $ly^3$=$31$$ly^3$. In order to find the average distance between stars we take a cube root of 31 to get that the distance is about 3.2 light years.

3. Mass of lead $m = 1 t = 1000 kg = 1\times 10^6 g$. Density of lead $11.34$g/cm$^3$. Thus, the volume of 1 tonne of lead is $V =(1\times 10^6) /(11.34)$ cm$^3$ = 88180 cm$^3$.

We can imagine this volume as a cube with side length 44.5 cm or as a ball with diameter $55.2 cm$, for a comparison the diameter of football ball is 14-16 cm.

4. Firstly, lets find a mass of a roll of aluminum kitchen foil. A thickness of foil is about 20 micrometes, lenght is 10 meters and width is about 370 milimetres. Hence, the volume of a roll $V = 20\times 10^{-6} \times 10\times 370\times 10^{-3} m^3= 7.4\times 10^{-5} m^3$. The density of aluminum is $7874 kg/m^3$. The mass of this roll is $7.4\times 10^-5\times 7874 = 0.58 kg$. It takes around 2 kg of bauxite (aluminum ore) to make 1 kg of pure aluminum metal. Thus, we need about 1.2 kg of aluminum ore.

5. One AA zinc-carbon battery has a capacity of 1100 mAh. For example one AA Alkaline battery has much bigger capacity of 2700 mAh. A power being used by a laptop is about 60 watts and AC adapter changes voltage from 240 volts to average 20 volts. The formula which relates power with the voltage and the current is $P = I\times U$ where P is power, I the current and U is the voltage. The electric current is a flow of electric charge. Thus, the charge required is $60/20\times 1 = 3$Ah. We can conclude that we need about three AA zinc-carbon batteries or one good Alkaline battery to run a laptop for one hour.

6. A mass of one staple is about 30 mg. A staple is made of stainless steel which consists mostly of iron. The molar mass of iron is $M = 55.85 g/mol$ and** ** Avogadro's number is $6.02\times 10^23$ 1/mol. Hence, the number of atoms in a staple is $N = 6.02\times 10^{23}\times 0.03/55.85 = 3.2\times 10^{20}$

7. Energy stored in a staple is $E = mc^2$ where mass of a stample is $m = 30$ mg and the speed of light is $c = 3\times 10^8$ m/s. So, $E = 2.7\times 10^{15}$ J. Power is the rate at which energy is transferred. If we assume that a laptop use 60 W power then it could run for $2.7\times 10^{15} : 60$ s= $4.5\times 10^{13}$ s = 1.4 million years.

8. Suppose a volume of room (class) is $10\times 6\times 3 m^3= 180 m^3$. The specific heat capacity of air when a pressure is constant $c = 1012$ $J/(kg\times K)$. The energy required to raise the temperature by T = 1$^\circ$C could be calculated by $E = mcT$ where m is the mass of air in the room. A density of air is $1.2 kg/m^3$). Thus, $E = 1.2\times 180\times 1012\times 1 = 2.2\times 10^5$ J. The heat of combustion is the energy released when a compound undergoes complete combustion. For the natural gas it is $45 MJ/kg$. The mass of natural gas required to raise the temperature by T = 1$^\circ$C is $2.2\times 10^5 : 45\times 10^6$ = 5 grams. The density of natural gas is about $0.8 kg/m^3$. Thus, we need $V = 6.25\times 10^{-3} m^3$ of natural gas. Most heating isn't perfect however, and supposing it is 50% efficient we actually need twice as much gas! That's $1.25\times10^{-2} m^3$. In Britain gas is billed according to energy, not by volume, and is priced per kilowatt-hour ($kWh$). A typical price could be £$0.038$ per $kWh$. On average we expect gas to have a calorific value of $40 MJ/m^3$, which means that one metre cubed of gas produces $40 MJ$ of energy. Thus our $1.25\times10^{-2} m^3$ of gas holds roughly $0.5 MJ$ of energy. Convert this to $kWh$ using the factor $1 kWh = 3.6 MJ$ to get that we need $0.139 kWh$ of gas. Using the price above we find that this costs merely half a penny!