Big and small numbers in physics
Problem
Physics makes use of numbers both small and large. Try these questions involving big and small numbers. You might need to use pieces of physical data not given in the question. Sometimes these questions involve estimation, so there will be no definitive 'correct' answer; on other occasions an exact answer will be appropriate. Use your judgement as seems appropriate in each context.
- It is known that the value of $g$ on the moon is about one-sixth that on earth. How high do you think that you would be able to jump straight up on the surface of the moon?
- The mass of an atom of lead is $3.44\times 10^{-22}\textrm{ g}$. Lead has a density of $11.35\textrm{ g cm}^{-3}$. How many atoms of lead are found in a single cubic centimetre of lead?
- The earth orbits the sun on an almost circular path of average radius about $149\,598\,000\,000\textrm{ m}$. How fast is the earth moving relative to the sun?
- The tallest buildings in the world are over $800\textrm{ m}$ high. If I dropped a cricket ball off the top of one of these, estimate how fast it would be moving when it hit the ground.
- What weight of fuel would fit into a petrol tanker?
- The charge on a proton is $1.6\times 10^{-19}\textrm{ C}$. What is the total sum of the positive charges in a litre of hydrochloric acid of $\mathrm{pH}$ $1.0$?
- What is the mass of a molecule of water?
- How many molecules of water are there in an ice cube?
- Around $13.4$ billion years ago the universe became sufficiently cool that atoms formed and photons present at that time could propagate freely (this time was called the surface of last scattering). How far would one of these old photons have travelled by now?
- How much energy is contained in the matter forming the earth?
An obvious part of the skill with applying mathematics to physics is to know the fundamental formulae and constants relevant 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
Note that some of these questions involve making estimations and assumptions so these answers are not definitive!
1. Required extra data:
Typical height a person can raise their centre of mass through when jumping on the Earth $h \approx 0.5 \textrm{ m}$.
$\therefore$ under a gravitational field of one-sixth of the strength height reached $h \approx 3.0 \textrm{ m}$
2. Required extra data: None
$ \rho_{Pb} = 11.35\ \textrm{g cm}^{-3} \Rightarrow 1 \textrm{ cm}^3$ of lead has mass $m = 11.35\ \textrm{g}$
Mass of one lead atom $m_{A:Pb} = 3.44 \times 10^{-22}\ \textrm{g}$
$\therefore$ number of atoms in $1 \textrm{ cm}^3$
\[n = \frac{m}{m_{A:Pb}} = \frac{11.35}{3.44 \times 10^{-22}} = 3.30 \times 10^{22}\]
3. Required extra data:
$1$ year = $365.25$ days
$1$ year is one full rotation around Sun, therefore period of orbit in seconds is:
$$T = 365.25 \times 24 \times 60 \times 60 = 31\ 557\ 600\ \textrm{s}$$
The angular velocity of the Earth's orbit is therefore:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{31\ 557\ 600} = 1.99 \times 10^{-7}\ \textrm{rad s}^{-1}$$
The speed of the Earth relative to the Sun is therefore:
$$ v = \omega R = 1.99 \times 10^{-7} \times 149\ 598\ 000\ 000 = 29\ 800\ \textrm{m s}^{-1}$$
4. Required extra data:
Acceleration due to gravity $g = 9.81\ \textrm{m s}^{-2}$
Assuming constant acceleration (i.e. ignoring drag effects), $u = 0 \textrm{ m s}^{-1};\ s = 800\textrm{ m};\ a =g$
$$v^2 = u^2 + 2as \Rightarrow v^2 = 0 + 2(g)(800) \Rightarrow v = 40 \sqrt{g} = 125 \textrm{ m s}^{-1}$$
Note air-resistance is likely to cause the ball to tend towards a terminal velocity below this, typically around $90 \textrm{ m s}^{-1}$
5. Required extra data:
Density of fuel (assume petrol) $\rho_{fuel} \approx 700\ \textrm{kg m}^{-3}$
Typical diameter of fuel tank $D = 4\ \textrm{m}$
Typical length of fuel tank $\ell = 15\ \textrm{m}$
Volume of fuel tank
$$V = \pi \left(\frac{D}{2}\right)^2 \ell = \frac{\pi \times 4^2 \times 15}{4} = 60 \pi\ \textrm{m}^3$$
$\therefore$ weight of fuel in tank
$$W = V \rho_{fuel} = 60 \pi \times 700 = 132\ 000\ \textrm{kg}$$
This seems to be an overestimate as typical values for capacity of a petrol tanker are between 20000 and 40000 litres. What values might be more sensible for the diameter and length?
6. Required extra data:
Hydrochloric acid is a strong acid (complete dissociation) $\Rightarrow \textrm{pH} = -\log_{10} [\textrm{H}^+]$
Avogadro's constant $N_A = 6.022 \times 10^{23}\ \textrm{mol}^{-1}$
$$\textrm{pH} = 1.0 \Rightarrow [\textrm{H}^+] = 10^{-1.0} = 0.1 \textrm{ mol dm}^{-3}$$
$$\therefore 0.1 \textrm{ mol in } 1 \textrm{ dm}^3 (1\ \ell) \textrm{ which is } 0.1N_A = 6.022 \times 10^{22}\textrm{ hydrogen ions}$$
$$\therefore \textrm{ total sum of positive charges } = 6.022 \times 10^{22} \times 1.6 \times 10^{-19} = 9635 \textrm{ C}$$
7. Required extra data:
Relative molecular mass of water $m_{R} = 18.02\ \textrm{u}$
Avogadro's constant $N_A = 6.022 \times 10^{23}\ \textrm{mol}^{-1}$
1 mole of water has a mass $m = 18.02\ \textrm{g}$
There are $N_A$ particles in 1 mole of a substance $\therefore$ one molecule of water has mass
$$m = \frac{18.02}{6.022 \times 10^{23}} = 2.992 \times 10^{-23}\ \textrm{g}$$
8. Required extra data:
Answer to Q7. (mass of one molecule of water) $m = 2.992 \times 10^{-23}\ \textrm{g}$
Typical volume of an ice cube - assume dimensions $2\ \textrm{cm} \times 2\ \textrm{cm} \times 3\ \textrm{cm} \Rightarrow V = 12\ \textrm{cm}^3$
Density of ice $\rho_{ice} = 0.917\ \textrm{g cm}^{-3}$
Mass of ice cube
$$M = \rho_{ice}V = 0.917 \times 12 = 11.004\ \textrm{g}$$
$\therefore$ number of molecules in one ice cube is
$$n \approx \frac{11.004}{2.992 \times 10^{-23}} = 3.7 \times 10^{23}$$
9. Required extra data:
Speed of electromagnetic radiation (in free space) $c = 299\ 792\ 458\ \textrm{m s}^{-1}$
$$T = 13.4\ \textrm{billion years} = 13.4 \times 10^9 \times 365.25 \times 24 \times 60 \times 60 = 4.229 \times 10^{17}\ \textrm{s}$$
$$ \therefore d = cT = 299\ 792\ 458 \times 4.229 \times 10^{17} = 1.27 \times 10^{26}\ \textrm{m}$$
10. Required extra data:
Mass of Earth $M_E = 5.9742 \times 10^{24}\ \textrm{kg}$
Speed of electromagnetic radiation (in free space) $c = 299\ 792\ 458\ \textrm{m s}^{-1}$
Einstein's famous formula relating energy to mass is:
$$E = mc^2 \therefore E_{Earth} = M_E c^2 = 5.9742 \times 10^{24}\times 299\ 792\ 458^2 = 5.369 \times 10^{41}\ \textrm{J}$$