Weekly Challenge 13: Comet Catastrope
This problem involves momentum which we calculate through the
formula $p=mv$. We need to use standard units (or, at least, the
same units) throughout. We will keep a sensible number of places
for accurary.
Volume of comet
$V = \frac{4}{3}\pi \left(\frac{5000}{2}\right)^3=
6.54\times 10^{10} \mbox{m}^3$
The density $\rho$ of the comet in kg m$^{-3}$ is
$300 < \rho < 700$
Using $m = \rho V$ we can therefore estimate the mass $m$ of the
comet, in kg, to be bounded as
$1.96\times 10^{13}< m< 4.58\times 10^{13}$
The momentum of the comet, measured in kg m s$^{-1}$, can now
be estimated as
$1.18 \times 10^{18}< p < 2.75 \times 10^{18}$
If we assume that all of this momentum is transferred to the earth
on impact then the velocity change $\Delta V$, measured in m
s$^{-1}$, that this would impart to the earth is found by
$\Delta v = p/M$ where $M$ is the mass of the earth.
$1.97 \times 10^{-7} < \Delta v < 4.60 \times
10^{-7}$
Taking the centre of the range we can estimate a velocity change
for the earth of about $2.3\times 10^{-7}$ ms$^{-1}$.
In every-day units this is about $8.4\times 10^{-7}$ km
h$^{-1}$, which is tiny. (Note: the earth's velocity around the sun
is around $30$ km s$^{-1}$)