Weekly Challenge 13: Comet catastrope

7055 /www/nrich/html/content/id/7055/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/5682&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/6626&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/z/node/6792">Discuss and explore</a> </li> <li> <a href="http://www2.jpl.nasa.gov/sl9/">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7053&part=solution">Last week's solution</a> </li> </ul> <div> In July 1994 the Comet Shoemaker-Levy 9 struck the planet Jupiter at a speed of around 60km s$^{-1}$ . Although it broke up before impact, the core of the original comet was around 5 km diameter. Its estimated density was $0.3 - 0.7$ g cm$^{-3}$ . Imagine that such a comet had struck the earth. How much would it have changed the earth's velocity relative to the sun? (Note that the mass of the earth is about $6\times 10^{24}$ kg). </div> <div class="framework"> Did you know ... ? The mathematics of the orbits of comets and planets and cricket balls is identical and forms a large part of university applied mathematics courses.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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} 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$^{-1}$) </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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} 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$^{-1}$) </mdoxml> 2 3 0 0 0 1 1 Weekly Challenge 13: Comet catastrope A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Stage 5 Mechanics mapping document Collisions M2