To find our weight on 67P/C-G we need to know the gravitational field strength of the comet. Given that the comet is not really close to spherical it does make the problem considerably more difficult as one would have to model the gravitational field profile for the comets double lobe shape, which has been done
here. To make this a more digestible problem we will assume the comet has spherical symmetry and treat it has having a point mass so we can easily use Newton's Law of Gravitation. So to start with, to find the gravitational field strength of an object on the comet, we can equate the forces of weight with the force between two point masses (Newton's Law of Gravitation). The equation for weight is \[F=mg_{67P}\] where \(F\) is the force of the object's weight on the surface of the comet, \(m\) is the object's mass and \(g_{67P}\) is the gravitational field strength of the comet. The equation for the force between the object's mass and the comet's mass is \[F=\frac{Gmm_{67P}}{r^2}\] where \(G\) is the gravitational constant, \(m\) is the object's mass, \(m_{67P}\) is the comet's mass and \(r\) is the radius of the comet. Equating these two equations gives \[mg_{67P}=\frac{Gmm_{67P}}{r^2}\]The object's mass \(m\) is common on both sides and so cancels giving \[g_{67P}=\frac{Gm_{67P}}{r^2}\]We now have a formula for the gravitational field strength of comet 67P/C-G. Notice that this does not depend on the mass of whatever object might come into the influence of the comet's gravitational field. It is only dependant on the Gravitational constant (which itself is constant), the mass of the comet (which in this case is also constant) and the radius (which is how far we are from the centre of the sphere when standing on the surface). To find a value for \(g_{67P}\) we need values for all three. The Gravitational constant is \(G=6.67×10−11\rm Nm^{2}kg^{−2}\) and the comet's mass is quoted as approximately \(m_{67P}=9.98\times 10^{12}\rm kg\). The value for the radius takes a little more work as we have essentially imagined the comet being melted down and reformed as a perfect sphere. To find the radius we can work backwards from the volume of a sphere \[V=\frac {4}{3}πr^3\]and rearrange for the radius \[r=\sqrt[3]{\frac{3V}{4\pi }}\]Now we just need a volume estimate of our reformed comet, which comes in at \(V=1.87\times 10^{10}\rm m^3\). In turn this gives us a rough value for the radius of \[r=\sqrt[3]{\frac{3\times 1.87\times 10^{10}}{4\pi }}=1646\rm m\]We now have what we need to find the gravitational constant of \(g_{67P}\). Finally putting this all together using all of our three values, we get \[g_{67P}=\frac {6.67\times 10^{−11}\times 9.98\times 10^{12}}{\left (1646 \right )^{2}}=2.46\times 10^{−4}\rm ms^{−2}\]What does this tell us? Well, by comparing with the gravitational field strength on Earth \(g_{\bigoplus }=9.81\rm ms^{−2}\) we can see that comet 67P/C-G has a gravitational field strength that is smaller by a factor \[\gamma =\frac {9.81}{2.46×10−4}=39878\]So, as for my weight on the comet, i'd be about 40000 times lighter. For a \(70\rm kg\) man, like myself, that would be equivalent to \[\frac {70}{39878}=1.76\times 10^{−3}\rm kg=1.76g\]In short, really not a whole lot. To compare, a 5 pence coin on Earth weighs more, at \(3.25\rm g\). So my weight would be a little more than half of a 5 pence coin. Theoretically i could stand on the comet but the tiniest movement of the muscles in my legs, which are easily capable of exerting forces tens of thousands of times that of the weight of a 5 pence coin, will cause me to leave the comet's surface. Walking would be an incredibly delicate process.
