I am asking what might be the weight (not mass!) of the earth. My weight is the amount of force exerted on me by Earth (say 600N). So, weight of earth should be amount of force exerted on Earth by me (600N)? And it also seems to be changing from person to person. Where am I wrong?
[Physics] the ‘weight’ (not mass!) of the Earth
earthmassnewtonian-gravityweight
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If we look at the $x$- and $y$-components of the forces, we see that $$ \begin{align} W_x - N_x &= m\omega^2 R\cos\theta,\\ W_y - N_y &= 0. \end{align} $$ The normal force $\vec{N}$ is a radial force, so if we assume that the Earth is spherically symmetric, we indeed have $N_x = N\cos\theta$ and $N_y = N\sin\theta$, so that $N=N_R$. But the weight force $\vec{W}$ is not purely radial. It also has a tangential component, precisely because of the rotation of the Earth: $$ \begin{align} W_R &= W_x\cos\theta + W_y\sin\theta,\\ W_\theta &= -W_x\sin\theta + W_y\cos\theta. \end{align} $$ So in general $W_x\neq W\cos\theta$. For the normal force, we have $$ \begin{align} N_R &= N_x\cos\theta + N_y\sin\theta = N\cos^2\theta + N\sin^2\theta = N,\\ N_\theta &= -N_x\sin\theta + N_y\cos\theta = 0. \end{align} $$ From $$ \begin{align} W_x\cos\theta - N_x\cos\theta &= m\omega^2 R\cos^2\theta,\\ W_y\sin\theta - N_y\sin\theta &= 0, \end{align} $$ we find the centripetal force $$ W_R - N = m\omega^2 R\cos^2\theta, $$ but note that there's also a tangential force $$ W_\theta = -m\omega^2 R\sin\theta\cos\theta. $$
As you rightly pointed out, the fact that 67P is oddly-shaped should alter its gravitational attraction on various parts of the comet.
That said, if we were to go by Wikipedia in a rather off-hand manner, we find that the lander is $100 kg$ (as @fibonatic rightly pointed out) and 67P has an acceleration due to gravity of $\textbf{g'} = 10^{-3} m/s^2$. Its weight $W$ would therefore be simply a calculation of $W = m\,\textbf{g'}$, giving us $W = \frac{10^2}{10^3} kg = 0.1 kg$ or $100 \,\,\verb+earth+\, g$.
[P.S. I will update this answer with better sources than Wikipedia as soon as I find time.]
Edit 1: This ESA webpage seems to confirm the figures.
Edit 2: Calculating $\textbf{g'}$
I made some calculations: Using the formula $\textbf{F} = M\,\textbf{g'} = \frac{GmM}{r^2} \Rightarrow \textbf{g'} = \frac{GM}{r^2}$ we can calculate the acceleration due to gravity on 67P (m being the comet's mass and M that of our lander). The above ESA page gives us this figure:
Seeing how the dimensions vary wildly, I decided to consider a mean of, say, 3.5km as diameter and 1.75km as $r$. 67P's mass is, of course, $10^{13}\,kg$ which gives us, $$ \textbf{g'} = \frac{6.67 \times 10^{-11} \times 10^{13}}{\left( 1.75 \times 10^3 \right)^2} \approx 10^{-3} ms^{-2} $$ A more precise answer, is, of course, $0.217 \times 10^{-3} ms^{-2}$ but since we have been very liberal in our assumptions of mass and radius, I think we ought to simply consider the order of magnitude, $10^{-3} ms^{-2}$. This pdf file contains some simulation data that agrees with our result.
Best Answer
It seems you are trying to apply weight inconsistently.
We generally define weight as "force on an object due to gravity."
You cannot really define weight of an individual object. When defining weight we usually are implicitly talking about the objects weight on Earth. We can also find the weight of objects on the moon for example. Objects with the same mass will obviously have different weight in different gravity.
This is because gravity is an attractive force between two objects. The question of "What is the weight of Earth?" has no obvious answer, because generally when we don't specify where, we mean on Earth. The question would then read "What is the weight of Earth on Earth?", which is obviously nonsense.
In your example, you could ask the question "What is the Earth's weight on me?", and then it would be 600 N. The weight of the Earth for someone heavier would be more.
It's not really standard to talk about the Earth's weight on small objects either, which is why it may seem odd to consider that.