The equation "m*g=m*a_g-m*w^2*R" is only true for the case that the object is on the equator. For this case, what I need is to find the centripetal acceleration "w^2*R". When the object is not on the equator (e.g. the object is on the surface of the Earth at north latitude of thirty degrees)...
Hi, Astrum
Thanks for the reply.
Here a_g means the gravitaional acceleration. The corresponding gravitational force is F=G*m*M/R^2, where M is the mass of the earth, m is the mass of the object, G is the gravitational constant. Thus a_g=G*M/R^2
If you do not consider the rotation of the earth...
Hi, experts
I got a naive question about the relation between the free-fall acceleration vs gravitational acceleration.
When you consider the rotation of the Earth around the axis which runs through the north and south poles, the free-fall acceleration "g" of an object with mass m is different...
Thanks for the replies. I think I got the answer. Say the four vector of the incoming proton in the lab frame is p1=(E_lab,0,0, p_z), the four vector for the rest proton is p2=(M,0,0,0), then s should be equal to (E_lab+M)^2-(p_z)^2=2M^2+2E_lab*M. Thanks a lot.
Hi, experts
I got a very naive question.
I read the following sentence in a paper: "New results on the production of charged pions in p+p interactions are presented. The data come from a sample of 4.8 million inelastic events obtained with the NA49 detector at the CERN SPS at 158 GeV/c beam...