Recent content by Siebevp

it is not about measuring g

something doesn't add up for me, and I will have to think how to formulate the question, which I'm puzzleing about

i know that it's not similar, to calculate g you ofcourse need the equation g=GM/r^2, but every time on one side of the equation there is just one variable of time. so we keep saying that to calculate g which is in a distance devided by time square, is equal to a constant called G what is in...

Doctor, I'm from holland and I'm 18 years old, I will be going to study physics in holland, or where ever. And do not start about articulating because english is not my first language, and for that I will never be as good in articulating *** someone who was born with this language and learned...

well you still did not answer the question for me... because you probably do not understand what i ment,, srry to have bothered you and your precious time and I hope you will get a nobel price for what you did to get a phd title

would you mind if i asked you... what degree do you have?

never mind...

you don't need accelaration, but you will get one because of gravity, otherwise you wouldn't be able to calculate the gravitational force in Newtons

the point is... to calculate the force of gravity on a object (which is standing on the surface of the earth) we say it will have a movement towards the Earth (the accelaration of 9.81 m/s^2), and a certain mass, multiplie these two givins and you have calculated the force of gravity that is...

thats how you calculate the amount of force with an accelaration you just said it now yourselve, but you do not have an accelaration, nor does the Earth have one towards you. because you don not move ( the distance from your core of your mass, to the core of the mass of earth) so you do not have...

i understand that ofcourse the netforce is zero, meaning that the gravitaional force is pulling me down with a certain amount in n (Newtons), and that there is another force that is pushing me up with the same amount of n (Newtons), so you have two different vectors, two different forces with a...

but then the force would be equal and opposite to each other, still showing me that there is no accelarations of the objects toward each other, let's say i weigh in about a 100 kilos, that the gravitational force is 981 Newtons, this force accounts for both of the objects. both of the objects...

i understand that ofcourse the netforce is zero, meaning that the gravitaional force is pulling me down with a certain amount in n (Newtons), and that there is another force that is pushing me up with the same amount of n (Newtons), so you have two different vectors, two different forces with a...

and btw time continues so that is not going to be zero, if we measure my "falling accelaration" the only thing that is going to be zero is my traveled distance...

you say that gravitational formula only accounts for falling objects equal and opposite to each other, right.. how do you know that f=m x a because you can also known in its other form as fz=m x g you can get that with putting these formulas as shwon before equal m x g = G x (m1 x...