I've been trying to come up with an equation to fit the path of a thrown object. I know many books give a very simple formula, but his just doesn't seem right to me, if anyone thinks that they know this equation and wants to tell me why it is true, then immediately post and do not read the rest of this.

Anyways, I have been able to reduce the equation for motion down to:

y=v₀cosqò1/(v₀sinq-((2G m/h)^1/2-(2G m/(h+x))^1/2))dx

where q is the angle of launch with respect to the x-axis, x and y start at 0, and where h is the launching position vector. How can we simplify this, namely simplify the integral part?
Thanks,

Brad

I think you're using Newtons Law of Universal Gravitation, given all the (big) Gs floating around. The simple formulae in text books just assume that the gravitational potential is a constant. i.e. The gravitational force doesn't vary with height. Sorry if I'm way off the mark.

From looking at your formula it seems that you have used -GmM/x as the gravitational potential energy. Although this is strictly correct, it isn't particularly appropriate for things like throwing objects on the Earth and we can take the potential energy to be mgh, where h is the height and g is the acceleration due to the Earth's gravitation field (different from the G above). This is an approximation that is always used and it gives you a nice answer that you can integrate (try it).

Here is where the approximation comes from:

potential energy(PE) = -mass(m) xconstant(k) x1/x

where x is the distance to the centre of the Earth. Write x = R + h, where R is the radius of the Earth and h is the distance of the object form the ground. So

PE = -mk/(R+h) = -mk/R x1/(1+h/R)

But R < < h (this is the approximation) and then h/R < < 1. (< < means much less than). And there is result that when s is small then 1/(1+s) is appoximately 1-s (this is from Taylor's theorem, check it for a few small values of s if you want). So

PE = -mk/R x(1-h/R)

now the first term is just a constant which we don't care about when dealing with energy. We regroup the other terms into a single constant calling it g to get

PE = mgh

If you use this result you will get the nice answers that are in the books.

Sean