# Homework Help: How fast to overcome gravity?

1. Aug 4, 2010

### zephyrous

1. The problem statement, all variables and given/known data
A question I made up myself:
With what velocity would an object need to be launched at to totally overcome gravity of the earth (to have a constant velocity out in space somewhere.) Neglecting other planets, atmosphere friction, etc.

2. Relevant equations
Newton's Second Law
Newton's Universal law of gravitation

3. The attempt at a solution
I figured mass would be negligable if I combined N2L and NGrav and said:

F=ma=G*m*(earth's mass)*(r^-2)

So acceleration is -G(mass earth)(r(t)^-2)

But the thing is that I want to launch this baby so fast that "R" is a function of time. So when I integrate over a time differential I run into problems, because I want to integrate from t=0 to t="t" so what pops out for velocity is g(mass earth)/(r(t))- g(mass earth)/(earth's radius) plus initial velocity if you will.

I don't even know if this is correct, and I know if I take the next step to get to position I will get into natural logarithms and I'd really not trust myself then. I guess whenever I get to my position equation then that becomes my "r(t)" and I just have myself a differential equation of some sort.

I'm sure this problem has been done before and I'm almost certain it takes diffeq even at the most simple level. Does anyone know of a basic explanation of how to calculate this that only uses relatively simple undergraduate calculus and physics? I know it's connected to the topics of trajectory motion but I'm guessing the solution is more complicated than the typical "throw the ball in the air how high will it go?" brand of problems.

Thanks

2. Aug 4, 2010

### diazona

Are you familiar with potential energy? The problem is trivially easy to do if you use energy conservation - no calculus required. For inspiration, take a look at the Wikipedia article on escape velocity.

Alternatively, you can do it by solving the differential equation you obtained. It's a little more complicated that way, but it's an interesting exercise (and it's necessary if you want to know how much time it takes the object to reach a certain height).

3. Aug 4, 2010

### zephyrous

I figured it out!!
I guess it's a concept called escape velocity and it can be derived a few different ways.
Very Cool!

4. Aug 4, 2010

### zephyrous

Yep! very cool