Recent content by Phys_Boi

I'm sorry, I still don't understand

I understand now... However, I don't know how to make it a function of x. And is this right? $$\frac{-v^{2}}{2}$$

Velocity is a function of x. As the object starts off (with a large x value) the velocity is very small. However, as the object gets closer to Earth, the velocity is very great... The object is traveling to the left, toward the Earth.

I added the negative just because the drawing that I drew had the object moving left. I'm trying to find the total time it takes the object to move from (x = R+D) to (x=R). I realized the mistake and the new equations are: $$-GM\int_{R+D}^{R} \frac{1}{x^{2}} dx = \frac{v^{2}}{2}$$ $$v =...

At time = 0, r = R + D. I was going to use conservation of energy, but hesitated due to the fact that you use "g", an approximation. In the first integral, would you use "R" as the top bound instead of "R + D"?

a = GM/r^2 If you treat Earth at the x value of 0 then r is a function of x..? r(x) = x? I do understand D>0 causing a problem, but I don't know where I went wrong.

Homework Statement Find the total time, t, that an object takes to reach the surface of the Earth from a distance, D, using the Law of Gravitation: $$F_{g} = \frac{GMm}{x^2}$$ R is radius of Earth D is distance from surface R+D is total distance from center of masses ****** One Dimension...

So how do you integrate over a time interval? That is to say, how do you find the velocity over the interval [0, t]?

So is the following correct? $$v dv = \frac{-MG}{x^2} dx$$

But doesn't the sign just determine direction?

So what does that equation mean?

Thanks for this.. how did you get this though? I'm only half way through calc one..

Positive

Yes