Vertical throw in earth's field of gravity

[covers]

Homework Statement

A mass m is thrown vertically from the surface of the Earth with a velocity v0. Find a function that describes the velocity v of the mass m in dependence of its distance z from the center of the earth.

Homework Equations

m*z'' = -G* m*M / z^2
z'' = dv / dt

G = gravitational constant
M = mass of earth


3. the solution and my problem with it

z'' = -G * M * 1/z^2
z'' = dv/dt = v * dv/dz

==> v * dv = -G * M * dz/z^2

Now my Problem: In the next part both sides of the above equation are to be integrated. The left side from v0 to v, and the right side from R to z. Like that:

integral(v0..v, v*dv) = -G * M * integral(R..z, dz/z^2)

Unfortunately, I don't understand that. Why can I inegrate both sides of the equation like that, and why is it still the same afterwards? What exactly does the term on the left 'integral(v0..v, v*dv)' and the term on the right '-G * M * integral(R..z, dz/z^2)' side of the above equation mean?

It would be great if someone could provide a little help!

 

[Hootenanny]

What you have done is known as separation of variables. The solution has missed out a step which may seem confusing if you have not met differential equations before. This is actually what is happening;

We have;

\left. \begin{array}{r}<br /> z&#039;&#039; = - GMz^{-2}\\<br /> z&#039;&#039; = v\cdot dv/dz<br /> \end{array}\right\}\Rightarrow -GM\frac{1}{z^2} = v\frac{dv}{dz}

Now, what we do here (and what the solution excludes) is integrate both sides with respect to dz i.e;

\int -GM\frac{1}{z^2}\cdot dz = \int v\frac{dv}{dz}\cdot dz

Notice that the two differentials dz on the RHS cancel leaving;

-GM \int \frac{dz}{z^2} = \int v\; dv

Does that make more sense?

 

[covers]

Notice that the two differentials dz on the RHS cancel leaving;

-GM \int \frac{dz}{z^2} = \int v\; dv

Does that make more sense?

Yes, it does make more sense, thanks! But there's still something I don't fully understand. If you cancel out the dz's on both side of the equation, you are left with two integrals, one with respect to v, and the other with respect to z. How do I know that when I integrate one side from v0 to v it is equal to the integration of the other integral from R to z? I know, v0 is the velocity at the height R, and v the velocity at a certain height z, but still, it somehow confuses me...

By the way, do you use some kind of tool to produce the equations you have written(so that they are displayed in such a readable way)?

 

[Hootenanny]

Okay, let's start by integrating both side wrt the same limits;

-GM\int^{z=z_{1}}_{z=R}\frac{dz}{z^2}=\int^{z=z_{1}}_{z=R}v\;dv

Now, we can't find the definite integral of the RHS directly since the limits do not match the variable, so we must change the limits to match the variable. Now, we know that acceleration is the rate of change of velocity, i.e.;

z&#039;&#039; = \frac{dv}{dt}\Rightarrow \int z&#039;&#039;\;dt = \int \frac{dv}{dt}\;dt\Rightarrow z&#039; = v

Now, rewriting our final result;

v = \frac{dz}{dt}

Note that the above result is a velocity.

So, taking our limits;

\frac{dR}{dt}:=v_{0}\hspace{1cm}\frac{dz_1}{dt}:=v_{1}

Note that both of which are velocities and therefore we can substitute them into our integral on the RHS as limits; giving;

-GM\int^{z_{1}}_{R}\frac{dz}{z^2} = \int^{v_{1}}_{v_{0}} v\; dv

I hope that make sense (I'm a little rushed because I'm working while posting - on a Sunday )

I used \LaTeX to format the math, which is a markup language. There is a tutorial available in the tutorials section.

 

[covers]

Thanks for the detailed explanation, it did make sense and it did help me!
I wish you a happy new year.

 

[Hootenanny]

I wish you a happy new year.

All the best for the coming year and those that follow