Vertical projectile mathematics

1. Mar 27, 2014

Bob__

Hello
I have been struggling with this problem for a whole day, and finally admitting that i can't solve it, so i ask for help
I am familiar with the escape velocity , and the speed needed to brake into interplanetary space - the newton cannonball thought experiment. But i was wonder about launching a cannonball perpendicular to the gravitational field - vertical lunch. Ignoring air resistance, and other celestial bodies , what is the initial speed the cannonball must have in order to break trough gravitational influence of the earth and head to infinity?
I tried , but i got just mathematical understanding of calculus, i never learned it's applications in physics.
I know only that this values change over time like this:
v(t) = Vo-g*t g= M*G/r^2 , and R=ro+V(t) H= Vo*t-(g*t^2)/2 or after brief calculation, H=(Vo^2)/2*g
where M is the mass of earth
t-time
ro initial semi major axis of earth
vo- initial velocity
H-max height in homogeneous g-field
because speed tends to reduce for g each second , and g tends to reduce by inverse square of r
there should , in my opinion be a sufficiently large initial velocity with which a verticaly launched projectile would leave the earth to infinity, as R-> infinity g ->0 , LimV= Vmin and LimH=infinity
How to mathematically express that kind of solution, if there is one?

Last edited: Mar 27, 2014
2. Mar 27, 2014

AlephZero

Your formula v(t) = Vo-g*t is only true when g is constant, so it is wrong for this question.

Probably the easiest way to do the physics is to use conservation of energy. The work required to raise the ball to height h above the center of the earth is the integral of (force x distance). And force = mass x acceleration, so
work = $\int_{r_0}^h \frac{mMG}{r^2}dr$ where $r_0$ is the earth's radius.

So the work to reach height h is $mMG(\frac{1}{r_0} - \frac{1}{h})$.

If the initial kinetic energy $\frac{mv^2}{2}$ is greater than $\frac{mMG}{r_0}$ there is enough energy to reach any height h, so the ball can escape from the earth's gravitational field.

You can also solve this problem by integrating the "force = mass x acceleration" equation, but this is tricky because the force is a given as a function of the distance, but acceleration is a derivative with respect to time not distance.

3. Mar 27, 2014

Bob__

$\sqrt{2GM/r}$ -and it is the escape velocity :surprised