1. ## Gravitational Law

An object is fired vertically upwards from the surface of a planetary body; it moves under the action of Newton’s Gravitational Law, without
resistance, so the equation is z'' = -gR^2 / (z + R)^2 . Find the relation between v = z' and z and use this model, and the relation that you have
found, to obtain a numerical estimate for the escape speed on the surface of the Earth.

Its the wording of the question I don't get? I assume I integrate z'' twice to find the velocity and distance respectively, but the escape speed to escape the surface of the earth? is this like when g (9.81) begins to decrease as it moves from the planetry body? I havn't got a clue at all. Any help appreciated

2. The escape velocity is the initial velocity necessary to "escape" to infinity. But you don't need to integrate twice. Just integrate once to get the formula for the velocity. The escape velocity is the smallest initial velocity such that the velocity is never 0 for any finite t.

3. How can I integrate this though? I have z'' on the LHS and z on the RHS, but I can't integrate z because I don't know it? I know I can write z'' as vz' using the chain rule. Anymore help appreciated, thanks

4. Originally Posted by mitch_nufc
How can I integrate this though? I have z'' on the LHS and z on the RHS, but I can't integrate z because I don't know it? I know I can write z'' as vz' using the chain rule. Anymore help appreciated, thanks
Substitute $\displaystyle \frac{d^2 z}{dt^2} = v \frac{dv}{dz}$ where $\displaystyle v = \frac{dz}{dt}$.

5. Originally Posted by mr fantastic
Substitute $\displaystyle \frac{d^2 z}{dt^2} = v \frac{dv}{dz}$ where $\displaystyle v = \frac{dz}{dt}$.
So you have $\displaystyle v \frac{dv}{dz} = f(z) \Rightarrow \frac{1}{2} v^2 = \int \! f(z) \, dz + C$.

Solve for v as a function of z: v = v(z).

Now solve $\displaystyle v(z) = \frac{dz}{dt} \Rightarrow \int \, dt = \int \! \frac{1}{v(z)} \, dz$.

6. I sort of get what you're saying

so I'd have 1/2v^2 = RHS? Or not... Hmm I used to be good at applied questions too

7. Originally Posted by mitch_nufc
I sort of get what you're saying

so I'd have 1/2v^2 = RHS? Or not... Hmm I used to be good at applied questions too
f(z) = -gR^2 / (z + R)^2. This is what you first have to integrate.

8. so I get (gR^2)/(z+R) = (v^2)/2

(2gR^2)/(z+R) = v^2

Not sure what this is telling me though

9. ## The missing element

Your mathematics is correct however what 's missing from the discussion

is that vdv is integrated from v0 to 0 the initial velocity to a velocity of 0

at z = infinity

The Rhs is integrated from 0 to infinity i.e. we want to integrate z over a range where it completely escapes the gravitational pull of the earth which is at infinity.

We obtain v0^2 = 2gR

or v0 = (2gR)^(1/2)

Make sure to convert g to km/s^2 by dividing 9.8/1000 and use R =6400km

You obtain 11.2km/sec which is the accepted value