What is the Escape Speed for an Object Fired Vertically from Earth?

[mitch_nufc]

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 i already have a z on the RHS so i can't integrate, 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

 

[Cyosis]

The escape velocity is the velocity at which the kinetic energy is equal to the gravitational potential energy of the planet.

Using the chain rule you can write $$\frac{d^2 z}{dt^2}=\frac{d v}{dt}=\frac{dv}{dz}\frac{dz}{dt}=v\frac{dv}{dz}$$ After rearranging your equation you can now integrate.

How did you get the equation for z''?

 

[mitch_nufc]

it was given in the question, the equation for z'' that is

 

[Cyosis]

Odd that the mass of the planet isn't a part of the equation.

 

[mitch_nufc]

Does it help to know I'm doing a Maths degree and not a Physics one? We were told none of our questions would require knownledge of classical mechanics/physics, but just differential equations etc

 

[Cyosis]

Regardless of what they said this is somewhat of a physics problem, but it matters not. Just use the equation you were given.

 

[fatra2]

Hi there,

Your first equation: $$\frac{d^2z}{dt^2} = ...$$ is not Newton's gravitational law. It is the result of the Newton's gravitational law, considering a planetary body, like the Earth, the Sun, or the Moon. Newton's real gravitational law is define as the force that attracts two body of masses, and is express with : $$F_g = G\frac{m_1 \cdot m_2}{r^2}$$ where $$G$$ is Universal gravitational constant, $$m_i$$ are the mass of each body, and $$r$$ is the distance between the two body's center of mass.

Hope this helps you understand it more.

 

[mitch_nufc]

yeah i remember all this from A-level physics, but i don't know any masses or distances so i assume theyre kept as constants in the ODE

 

[jrlaguna]

There is no mention to the mass of the planet because it's inside g, the formula is correct (physically speaking). If this was merely a maths problem, you'd have to solve the ODE with the initial conditions z(0)=0, z'(0)=v and find out for which v z(infty)->infty. But this is too much work for a physicist... We know that there is a conserved quantity, energy:

$$E=z'^2 - {gR^2\over z+R}$$

and that for E=0, the particle will escape, so... we set z=0 in that equation and solve for z'