Question about Grav. Potential Energy and Escape Velocity

[Divergent13]

I am finding this question to be a bit challenging---

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed Vo. (< Vesc)

(Ignore air resistance and the Earth's rotation)

I know I should plug in numbers only until the end, but by using Gravitational potential energy, is it appropriate to begin with (1/2)m*v1^2 - [G(m)(Mearth)]/r = (1/2)m*v^2 - [G(m)(Mearth)]/r ?

The chapter is quite short --- Thanks for your help!

 

[W.A. McCormick]

I am not a physicist or even a scientist, but wouldn't it be merely a matter of 1G deceleration till no momentum is left, and then calculate the distance achieved in that time?

W.A. McCormick

 

[cookiemonster]

You want to use an energy argument.

The object initially has zero potential energy (if you define the PE at the radius of the Earth as zero) and some amount of kinetic energy. When it's at the maximum height, it has zero kinetic energy and some amount of potential energy. Equate and solve.

cookiemonster

 

[Doc Al]

I know I should plug in numbers only until the end, but by using Gravitational potential energy, is it appropriate to begin with (1/2)m*v1^2 - [G(m)(Mearth)]/r = (1/2)m*v^2 - [G(m)(Mearth)]/r ?

Yes, that's essentially it. But set it up carefully as a statement of energy conservation. KE₁ + PE₁ = KE₂ + PE₂. Of course, maximum height means that KE₂ = 0.

$$\frac{1}{2}mv_0^2 - \frac{GmM_e}{r_1} = - \frac{GmM_e}{r_2}$$

The height will be r_2 - r_1; r_1 = radius of Earth.