Gravitational Potential Energy of a rocket fired straight up

[genPhysics]

Homework Statement

A rocket is fired straight up from the surface of the Earth at half the escape velocity. How high will it go relative to the surface of the earth?
Neglect dissipative forces.

Homework Equations

Uinitial + Kinitial = Uf

Kf is zero

G*m*M/radius^2 + 1/2mv^2 = -G*m*M/(2R)

The Attempt at a Solution

I used the above equations and I can solve for the initial velocity. I am stuck on how to find the height. Is it just 2R?

Many thanks...physics is a struggle for me.

 

[rl.bhat]

On the surface of the earth, PE = -GMm/R and KE = 1/2*m*v^2
At a certain height, KE = 0, and PE = -GMm/(R + h)
What is the expression for the escape velocity?
From that find the initial velocity.

 

[tomcue]

Firstly, it is important to note that the formula provided in the question is not the correct formula for calculating the gravitational potential energy of a rocket. The correct formula is U = mgh, where m is the mass of the rocket, g is the acceleration due to gravity, and h is the height of the rocket relative to the surface of the Earth.

To find the height that the rocket will reach, we can use the conservation of energy principle, which states that the total energy of a system (in this case, the rocket) remains constant. This means that the initial potential energy of the rocket (when it is at the surface of the Earth) is equal to its final potential energy (when it reaches its maximum height).

Using the formula U = mgh, we can set the initial potential energy equal to the final potential energy and solve for h. We know that the initial velocity is half the escape velocity, so we can use the formula for escape velocity (v = √(2GM/R)) to find the initial velocity.

Substituting this into the equation, we get:

mgR/2 = mg(h + R)

Solving for h, we get h = R/2.

Therefore, the rocket will reach a height of R/2 relative to the surface of the Earth.

I hope this helps and keep up the good work in your physics studies!