Calculating Projectile Height Using Law of Gravitation

[majormaaz]

Homework Statement

At the Earth's surface a projectile is launched straight up at a speed of 8.1 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

Homework Equations

F = GMm/r²
U_g = -GM/r²
G = 6.67 e -11 whatever units it happens to be
That's all I'm 'given' for this problem.

The Attempt at a Solution

Well, I figured that the kinetic energy that the rocket/thing starts off with converts into gravitational energy.
1/2 mv² = GM/x²
where x is the distance between the centers of the rocket and the earth.
So that would mean x = r + h?
So if we rearrange terms to solve for x, we get:
x = ±2GM/v² - r
However, that doesn't seem to work, either sign for the potential energy.
Am I missing something?

 

[rock.freak667]

Your initial equation should be

½v²= GM/x


But in any event, you did the rearranging correctly. Why doesn't it work?

 

[haruspex]

1/2 mv² = GM/x²

That's not right, but you seem to have (almost) used the correct version below.

where x is the distance between the centers of the rocket and the earth.
So that would mean x = r + h?
So if we rearrange terms to solve for x, we get:
x = ±2GM/v² - r

Looks like you originally had x², but when you corrected it you forgot to get rid of the ±. And you've left out the PE it started with.

 

[majormaaz]

1/2 mv2 = GM/x2
That's not right, but you seem to have (almost) used the correct version below.

Terribly sorry about that. I used the formula U = -GMm/x for that, where x still is what I defined it to be previously.

And you've left out the PE it started with.

So you're saying that the equation looks like:
1/2 mv² + GMm/r = GMm/x ...? where r is the radius of the earth?
Let me try this out... looks promising.
But here's another question.

A long long time ago I learned that there was kinetic energy and potential energy, mgΔh. This type of potential energy is gravitational and so is GMm/r, so would the problem also be solved by replacing GMm/r with mgΔh?

 

[haruspex]

A long long time ago I learned that there was kinetic energy and potential energy, mgΔh. This type of potential energy is gravitational and so is GMm/r, so would the problem also be solved by replacing GMm/r with mgΔh?

No, that's only valid for constant g.

 

[majormaaz]

No, that's only valid for constant g.

aahhh... makes sense. It's similar to the reason why we don't use kinematics for equations with velocities accelerating at increasing/decreasing rates.

 

[majormaaz]

Got it. Thanks!