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Distance and gravitation problem.

  • #1

Homework Statement


A planet has a radius of 500km, and the acceleration due to gravity on the surface is 3.00 m/ s^2. An object is thrown up with an initial velocity of 1000 m/s. What is the maximum height the object will go? (Note: acceleration due to gravity is not constant).

So my teacher said the easy way to solve this is using conservation of energy. But then I asked if there is another way and he said there is, and he challenged the entire AP Physics class to show how to solve the problem another way without using Energy.


Homework Equations



F = GMm / r ^ 2

Pretty much it unless I'm missing something else

The Attempt at a Solution



So I went with calculus on the acceleration of gravity.

Force of gravity = GMm / r^2.
acceleration due to gravity = GM / r^2 <-- acceleration changes with the radius

So I took the derivative of the acceleration equation with respect to r (radius)
da / dr = -2GM / r^3

Then integrated it back again, this time plugging in my limits as from r to r + h. Where h is the height.
so:

acceleration total = [tex]\int[/tex] (from r to r + h) (-2GM / r ^ 3) dr

Which became
acceleration total = GM / (r+h) ^2 - GM / r^2

There is where I ran into my first stumble, since this is the magnitude of the acceleration, I have to put acceleration into a vector. I'm mostly sure the acceleration is negative so I switched the signs.

acceleration total = GM / r^2 - GM / (r+h)^2 <--- This looked quite odd so I wasn't sure.

Then to find the height, I have to keep integrating. So acceleration - > velocity -> distance.

V = [tex]\int[/tex] a dt = [tex]\int[/tex] (GM / r^2 - GM / (r+h)^2) dt
V = GMt / r^2 - GMt / (r+h)^2 + v_initial

Now I integrate Velocity to get distance (which is the height).

h = [tex]\int[/tex] v dt = [tex]\int[/tex] (GMt / r^2 - GMt / (r+h)^2 +v_initial) dt

h = (1/2)GM * t^2 / r^2 - (1/2)GM * t^2 / (r + h)^2 + v_initial * t

So I have two equations, one for velocity final and for height. I set the equation for velocity final equal to 0 because when the object is at max height the velocity is 0.

0 = GMt / r ^ 2 - GMt / (r+h)^2 + v_initial. I need to solve for time because I need to eliminate the time variable

Solving for t I get this:

t = v_initial / (GM / (r+h)^2 - GM/r^2)

Then I need to plug t it into this equation:

h = (1/2)*GM * t^2 / r^2 - (1/2) * GM * t^2 / (r + h)^2 + v_initial * t

Which seems kind of crazy, but the t contains only h and constants, and plugging it into the equation for h has h equals to some constants and some h. Which seems right because using algebra I should be able to solve for h right? Can anyone verify it?

Was using calculus on the acceleration right in the first place? Was there a different way to solve it without using energy OR calculus?

Sorry I suck at using symbols for math equations.

Thank you.
 

Answers and Replies

  • #2
Filip Larsen
Gold Member
1,237
169
I must admit I have not read your entire post. You are initially on the right track, but you should probably first think about what it is you need to integrate. Try think if you can use the kinematic relation a = dv/dt and v = dh/dt to something.
 

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