Gravitational Energy Homework: How High Does it Go?

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In summary, someone provided an equation to calculate the height of a projectile shot up from the Earth's surface at a speed of 1.40×10^4 km/hr. The equation yielded an incorrect result of 12,700km.
  • #1
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Homework Statement


A projectile is shot straight up from the Earth's surface at a speed of 1.40×10^4 km/hr.

How high does it go?

Homework Equations


[tex](1/2)m(v_0)^2-((G*M*m)/(R+y_0))=(1/2)mv^2-((G*M*m)/(R+y))[/tex]

The Attempt at a Solution


I assumed this is a gravitational energy problem instead of a flat Earth energy problem because of the speed for which it launched. Converting to m/s I get 3889 m/s. Using the equation above, the mass of the projectile cancels and is not needed. That provides the equation:

[tex](1/2)(v_0)^2-((G*M)/(R+y_0))=(1/2)v^2-((G*M)/(R+y))[/tex]

Since y_0 is 0 and assuming the max height is being asked for, v=0 m/s.

[tex](1/2)(v_0)^2-((G*M)/(R))=-((G*M)/(R+y))[/tex]

Solving for y:
[tex]x=-((4*G^2*M^2)/((R*(v_0)^2-2*(G*M)))*(v_0)^2)-R-((2*G*M)/((v_0)^2))[/tex]

The answer I get is 12,700km which is wrong. When I do this problem like an ordinary energy problem I get 770km for the height. Any help?
 
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  • #2
I would work this problem by integrating the force (due to Earth's gravitational field) function with respect to displacement. That will yield a change in potential energy, which should be equal but opposite to the original energy (kinetic) of the object in question.

Someone please feel free to correct me if I'm wrong.
 
  • #3
Huh? I don't understand. How does that get me the height?
 
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  • #4
The upper limit of the integral is height, the lower the Earth's radius. The integral = Vi^2/2. Sort of like a kinematics soln where sqrt(2a*y)=v^2 but you cannot assume costant a as that varies with inverse square.
 
  • #5
Where did you get Vi^2/2? I don't know what you mean about the kinematics soln either. This seems WAY more complicated than the book and the professor's lectures.
 
  • #6
let me try again; potential energy gained=kinetic lost
for linear accelaration and a vertical shot like the one you descibe,h is height

m(9.80)h=1/2mv^2

But this is like a mach 10 shot w/o drag--gonna go real high so we need to aaccount for the change in g, so one ends up with an integral from the surface of the Earth to the peak altitude with the force of gravity varying as a function of height. At least I believe that was what Capt Zappo was referring to. I believe the soln is in natural logs.
 
  • #7
Ok I understand the explanation but they way that is being suggested to solve the problem is foreign to me. It seems like this method is a new method to me by comparing my notes and textbook. While I don't doubt that this way is correct and valid I just don't want to learn another method when the methods presented to me are troublesome enough.
 
  • #8
actually the methodology is the same, just expressed a bit differently. The eqn you cited lumps all the energy both potential and kinetic on each side of the eqn instead of the difference in kinetic energies being equal to the gain of potential.

I believe if you let Yinit=0, and collect all the terms,
Y(distance from the center of the earth) and consider the rocket at rest at apex:

2GM/R-2GM/h=V^2, solve for h, and then subtract R from it as again, is referenced to the Earth's center.
 
  • #9
My equation is perfectly valid. Energy is being conserved throughout the problem because gravity is the only force acting on the projectile sans drag. No external forces are acting on the system. This allows the starting and ending energies to be equal. All I did was to forget to subtract the Earth's radius from my answer. Thanks for the help.
 
  • #10
never said it wasn't valid. just diffeent, glad I was of assistance.
 

1. What is gravitational energy?

Gravitational energy is the potential energy stored in an object due to its position in a gravitational field. It is the energy that is required to move an object against the force of gravity.

2. How is gravitational energy calculated?

The equation for gravitational energy is E = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. This equation can be used to calculate the amount of energy an object possesses due to its position in a gravitational field.

3. What factors affect the amount of gravitational energy an object has?

The amount of gravitational energy an object has depends on its mass, the strength of the gravitational field it is in, and its height above the ground. The greater the mass and height of the object, and the stronger the gravitational field, the more gravitational energy it possesses.

4. How high will an object go if all its gravitational energy is converted into kinetic energy?

If all of an object's gravitational energy is converted into kinetic energy, it will reach a height equal to the initial height from which it was dropped. This is because the total energy of an object remains constant, and the potential energy is converted into kinetic energy as the object falls.

5. How can gravitational energy be used in everyday life?

Gravitational energy is used in many everyday activities, such as riding a rollercoaster or using a playground swing. It is also harnessed in hydroelectric power plants, where the potential energy of water stored in a reservoir is converted into kinetic energy to generate electricity. Additionally, gravitational energy plays a crucial role in the orbits of planets and satellites in our solar system.

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