Conservation of Energy/Gravity Problem

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SUMMARY

The discussion centers on calculating the maximum height an object reaches when projected upward from Earth's surface with an initial speed of 3.37 km/s. The gravitational constant (G) is defined as 6.673x10^(-11) m^3/(kg s^2), and the mass of Earth (M) is 5.9742x10^24 kg. The correct approach involves using the initial potential energy from the center of the Earth, leading to the maximum height of 6.37×10^5 m, rather than the incorrect calculation of 56700493 m based on flawed assumptions about potential energy.

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Homework Statement



An object is projected straight upward from the surface of Earth with an initial speed of 3.37 km/s. What is the maximum height it reaches?

Homework Equations



G = gravitational constant = 6.673x10^(-11) m^3 / (kg s^2)
M = mass of the Earth = 5.9742x10^24 kg
R = radius of the Earth = 6.378x10^6 m
r = max altitude reached (to be determined)
v = initial velocity of object = 3.37 km/s = 3370 m/s
m = mass of the object (numerical value not needed since it will drop out)

conservation of energy

The Attempt at a Solution



Initial Potential Energy = 0 (height is zero)
Initial Kinetic Energy = 1/2 (3370)^2

Final Potential Energy = r * G (M/(R+r)^2)
Kinetic Energy = 0 (velocity would be zero at maximum altitude)

Solving for this, I get the answer as 56700493 m. However the answer is 6.37×10^5 m, what am I doing wrong?
 
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Initial potential energy = 0 (height is zero)

This is not correct. PE of the Earth is measured from the center of the earth.
So the initial potential energy is

[tex]PE_o = \frac{GMm}{R}[/tex]
 
Ok that makes sense, thank you!
 

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