What is the Maximum Height Attained by an Object Launched from Earth's Surface?

In summary, an object is fired vertically upward from the surface of the Earth with an initial speed comparable to but less than the escape speed. It can be shown that the object will reach a maximum height given by h = (REvi^2)/(vesc^2 - vi^2), where RE is the radius of the Earth, vi is the initial speed, and vesc is the escape speed. This can be derived using the conservation of energy equation (K + Ug)f = (K + Ug)i, where K is the kinetic energy and Ug is the gravitational potential energy. The equation ultimately simplifies to (1/2)(vi^2) - (1/2)(vesc^2) = -(1/2
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GOsuchessplayer
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Homework Statement


An Object is fired vertically upward from the surface of the Earth ( of radius RE ) with an initial speed vi that is comparable to but less than the escape speed vesc

Show that the object attains a maximum height h given by h=[tex]\stackrel{R_{E}v^{2}_{i}}{v^{2}_{esc}-v^{2}_{i}}[/tex]

Homework Equations


The Attempt at a Solution


I have the solution but I am confused with a step.
Here is the solution provided up until the step on which I am confused:

[tex](K+U_{g})_{f}[/tex]=[tex](K+U_{g})_{f}[/tex]

[tex]\stackrel{1}{2}[/tex]m[tex]v^{2}_{i}[/tex] - [tex]\stackrel{GmM_{E}}{R_{E}}[/tex] = 0 - [tex]\stackrel{GmM_{E}}{R_{E}+h}[/tex]

where [tex]\stackrel{1}{2}[/tex]m[tex]v^{2}_{esc}[/tex]= [tex]\stackrel{GmM_{E}}{R_{E}}[/tex]

Then[tex]\stackrel{1}{2}[/tex][tex]v^{2}_{i}[/tex] - [tex]\stackrel{1}{2}[/tex][tex]v^{2}_{esc}[/tex]= -[tex]\stackrel{1}{2}[/tex][tex]v^{2}_{esc}[/tex] ([tex]\stackrel{R_{E}}{R_{E}+h}[/tex])

So the jump I am confused about is the equation that follows the "Then" Statement. I'm not sure why this is true.

(sorry for the poor look of the equations. Hopefully its readable, I wasn't sure how to make fractions.)
 
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\frac x y

[tex]\frac x y[/tex]

And if not absolutely necessary don't put LaTeX inline, it looks better in its own lines.
 

FAQ: What is the Maximum Height Attained by an Object Launched from Earth's Surface?

1. What is a satellite?

A satellite is any object that orbits around a larger object, such as a planet or a star. Artificial satellites, which are man-made, are often used for communication, navigation, and scientific research.

2. How does a satellite stay in orbit?

A satellite stays in orbit due to a balance between its speed and the force of gravity from the larger object it is orbiting. This balance is known as the centripetal force and is necessary to maintain a circular orbit.

3. What is escape velocity?

Escape velocity is the minimum speed an object needs to achieve in order to escape the gravitational pull of a larger object. For example, the escape velocity from Earth is about 11.2 kilometers per second.

4. How is escape velocity related to satellites?

Satellites in orbit around a planet or other large object must maintain a speed that is lower than the escape velocity to remain in orbit. If a satellite's speed exceeds the escape velocity, it will escape the gravitational pull and move away from the larger object.

5. How is escape velocity calculated?

Escape velocity can be calculated using the formula v = √(2GM/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the larger object, and r is the distance from the center of the larger object to the object attempting to escape its gravitational pull.

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