Energy Conservation applied to Earth's surface

AI Thread Summary
To determine the initial speed required for an object to reach a height equal to Earth's radius, energy conservation principles are applied. The kinetic energy at the surface equals the gravitational potential energy at the maximum height, leading to the equation \( \frac{mv^2}{2} = mgh \). However, since the height involved is significant, the standard gravitational potential energy formula is inadequate; the general formula considering Earth's mass and distance must be used. The correct calculation yields an initial speed of approximately 7.91 x 10^3 m/s. This highlights the importance of using the appropriate gravitational equations for large distances.
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Homework Statement


With what initial speed must an object be projected vertically upward from the surface of Earshot rise to a maximum height equal to Earth's radius? (neglect air resistance.) Apply energy conservation.

Homework Equations


##E_{k_1}+E_{g_1}=E_{k_2}+E_{g_2}##, however since ##E_{g_1}=0## and ##E_{k_2}=0##,
the equation is ##E_{k_1}=E_{g_2}##.
##E_{k_1}=\frac{mv^2}2##
##E_{g_2}=mgh##

The Attempt at a Solution


##\frac{mv^2}2=mgh##
##v=\sqrt{2gh}##
I know that the maximum height of the object is ##6.38*10^6 m##
I do not know if this is the right formula that i am using because i get the wrong answer.
The answer is supposed to be 7.91*10^3 m/s.
 
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garr6120 said:

Homework Statement


With what initial speed must an object be projected vertically upward from the surface of Earshot rise to a maximum height equal to Earth's radius? (neglect air resistance.) Apply energy conservation.

Homework Equations


##E_{k_1}+E_{g_1}=E_{k_2}+E_{g_2}##, however since ##E_{g_1}=0## and ##E_{k_2}=0##,
the equation is ##E_{k_1}=E_{g_2}##.
##E_{k_1}=\frac{mv^2}2##
##E_{g_2}=mgh##

The Attempt at a Solution


##\frac{mv^2}2=mgh##
##v=\sqrt{2gh}##
I know that the maximum height of the object is ##6.38*10^6 m##
I do not know if this is the right formula that i am using because i get the wrong answer.
The answer is supposed to be 7.91*10^3 m/s.
mgh for gravitational PE only applies to bodies close to the surface of the Earth. Since this problem involves a distance that is double the Earth's radius you should revert to Newton's general formula.
 
garr6120 said:

Homework Statement


With what initial speed must an object be projected vertically upward from the surface of Earshot rise to a maximum height equal to Earth's radius? (neglect air resistance.) Apply energy conservation.

Homework Equations


##E_{k_1}+E_{g_1}=E_{k_2}+E_{g_2}##, however since ##E_{g_1}=0## and ##E_{k_2}=0##,
the equation is ##E_{k_1}=E_{g_2}##.
##E_{k_1}=\frac{mv^2}2##
##E_{g_2}=mgh##

The Attempt at a Solution


##\frac{mv^2}2=mgh##
##v=\sqrt{2gh}##
I know that the maximum height of the object is ##6.38*10^6 m##
I do not know if this is the right formula that i am using because i get the wrong answer.
The answer is supposed to be 7.91*10^3 m/s.
garr6120 said:

Homework Statement


With what initial speed must an object be projected vertically upward from the surface of Earshot rise to a maximum height equal to Earth's radius? (neglect air resistance.) Apply energy conservation.

Homework Equations


##E_{k_1}+E_{g_1}=E_{k_2}+E_{g_2}##, however since ##E_{g_1}=0## and ##E_{k_2}=0##,
the equation is ##E_{k_1}=E_{g_2}##.
##E_{k_1}=\frac{mv^2}2##
##E_{g_2}=mgh##

The Attempt at a Solution


##\frac{mv^2}2=mgh##
##v=\sqrt{2gh}##
I know that the maximum height of the object is ##6.38*10^6 m##
I do not know if this is the right formula that i am using because i get the wrong answer.
The answer is supposed to be 7.91*10^3 m/s.

LCE: Ek1= Eg2
mv^2/2 = -Gmem/re + alt
Where m=mass of the object and
me=mass of the earth
Alt= distance above Earth's surface

You'll see that m cancels out so ur left with

V^2 = -Gme/re + re
U rearrange to get
V= √ 2Gme/2 re
 
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