Energy Conservation applied to Earth's surface

[garr6120]

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.

 

[gneill]

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.

 

[susu]

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.

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