Energy Conservation applied to Earth's surface

Click For Summary
SUMMARY

The problem involves calculating the initial speed required for an object to be projected vertically upward from the surface of Earth to reach a maximum height equal to Earth's radius, applying the principle of energy conservation. The relevant equations are the kinetic energy equation, \(E_{k_1} = \frac{mv^2}{2}\), and the gravitational potential energy equation, \(E_{g_2} = mgh\). The correct approach necessitates using Newton's general formula for gravitational potential energy at greater distances, leading to the final answer of 7.91 x 103 m/s for the initial speed.

PREREQUISITES
  • Understanding of energy conservation principles in physics
  • Familiarity with kinetic and gravitational potential energy equations
  • Knowledge of Newton's law of universal gravitation
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study Newton's law of universal gravitation in detail
  • Learn how to derive gravitational potential energy at varying distances
  • Explore energy conservation problems involving varying heights
  • Practice solving projectile motion problems in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of applying theoretical concepts to practical problems.

garr6120
Messages
42
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
4K
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
24
Views
4K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K