Gravitational Potential Energy of a rocket fired straight up

Click For Summary
SUMMARY

A rocket fired straight up from the Earth's surface at half the escape velocity will reach a maximum height of 2R, where R is the Earth's radius. The discussion utilized the conservation of energy principle, equating initial potential and kinetic energy to final potential energy at maximum height. The relevant equations included gravitational potential energy (U) and kinetic energy (K), specifically U = -GMm/R and K = 1/2mv^2. The key conclusion is that at the peak height, kinetic energy becomes zero, allowing for the calculation of maximum height using the derived equations.

PREREQUISITES
  • Understanding of gravitational potential energy (U = -GMm/r)
  • Familiarity with kinetic energy equations (K = 1/2mv^2)
  • Knowledge of conservation of energy principles
  • Basic understanding of escape velocity concepts
NEXT STEPS
  • Study the derivation of escape velocity formulas
  • Learn about the conservation of mechanical energy in gravitational fields
  • Explore advanced topics in gravitational potential energy
  • Investigate the effects of dissipative forces on projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to gravitational potential energy and escape velocity.

genPhysics
Messages
2
Reaction score
0

Homework Statement


A rocket is fired straight up from the surface of the Earth at half the escape velocity. How high will it go relative to the surface of the earth?
Neglect dissipative forces.


Homework Equations



Uinitial + Kinitial = Uf

Kf is zero

G*m*M/radius^2 + 1/2mv^2 = -G*m*M/(2R)

The Attempt at a Solution



I used the above equations and I can solve for the initial velocity. I am stuck on how to find the height. Is it just 2R?

Many thanks...physics is a struggle for me.
 
Physics news on Phys.org
On the surface of the earth, PE = -GMm/R and KE = 1/2*m*v^2
At a certain height, KE = 0, and PE = -GMm/(R + h)
What is the expression for the escape velocity?
From that find the initial velocity.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K