• Support PF! Buy your school textbooks, materials and every day products Here!

Sped of launchin rocket

  • Thread starter bcjochim07
  • Start date
374
0
1. Homework Statement
A rocket is launched straight up from the earth's surface at a speed of 15,000 m/s. What is its speed when it is very far away from earth?


2. Homework Equations
K=.5mv^2
U=-GMm/R


3. The Attempt at a Solution
Is this correct: I used conservation of energy and assumed very far was where U=0

(.5)(m)(15000m/s)^2 + -[(6.67*10^-11)(5.98*10^24)m]/[6.37*10^6m)=.5mv^2
v=9988 m/s

Also could someone explain to me why potential energy decreases with increasing height in this case as opposed to increasing U by lifting an object above the ground on earth.

Thanks
 

Answers and Replies

alphysicist
Homework Helper
2,238
1
Your equation looks okay to me.


Also could someone explain to me why potential energy decreases with increasing height in this case as opposed to increasing U by lifting an object above the ground on earth.
It doesn't; the potential energy is a negative number that get closer to zero with increasing height (i.e., distance away from the earth), which means it is increasing.
 
374
0
Oh yeah. I see exactly what you are saying. Thanks!
 

Related Threads for: Sped of launchin rocket

Replies
3
Views
29K
  • Last Post
Replies
5
Views
733
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
12K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Top