# Gravitation and velocity problem

1. Jan 3, 2012

### canicon25

1. The problem statement, all variables and given/known data

2. Relevant equations
2ΔEk=-ΔEp

3. The attempt at a solution

-Gm1m2(1/r1-1/r2)-Q=mv2

plug in given values solve for v.

Last edited: Jan 3, 2012
2. Jan 3, 2012

Any ideas?

3. Jan 3, 2012

### D H

Staff Emeritus
Watch your units, be careful to distinguish between altitude and radial distance, and most important of all, show your work. There's no way to tell you where you went wrong if you don't show your work.

BTW, the given answer is correct.

4. Jan 3, 2012

### alemsalem

its 400000m not 400m

5. Jan 3, 2012

### canicon25

1600v2=-G(1600)(5.98E24)/(6.38E6+400000) - (-G)(1600)(5.98E24)/ (6.38E6+100000) - 8.3E8

v=1407 m/s

still not getting the correct solution.

Last edited: Jan 3, 2012
6. Jan 3, 2012

### D H

Staff Emeritus
Showing your work does not mean spewing out an equation without units and without justification. What are the relevant equations?

7. Jan 3, 2012

### canicon25

i am stumped. i tried using

ΔEk+Q=-(0.5)ΔEp

where Ek= 0.5mv2

and Ep= -GMm/r

Q is heat energy. Given in problem.

Last edited: Jan 3, 2012
8. Jan 4, 2012

### alemsalem

why do you have a 0.5 in the energy conservation equation?

9. Jan 4, 2012

### canicon25

upon further reading i found that

ΔEk+Q=-0.5ΔEP for CIRCULAR ORBITS ONLY

am i correct in assuming that

ΔEk+Q=ΔEp would apply for a object falling to Earth

10. Jan 4, 2012

### canicon25

so i tried it and did get correct answer

0.5mv2+Q=-GMm(1/r1-1/r2)

when i solve for v i get 2133 m/s

Last edited: Jan 4, 2012
11. Jan 5, 2012

### cupid.callin

I got 2096 m/s ≈ 2100 m/s

12. Jan 5, 2012

### BruceW

I think you've done it right, but there might be some rounding error, which means your answer isn't exactly the same as the given answer. good job!