# Homework Help: Work Done In space - Satellites

1. Jul 26, 2008

### Renren

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

"Calculate the work done on a 5000kg satellite that moves from an orbit with an altitude of 300km to one with an altitude of 600km"

mass of earth = 5.98x10^24 kg

mean radius of earth = 6.37x10^6 m

2. Relevant equations

Force between 2 objects in space
G x (m1 x m2)/((r of planet + altitude above surface)^2)

Gravitational Potential Energy (GPE)
-G x (m1 x m2)/(r of planet + altitude above surface)

where G = 6.67x^-11

Work Done
W= F x d

3. The attempt at a solution

Attempt 1:

W= F x d so...

Force W=F x 300 000m ( as displacement = 600km-300km)

F = G x (m1 x m2)/((r of planet + altitude above surface)^2) so using initial height (300 000 + 6.37x10^6)

F= 4482.76

therefor W = 300 000 x 4482.76
= 1.34x10^9 J

I don't think im doing that right.

Attempt 2:

(GPE at 300) - (GPE at 600)

so yeah using a calc i got

-2.9855...x10^10 + 2.8572...x10^10

= -1.289x10^9

-G x (m1 x m2)/(r of planet + altitude above surface)

Conclusion

To be honest i blindly went about doing it in different ways based on the things i knew and been taught. But since the answers are similar is it possible they are correct. But even if they are i want to know why i am and probably learn how to understand what the question is asking me (physics wise, i know what it asks me logically)

2. Jul 26, 2008

### Dick

Attempt 1 is conceptually wrong. The force isn't constant between 300km and 600km. So you can't just use F*d. Though the difference isn't much because the force doesn't change hugely. Attempt 2 is better. You got the work you do against gravity correctly. But a satellite in an orbit of 600km is moving slower than one in an orbit of 300km and hence has less kinetic energy. Shouldn't you account for that??

3. Jul 27, 2008

### Janus

Staff Emeritus
Dick is right, you have to take both the gravitational potential and kinetic energies of both orbits into account.

4. Jul 27, 2008

### Renren

But if i do consider K.E.'s and stuff.. ain't i moving away from my question?

Say if i do

W = F x d

and for F i use G x (m of ship x mass of earth)/((r of earth + 600x10^3)^2)

which would give me an answer of 2.86x10^11

then i would multiply that by 300 000

which gives me 8.58x10^16

5. Jul 27, 2008

### Janus

Staff Emeritus
No. The total energy of an object in orbit is equal to the sum of its kinetic and gravitational potential energies. To find the work done moving from one orbit to another you need to find the difference between the total energies of the object at the different orbits.

P.S. Your answer in the last post is off by many orders of magnitude.

6. Jul 29, 2008

### Renren

Ok dokey...

GPE of 300Km Satellite = -2.99x10^11
GPE of 600Km Satellite = -2.86x10^11

by mv^2/r = GMm/r^2

Velocity of 300Km Satellite =7733m/s
Velocity of 600Km Satellite =7564.8m/s

Hence:
K.E. of 300Km Satellite = 1.5x10^11
K.E. of 600Km Satellite = 1.43x10^11

Adding the two energies together and subracting from their palcement.

(GPE of 300Km Satellite + K.E. of 300Km Satellite )- (GPE of 600Km Satellite + K.E. of 600Km Satellite )
-2.99x10^11+1.5x10^11 - - -2.86x10^11 + 1.43x10^11

Which equals to

2.8x10^11 J

However i get a feelign i may have done the subtraction the wrong way around.

7. Jul 29, 2008

### Dick

Your feeling is right. -2.86-(-2.99)+1.43-1.50. The PE part is positive since it takes positive work to raise the satellite against gravity. The KE part is negative since we can offset some of that by using the lost kinetic energy. Your first expression will give you the energy difference - the numerical one may as well, but I don't know what '- - -' means in an equation. But the answer isn't 2.8x10^11J.

8. Jul 29, 2008

### Janus

Staff Emeritus
You need to take care with your signs and your grouping of terms. Remember, you are subtracting total energy from total energy.

if TE1 = KE1-GPE1
And
TE2 = KE2-GPE2

Then TE2-TE1 = KE2-GPE2- (KE1-GPE1) = KE2-GPE2-KE1+GPE1

9. Jul 29, 2008

### Dick

??? Total energy is kinetic energy PLUS potential energy. As Renren wrote before the signs got screwed up.

10. Jul 29, 2008

### Janus

Staff Emeritus
Sorry, I forgot to mention that in the equations "GPE" stands for the absolute value of the Potental energy. I was trying to short hand in order to keep it simple. Written full out, it would be:

E = mv2^2/2 - GMm/r2 - (mv1^2/2 - GMm/r1 ) = mv2^2/2 - GMm/r2 - mv1^2/2 + GMm/r1

Last edited: Jul 29, 2008