Work Done In space - Satellites

Click For Summary

Homework Help Overview

The discussion revolves around calculating the work done on a satellite transitioning between two orbital altitudes, specifically from 300 km to 600 km above Earth's surface. The problem involves concepts from gravitational physics, including gravitational force, gravitational potential energy, and kinetic energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods to calculate work done, including using force and displacement, as well as gravitational potential energy differences. Some question the validity of assuming constant force over the altitude change.

Discussion Status

There is ongoing exploration of the relationship between gravitational potential energy and kinetic energy in the context of orbital mechanics. Some participants suggest that both forms of energy must be considered to accurately determine the work done. Multiple interpretations of the calculations are being examined, with no clear consensus reached yet.

Contextual Notes

Participants express uncertainty about the correct approach to the problem, particularly regarding the treatment of kinetic energy and the signs in their calculations. There is acknowledgment of potential errors in previous attempts, and the need for careful attention to detail in the calculations is emphasized.

Renren
Messages
9
Reaction score
0

Homework Statement



"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

Homework 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


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 I am 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)
 
Physics news on Phys.org
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??
 
Dick is right, you have to take both the gravitational potential and kinetic energies of both orbits into account.
 
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
 
Renren said:
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

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.
 
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.
 
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.
 
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
 
Janus said:
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

? Total energy is kinetic energy PLUS potential energy. As Renren wrote before the signs got screwed up.
 
  • #10
Dick said:
? Total energy is kinetic energy PLUS potential energy. As Renren wrote before the signs got screwed up.

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:

Similar threads

Replies
5
Views
3K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
3K
Replies
5
Views
2K
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
37
Views
4K