Calculating Work to Place Satellite in Synchronous Orbit

  • Thread starter Thread starter six789
  • Start date Start date
  • Tags Tags
    Satellites
Click For Summary

Homework Help Overview

The discussion revolves around calculating the work required to place a 650 kg satellite into synchronous orbit around Earth. Participants are analyzing gravitational potential energy and kinetic energy in the context of orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work using gravitational potential energy and question the accuracy of significant digits. There are inquiries about the types of energy involved in orbital mechanics, including potential and kinetic energy.

Discussion Status

The discussion is active, with participants exploring different energy calculations and their implications. Some guidance has been provided regarding the relationship between work and energy changes, although there is no explicit consensus on the calculations presented.

Contextual Notes

Participants are considering the effects of significant digits in their calculations and the assumptions related to energy states in orbit. There is an emphasis on understanding the differences between kinetic and potential energy in the context of a satellite in a circular orbit.

six789
Messages
127
Reaction score
0
here is another problem...
a 650kg satellite is to be placed into synchronus orbit around Earth.
What amount of work must be done onthe satellite to raise it into synchronous orbit?

i did..
W=GMm(1/r of Earth - 1/r of synchronous satellite)
= 6.673x10^-11N m^2/kg^2 (5.98x10^6kg)(650kg)(1/6.38x10^6m - 1/4.22x10^7m)
W = 3.442998737X10^10 J

the right answer is 3.7x10^10 J... is ther something wrong with the significant digits?? can anyone tell me the mistake that i did?
 
Physics news on Phys.org
six789,

What kind of energy did you calculate for the satellite? Does it have any other kind when it's in orbit?
 
a) the gravitational potential energy is -4.1x10^10 J
b) the total energy is -3.1x10^9 J

d) the kinetic energy is 3.1x10^9 J
 
The kinetic energy in this case is not the same as the potential energy. The only time kinetic energy = Potential energy is when the change in kinetic energy is caused by a change in potential energy. In this case, the potential energy is always cosntant as in a circular orbit, the satelite is always the same distance from the earth.
 
i get the right answer from doc al... he said that...
change of energy = Work-Energ(total)

change=work
therefore.. Work = Work-E
=3.442998737X10^10 J - (-3.1x10^9 J)
=3.75x10^10J..
so the answer is right!

thanks again for the help...
 

Similar threads

Replies
5
Views
3K
Replies
5
Views
2K
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
37
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
2K