Work done by a force changing the distance of a satellite's orbit

  • Thread starter kmb11132
  • Start date
  • #1
13
0

Homework Statement



A 5530-kg satellite is in a circular earth orbit that has a radius of 1.81 × 10^7 m. A net external force must act on the satellite to make it change to a circular orbit that has a radius of 8.01 × 10^6 m. What work must the net external force do?

Homework Equations



Orbital Speed:
v = sqrt( G * M_e / r )
where G is the gravitational constant and M_e is earth's mass

Work-Kinetic Energy Theorem:
W = K_f - K_i = .5*m*v_f^2 - .5*m*v_i^2

The Attempt at a Solution



v_i = sqrt( 6.674e-11 * 5.98e24kg / 1.81e7m )
v_i = 4695.74m/s

v_f = sqrt( 6.674e-11 * 5.98e24kg / 8.01e6m )
v_f = 7058.74m/s

W = .5*m*v_f^2 - .5*m*v_i^2
W = .5*5530kg*7058.74m/s^2 - .5*5530kg*4695.74m/s^2
W = 7.68002e10J
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Work = change in energy.
 
  • #3
13
0
Work = change in energy.
I incorrectly found the work to be the difference between the initial and final kinetic energies above.
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Are there other forms of energy present?
 

Related Threads on Work done by a force changing the distance of a satellite's orbit

Replies
4
Views
14K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
9
Views
1K
Replies
1
Views
421
Replies
5
Views
119
Replies
5
Views
597
Replies
8
Views
1K
  • Last Post
Replies
8
Views
847
Top