1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanical Energy- Orbit

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the total mechanical energy of a 380kg satellite in a circular orbit 3.0 Earth radii above the surface?


    2. Relevant equations
    W= E2- E1

    E= 1/2 mv2- [itex]\frac{GmM}{r}[/itex]


    3. The attempt at a solution

    I'm not sure if the equations above are suitable to solve this or I just don't understand how to start the question.
     
  2. jcsd
  3. Oct 31, 2013 #2
    Mechanical Energy is the sum of the potential energy, and the kinetic energy.

    Emechanical = Ekinetic + EPotential

    Since you're not close to the surface of the Earth, equating the potential energy to (mgh) is not applicable.

    Epotential = (-GMm)/r

    - r is the distance from the two objects centers,
    - G is the gravitational constant of Earth
    - M is the mass of Earth
    - m is the mass of the satellite

    Em = (1/2)mv2 - (GMm)/r

    You have all the variables for the potential energy, and for the kinetic energy, you have the mass. So you need to solve for the velocity.

    Since the mass of the satellite is just about negligible in relation to the earth, you can use the equation,

    v = √(GM/r)

    With that, you should be able to solve for the total mechanical energy.
     
  4. Nov 1, 2013 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The equation is good, you need to find the speed of the satellite.


    The satellite travels along a circle with speed v. What is the radius of the orbit? What is its centripetal acceleration? What force provides the centripetal force?

    ehild
     
  5. Nov 1, 2013 #4
    so,

    Fc= mac
    (GmM)/RE2=(mv2)/RE
    v2=(GM)/RE

    where RE is the radius of the earth


    then,

    E= K+ UG where K=1/2(mv2) and UG=-(GmM)/r2 and r= 2.55x107m (equal to 4 earth radii)

    so,

    E= 1/2(mv2) - (GmM)/r2
    E= 1/2(GM)/RE) - (GM)/r2
    E= 15633739.23 J
     
  6. Nov 1, 2013 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Why do you calculate the speed of the satellite with the radius of Earth?? You know that the radius of the the circular orbit is 4 Earth-radius.

    ehild
     
  7. Nov 1, 2013 #6
    so it would just be

    E= 1/2(GM/r) - (GM)/r
    E= -7819322.353 J
     
  8. Nov 1, 2013 #7

    gneill

    User Avatar

    Staff: Mentor

    You haven't included the mass of the satellite. So as it stands so far, what you've calculated is the Specific Mechanical Energy (energy per kg).
     
  9. Nov 1, 2013 #8
    would it be,

    E= K + UG
    E= 1/2 (mv2) - (GmM)/r
    E= 1/2 m(GM/r) - (GmM)/r
    E= 2.377074x1010 J
     
  10. Nov 1, 2013 #9

    gneill

    User Avatar

    Staff: Mentor

    No. The formula's okay, but something went wrong in the execution.

    Your previous value for the specific energy was good. Just multiply that by the mass of the satellite!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mechanical Energy- Orbit
  1. Orbits and Energy (Replies: 7)

  2. Energy of orbits (Replies: 1)

  3. Orbit and Energy (Replies: 20)

Loading...