1. Limited time only! Sign up for a free 30min personal 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!

Energy problem and rotation

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    A satellite of mass 190kg is placed in orbit 4.5 * 105 m above the surface of the Earth. Find the energy required to put this satellite into this orbit. Ignore air resistance but include Earth's daily rotation in your calculations.


    2. Relevant equations
    r^3/T^2 = GM/(4Pi^2)

    3. The attempt at a solution
    I calculated the change in potiential energy (F=-GmM/R) which yielded 1.68*1011. However I don't think is correct since it doesn't include the rotation.


    Thanks
     
  2. jcsd
  3. Apr 20, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi minifhncc! :smile:
    How about the change in kinetic energy ? :wink:
     
  4. Apr 20, 2010 #3

    Filip Larsen

    User Avatar
    Gold Member

    You need to consider the change in both potential and kinetic energy. For change in kinetic energy it seems you need to make some assumptions about where on the earth the satellite is launched from (for instance, near equator) and into which orbit it is placed (for instance, an east-ward equatorial orbit).
     
  5. Apr 20, 2010 #4
    Hi, the change in kinetic energy is the work done, but the change in gravi. potiential energy is also work done... but obviously they would yield different values. What's the difference between these values? (ie. change in K.E. and change in G.P.E)

    Thanks
     
  6. Apr 20, 2010 #5
    Hi again,

    I've been trying to get my head around this question. But I don't know how I can include the rotation of the Earth.

    Also, what velocity would I use? I worked out the velocity that the satellite should have if it was orbiting at that radius (it was about 7000 m/s, using Kepler's law of periods), but wouldn't that be tangential to the Earth?

    Furthermore, how would I consider both the G.P.E. and K.E. ? I really don't get this part.

    Thanks again
     
  7. Apr 20, 2010 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi minifhncc! :smile:
    Yes, PE is defined as (minus) the work done (by a conservative field, such as gravity).

    But how are you going to calcuate the work done in manoeuvring the rocket into orbit?

    Just use the definition of KE, not some long short-cut. :wink:
    ∆KE is 1/2 mv22 - 1/2 mv12, and PE you already know.

    v2 you can calculate from Newton's law of gravitation, the value of g at the Earth's surface, and the formula for centripetal acceleration

    v2 you can calculate from the radius of the Earth and the length of a day (assuming that the launch is, as Filip Larsen :smile: says, from the equator).
     
  8. Apr 20, 2010 #7
    Hi again,

    I understand what K.E. is, but I'm not sure how I'd use the value I get from the change in G.P.E. and the value I get from K.E.

    Would I add them together?

    Wouldn't you mean v1?

    Thanks
     
  9. Apr 20, 2010 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    yes … that's the great thing about the different forms of energy, they're interchangeable, you just add them. :biggrin:
    oops! :redface: :rolleyes:
     
  10. Apr 20, 2010 #9
    Okay thanks for that.

    So, I'd find the orbital speed of the mass at the surface (v=(GM/r)^(1/2)) and then use this as v1? But then how would I account for the length of a day using this method?

    Thanks
     
  11. Apr 20, 2010 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No!! the rocket isn't in orbit on the surface, it's just going with the flow, like you and me. :wink:
     
  12. Apr 20, 2010 #11
    Umm okay, then does that mean v1 = 0?

    Sorry for all these questions that may appear elementary to you, I'm usually okay at Physics, but this one is really doing my head in :(
     
  13. Apr 20, 2010 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No, if it's on the equator, it'll be going round the circumference of the Earth every 24 hours.
     
  14. Apr 21, 2010 #13
    Sorry I'm really not getting this then... :( If I'm not supposed to use v=sqrt(GM/r) as the orbit speed of the rocket at the surface of the Earth, then what value of v1 am I supposed to use?

    How would I take into account the daily orbit of the Earth's rotation (which I'd guess that you'd need to somehow need to use Period, T = 24 * 60 * 60 seconds).

    Again I apologise if this seem really elementary to you, it's really starting to annoy me :(

    EDIT: I've realised that you probably meant, to find v1 that I use v^2/r = 9.8 (where r is the avg. radius of the Earth)... but the question still remains as to how I should take into account the Earth's daily motion...
     
  15. Apr 21, 2010 #14

    Filip Larsen

    User Avatar
    Gold Member

    As seen from outside the earth (from an inertial reference frame not rotating with the earth - like you watching a rotating desktop earth globe) the earth is rotating once per sidereal day (almost 24h) and anything standing still on a point on the equator will make a uniform circular motion. Knowing the rotation rate and circumference (or radius) of this motion you can calculate the tangential speed of the satellite before launch.
     
  16. Apr 21, 2010 #15

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    For the PE you use the difference in GM/r (you've already done that).

    For the KE you use the difference in 1/2 mv2

    v2 is the final speed, which happens to be the orbital speed.

    v1 is the initial speed, which is when the satellite is in your garage.

    Your garage goes round in a large circle every 24 hours … what speed is that?
     
  17. Apr 21, 2010 #16
    Is it just v=(2*Pi*r)/T where r=avg radius of the Earth and T=24*60*60?
     
  18. Apr 21, 2010 #17

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yup! :biggrin:

    (though you may as well take the actual equatorial radius, if you're going to take it to the equator and launch it there :wink:)
     
  19. Apr 22, 2010 #18
    Hey again,

    I've got another question (which hopefully will be the last). For the given question, I got that the total energy required = 6.31*10^9 J could someone please confirm this answer?

    Also, for the speed of the satellite on the Earth, when I use mv^2/r = 9.8, I get v=573 m/s (m=190kg, r = 6.37 * 10^6) and when I use v=(2*Pi*r)/T I get 463 m/s (approx) (r=6.37*10^6, T=24*60*60). Which way is correct?

    I understand that it may be some errors dues to the rounding in the constants given... Could someone please clarify?

    Thanks again :D
     
  20. Apr 22, 2010 #19

    Filip Larsen

    User Avatar
    Gold Member

    I get a result very close to that value.

    The last one (the actual value is around 465 m/s because the rotation period of the earth relative to the stars is around 4 minutes less than 24 h).

    The first relation you mention does not make sense to me. I can't really guess how you derived it, but if you look at the physical dimension of the involved variables you can quickly conclude that is does not represent a physical valid relationship (unless the 9.8 has dimension energy per length :smile:).
     
  21. Apr 22, 2010 #20
    That's reassuring :D

    Ahh sorry. I messed up between centripetal acceleration (which is 9.8 m/s^2 at the surface), and force... In any case I don't think that applies here since the satellite at the surface is not in orbit.

    All good now. Thanks alot :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Energy problem and rotation
  1. Rotation problem (Replies: 3)

  2. Rotational Energy (Replies: 0)

Loading...