Results 1 to 9 of 9

Thread: Another AP Physics question

  1. #1
    Join Date
    Apr 2007
    Location
    Michigan -.-
    Posts
    1,357
    Mentioned
    2 Post(s)
    Quoted
    4 Post(s)

    Default Another AP Physics question

    Well here I am again, with another question because the people in my class are retarded and cant help me on our class forums lol...

    So here it is:

    Energy is required to move a 1090 kg mass from the Earth's surface to an altitude 2.01 times the Earth's radius (6.37 x 10^6 meters). What amount of energy is required to accomplish this move? Hint: G and Me can be eliminated in favor of parameter of the gravitational acceleration at the surface of the earth; i.e., the acceleration of gravity on Earth is 9.8 m/s^2. Answer in units of J.

    Any help on this would be appreciated since I have tried 4 times already and have 3 guesses left

    Thanks in advance!
    METAL HEAD FOR LIFE!!!

  2. #2
    Join Date
    Sep 2006
    Location
    West U.S.
    Posts
    2,172
    Mentioned
    0 Post(s)
    Quoted
    6 Post(s)

    Default

    oh man!

    this is at the top of my head!

    if only my AP Physics teacher emphasized more on the equations rather than the "concepts".... :doh:

    They are sisters...
    Runescape Classic

  3. #3
    Join Date
    Apr 2007
    Location
    Michigan -.-
    Posts
    1,357
    Mentioned
    2 Post(s)
    Quoted
    4 Post(s)

    Default

    Quote Originally Posted by legendaryhero90 View Post
    oh man!

    this is at the top of my head!

    if only my AP Physics teacher emphasized more on the equations rather than the "concepts".... :doh:
    Well lemme know if it comes to you then
    METAL HEAD FOR LIFE!!!

  4. #4
    Join Date
    Jul 2007
    Location
    St. Louis, Missouri, USA.
    Posts
    575
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    Working on it. Will edit in a little with the answer. (If I can figure it out)

    Edit:

    Actually we learned this at the beginning of the year.

    mgh = PEg (mass x gravity x height = Potential Energy of gravity)

    so
    1090kg x 9.8m/s x 12200700meters = approximately 1.3 x 10^11 J.

    Your welcome.
    -You can call me Mick-



  5. #5
    Join Date
    Dec 2007
    Location
    Somewhere in Idaho
    Posts
    480
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    Work = the integral of force over a distance. F = G*m1*m2/r^2 where G is the gravitational constant. So basicly the only thing that is changing here is the distance you are from the earth.

    so, (I wish I could do symbols) Work (energy) needed = G * m1 * m2 * integral from radius of the earth to 2.01 x the radius of the earth of dr / 1 / r^2 which gives us Energy = (-(G * m1 * m2) / r) from radius of the earth to the outer shell. Giving you Energy = (-G * m1 * m2) / (2.01 * radius of the earth) + (G * m1 * m2) / radius of the earth.

    Of course, G is the gravitational constant, m1 is the mass of the earth, m2 is the mass of your object. Make sure you plug in SI units (kg, meters, ect) Don't know what the hint is all about, but this should work (let me plug it into my calculator to give you an exact result)

  6. #6
    Join Date
    Dec 2007
    Location
    Somewhere in Idaho
    Posts
    480
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    Quote Originally Posted by mickaliscious View Post
    Working on it. Will edit in a little with the answer. (If I can figure it out)

    Edit:

    Actually we learned this at the beginning of the year.

    mgh = PEg (mass x gravity x height = Potential Energy of gravity)

    so
    1090kg x 9.8m/s x 12200700meters = approximately 1.3 x 10^11 J.

    Your welcome.
    This answer works if you are close to the surface of the earth, however, you are not (gravity will vary as you move farther out) my answer will be more exact.

    I get 6.89047244 × 10^10 J Sig. Figs puts it at 6.9 x 10^10 The formula I used is correct but I may have entered some numbers in wrong, so you'll have to check that.

    *edit, helps if you give the calculator the correct numbers*

  7. #7
    Join Date
    Jul 2007
    Location
    St. Louis, Missouri, USA.
    Posts
    575
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    Quote Originally Posted by boberman View Post
    This answer works if you are close to the surface of the earth, however, you are not (gravity will vary as you move farther out) my answer will be more exact.

    I get 1.37120402 × 10^15 Sig. Figs puts it at 1.4 x 10^15 The formula I used is correct but I may have entered some numbers in wrong, you you'll have to check that.
    Yeah, you're right. I was assuming gravity was constant. I have no idea how to solve it otherwise. But theres no way that as gravity decreases you would get a higher amount of energy to move it (i.e. my number 10^11 is smaller than 10^15)
    -You can call me Mick-



  8. #8
    Join Date
    Dec 2007
    Location
    Somewhere in Idaho
    Posts
    480
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    Quote Originally Posted by mickaliscious View Post
    Yeah, you're right. I was assuming gravity was constant. I have no idea how to solve it otherwise.
    The equation I used above will do it, You'll need to use Calculus to solve this problem though. (not hard calculus, just calculus). So it is the integral of the change of force over a distance that gives you work done.

    To get the force as a function of distance look up Gravitational Constant in the wikipedia. They tell you all you need to know.

    Oh, yeah about that, I was unsure that the difference would be that big either, I got my result by plugging the numbers into google (couldn't find my TI-86) so your answer could very well be in the right magnitude. But Im very certain mine will be more accurate (with correct numbers of course )

    Found it, It is actually quite a big difference, (my problem was that I did a 10^2 instead of 10^6 in one of my terms) so the REAL answer is 6.89047244 × 10^10 J *Edits first answer)

  9. #9
    Join Date
    Apr 2007
    Location
    Michigan -.-
    Posts
    1,357
    Mentioned
    2 Post(s)
    Quoted
    4 Post(s)

    Default

    thanks for the help guys, but i said screw it at about 11:00 and just went to bed =/
    METAL HEAD FOR LIFE!!!

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. AP Physics question :(
    By gerauchert in forum News and General
    Replies: 14
    Last Post: 12-03-2007, 03:06 AM
  2. Question?
    By SocCerSporTie in forum SRL Site Discussion
    Replies: 7
    Last Post: 10-28-2007, 02:55 PM
  3. Need help on a physics question
    By NxTitle in forum News and General
    Replies: 8
    Last Post: 09-15-2007, 12:11 AM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •