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Apparent Weight of Planet Problem

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 950.0 N on the earth weighs 919.0 N at the north pole of Planet X and only 855.0 N at its equator. The distance from the north pole to the equator is 1.889×104 Km, measured along the surface of Planet X.

    2. Relevant equations

    3. The attempt at a solution

    a)How long is the day on Planet X?

    Okay so first we need the mass of the astronaught which is we/ge=96.44 Kg. We can then use this to calculate gxequator and gxpole which equal 9.48 m/s^2 and 8.82 m/s^2 respectivly. using the formula g=g0-(v2/r) we can rearrange and solve for v. v=sqrt(gxpole-gxequator*rx) and I get a value of 3530.92 m/s. since v=2pir/T we rearange for T and I get a value of 33614.3 s. Now the answer is incorrect Could anyone please help determine what I'm missing from this question. As always any help is much appreciated.
  2. jcsd
  3. Nov 14, 2008 #2


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    Hi anubis01,

    I can't be sure without seeing the numbers you used, but I believe you used the wrong r value here. (Notice in the original problem they did not give you the radius of the planet.)
  4. Nov 14, 2008 #3
    The value that I Used for my r was 1.889x107m and I assumed that I could just use that value given the wording of the question. Okay so looking at the wording more closely were given a quater the distance around the planet, this is 1/4 the circumference so c=7.556x107. c=pi*2r. r=(c-pi)/2. r=3.778x107. Would this line of thinking be correct or am I still mistaken?
  5. Nov 14, 2008 #4


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    Your line of thinking is right, but:

    c= 2 \pi r \longrightarrow r = \frac{c}{2 \pi}
  6. Nov 14, 2008 #5
    sigh, silly mistakes are going to be the death of me. Thank you very much for all the help sir, your a life saver.
  7. Nov 14, 2008 #6
    Okay I got the answer for part a but I'm not sure about part b. which asks

    If a 4.400×104Kg satellite is placed in a circular orbit 3000 Km above the surface of Planet X, what will be its orbital period?
    alright since we know the velocity of the planet and its period we can use that to calcualte the the mass of said planet which I got to be 1.4315*10^24 Kg.

    Then using the the formula T=2pir2/3/sqrt(Gmplanet) I got the period to be 0.0392. Is this line of thing correct or am I missing something.
  8. Nov 14, 2008 #7


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    I don't think this mass is correct; can you show how you got it?
  9. Nov 14, 2008 #8
    I calculated the mass using the velocity of the planet which was found in the last part of the question v=2817.267m/s I used the formula (v^2*rplanet)/G=Mplanet
    Mplanet=(2817.267^2*1.203x10^7)/6.67x10^-11=1.4315x10^24 Kg

    I then used the formula T=2pir2/3/sqrt(Gmplanet) to find the orbital period of the satellite plugging in the values. T=(2pi(1.203X10^7+3x10^6)2/3)/sqrt(6.67x10^-11x1.4315x10^24)=0.0392 s
  10. Nov 14, 2008 #9


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    That V is the rotation of the planet, not something that is in orbit around the planet.

    You would do better to use the Astronaut's weight at the pole to figure GM.
  11. Nov 14, 2008 #10
    when I use the astronaughts weight at the pole I get following mass.


    I dont think the mass is right because the planet shouldn't have a heavier mass than the earth and yet exert a smaller fg on the astronaught.
  12. Nov 14, 2008 #11


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    It's not just the mass of the planet that effects fg, it's also the radius of the planet; and this planet is larger than the earth.
  13. Nov 14, 2008 #12
    Ah I got it now, thank you both for all the help.
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