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Finding the astronaut's weight on a planet's surface

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data

    A landing craft with mass 1.22×10^4 kg is in a circular orbit a distance 5.50×10^5 m above the surface of a planet. The period of the orbit is 5100 s . The astronauts in the lander measure the diameter of the planet to be 9.50×10^6 m . The lander sets down at the north pole of the planet.
    What is the weight of an astronaut of mass 84.1 kg as he steps out onto the planet's surface?

    2. Relevant equations
    circular motion speed: v = √GM/r
    g: GM/r2
    T = 2πr/v

    3. The attempt at a solution
    I tried to solve for the speed by using the period.
    5100 = 2π(5.50*10^5)/v
    v = 677.6 m/s
    and using the circular motion speed equation, I tried to solve for M, where the radius i used is the radius of the planet:
    677.6 = √6.67*10^-11*M/4.75*10^6
    M = 3.27*10^22

    and then i used GM/r^2 to solve for G.

    6.67*10^-11*3.27*10^22/(4.75*10^6) and i got a really small g. I think I'm missing something. Especially because I haven't used the space satellite's mass.
     
  2. jcsd
  3. Dec 11, 2016 #2

    TSny

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    In the equation T = 2πr/v, what does r represent? What does the distance 5.50×10^5 m represent?
     
  4. Dec 11, 2016 #3
    That distance represents how far the landcraft is hovering above the planet
     
  5. Dec 11, 2016 #4

    TSny

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    That's what the number 5.50 x 105 m represents. But the r in 2πr/v is a different distance.
     
  6. Dec 11, 2016 #5
    so the r should be the radius of the planet instead of the distance the landing craft is from the planet?
     
  7. Dec 11, 2016 #6

    TSny

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    No. Think about where the formula T = 2πr/v comes from. The craft moves at constant speed while in orbit. For something moving at constant speed, distance = speed x time. Or, time = distance/speed. Apply this to one orbit of the craft so that the time is equal to the period T:

    T = distance/v. Here, the distance is how far the craft travels in one circular orbit.
     
  8. Dec 11, 2016 #7
    so would the r be the planet's radius + landing craft's distance?
     
  9. Dec 11, 2016 #8

    TSny

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    Yes. r is the radius of the craft's orbit.
     
  10. Dec 11, 2016 #9
    can i use > 5100s = 2π(r1+r2)/√GmM/(r1+r2) to solve for M ?
     
  11. Dec 11, 2016 #10

    TSny

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    Yes, except why does m appear on the right side?

    Nit picky note: To avoid confusion with the square root symbol √, use parentheses to show what belongs inside the square root. If I write √ab, then it is not clear if both the a and the b are inside the square root. Most people would interpret √ab to mean that only the "a" is inside the square root. If you want to indicate that both a and b are inside, then you should write √(ab).
     
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