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Gravitation and planetes: exercise

  1. Sep 22, 2011 #1

    RHK

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    This is my first topic. I'm sorry for my english: that is not my language.
    On the planet A, sferically symmetric and with no atmosphere, an astronaut on one pole throws in vertical upwards, a little ball giving to it a certain velocity: this ball reaches the maximum height of 25 cm.
    The same thing is done on another planet, B, equal in mass to the planet A, and then the ball reaches the maximum height of 4 m.
    Requests:
    (i) to calculate the planetes radii ratio
    (ii) to calculate the planetes densities ratio, with the hypothesis that the planets are uniform.

    Successively, the astronaut make the same experiment on the B planet, this time on the equator, verifying that the ball reaches the height of 8 m.

    (iii) calculate the rotation period of the planet, supposing that the mass of the planet is 10^27 g and its radius is 5000 km.

    Temptative of solution:
    (i) We know that E=mgh, so h=E/mg. Writing down the same equation for the two planetes, the following ratio is obtained: (hA/hB)=(R_A/R_B)^2 = 1/16

    (ii) The density rho is writable as rho=M/V, where V=(4pi)/R^3, than (rhoA/rhoB)=(R_B/R_A)^3=64

    Is this right? And then, how can i proceed for the third point?
    Thanks a lot.
     
    Last edited: Sep 22, 2011
  2. jcsd
  3. Sep 22, 2011 #2
    the following ratio is obtained: (hA/hB)=(R_A/R_B)^2 = 1/16
    ......................

    chech the working in obtaining the above. I do not think it is correct if the densities of the two planets are not the same.
     
  4. Sep 22, 2011 #3

    Ryan_m_b

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    Staff: Mentor

    Thread moved to homework forums.
     
  5. Sep 22, 2011 #4

    RHK

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    What do you mean? How can density enter to play?
    Thanks
     
  6. Sep 22, 2011 #5
    If you show how you got the ratio of the heights then I can try to find if the method is correct.
     
  7. Sep 22, 2011 #6

    RHK

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    This is my temptative for the first point:

    E_A=m*g_A*h_A, so: h_A=E_A / m*g_A

    E_B=m*g_B*h_A, so: h_B=E_B / m*g_B

    Then it's possible to write: h_A / h_B = g_A / g_B

    where g=(G M)/R^2 and so: (hA/hB)=(R_A/R_B)^2 = 1/16
    What do you think about?
     
  8. Sep 22, 2011 #7
    Note that M is the mass of the planet. but there are TWO planets and you have to distinguish between them - planet A and planet B.
     
  9. Sep 22, 2011 #8

    RHK

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    I'm sorry: in the exercise text above the two planets masses are equal.
    I've fixed the text.
     
  10. Sep 23, 2011 #9
    Then I think that i and ii are correct.
    As regards iii note that the rotation of the planet causes the acc due to gravity to reduce from that at the pole.
    Use F = ma where a is the centripetal acc of a point on the equator towards centre of planet.
     
  11. Sep 23, 2011 #10

    RHK

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    Yes, i also thought at that.
    But i can not start the problem: F=ma for the centripetal acceleration, where a=R*w^2 and w is related to the period. But i can not proceed...
     
  12. Sep 23, 2011 #11
    Is not the centripetal acc the reason that the acc due to gravity reduced by half from that at the pole?
     
  13. Sep 23, 2011 #12

    RHK

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    Yes, of course it is.
    But i don't get the point: I have to build another system with two equation, one for the pole and another one for the equator?
     
  14. Sep 23, 2011 #13

    RHK

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    Is it right to suppose a new system, with other two heights (a system for the same planet, with different heights at the pole and the equator)?
    In this case i think that at the equator the acceleration is g+a_c isn't it?
     
  15. Sep 23, 2011 #14
    Since g[itex]_{equator}[/itex] = (1/2)g[itex]_{pole}[/itex]
    then g[itex]_{pole}[/itex]= 2R[itex]\omega[/itex][itex]^{2}[/itex]
     
  16. Sep 24, 2011 #15

    RHK

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    I don't understand just a thing: the acceleration at the equator is ONLY the centripetal one? Why?
     
  17. Sep 24, 2011 #16
    g[itex]_{equatir}[/itex] = GM/(R[itex]^{2}[/itex]) - R[itex]\omega[/itex][itex]^{2}[/itex]
     
  18. Sep 24, 2011 #17

    RHK

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    That's ok :)
    Thanks a lot
     
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