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Homework Help: Space Explorer/Orbits Question

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A space explorer comes across a spherical planet with a small moon in orbit around it. He observes that the moon completes one full orbit every 29 days and moves at a distance R = 330 million m from the center of the planet.

    2. Relevant equations

    so there are 4 parts.

    a) was to solve for the planet's mass. easy enough (3.38X10^24 kg)

    b) another satellite orbits in a third of the time the moon did. Find its radius. Again, easy enough, I got 158.6million m

    c) On earth he weighed 190pounds, there he weighs 164. what's the radius of the planet? Again, easy enough, 5170km.

    D) This is where I'm stuck. On the north pole his scale shows 105% what it originally showed. How long does it take to make one complete rotation? I have no clue. I've tried playing around

    3. The attempt at a solution

    Not even sure where to start.
    So Fg=GMm/r^2, so A=GM/r^2. I know for everything, so I can solve for A, which comes to
    Then if it's 105% the original weight, the acceleration due to gravity must be a * 1.05, or 0.0022 m/s^2. But that gets me to a dead end, as I need to solve for the period, T. I have equations relating period to GM (known) and R (known), but that doesn't seem to change at all between where on the planet I am and that's for a satellite orbiting, which doesn't seem to be the case here. Any help would be great. Thanks
  2. jcsd
  3. Dec 1, 2008 #2

    D H

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    What does the bolded part mean? I would presume his weight in part (C). Where did he land in part C? The equator?

    Hint: You weigh about 0.4% more at the Earth's north pole than at the equator. About 21% of that difference is because the Earth isn't a perfect sphere. What explains the other 79% of the difference?
  4. Dec 1, 2008 #3
    Yes, he landed near the equator in C.
    As for your hint, I'd assume it's because the equator is further away from the center of the earth, making R more and thus Fg (weight) less.
    But, without knowing the change in distance from the north pole to the equator, what can I do?
  5. Dec 1, 2008 #4

    D H

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    It's a good idea to supply all of the relevant information.
    No. That is the 21% I already told you about. What about the other 79%?
    You already know the distance. The planet is a perfect sphere per the problem statement.

    Another couple of hints:

    (1) A spring scale does not measure weight tautologically defined as mass*gravitational acceleration. It doesn't measure mass, either. What does a spring scale measure?

    (2) You are to determine the planet's rotation rate: The length of a sidereal day on that planet. What does the planet's rotation rate have to do with what a spring scale registers?
  6. Dec 1, 2008 #5
    My bad, I thought that info was originally supplied.

    [qupte]No. That is the 21% I already told you about. What about the other 79%? [/quote]
    Ah, it's got to be Centripetal force, F=mv^2/r, right? This is used at the equator but at the poles wouldn't be necessary. That's all I can think of. That force would then be subtracted from the gravitational force to give you your weight?

    The force acting on it; the tension in the spring?
    But mass*gravitational acceleration is force, so I'm not sure.
  7. Dec 1, 2008 #6

    D H

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    A spring scale directly measures the compression of a spring. The compression in turn provides a spring force. So you could say that a spring scale measures the force exerted by the spring(s) in the scale. For a person standing still on the scale, this spring force is equal to the normal force needed to keep the person stationary with respect to the planet. If the planet weren't spinning, that normal force would be equal to the gravitational force. The planet, however, is spinning. How does that change the normal force (and hence the reading of the scale)?
  8. Dec 1, 2008 #7
    Wouldn't you subtract the centripetal force from the gravitational force to get the normal force (which would be less at the equator where there's a larger centripetal force)?
  9. Dec 1, 2008 #8

    D H

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    You can arrive at this result the easy way by working in a planet-fixed frame (a frame rotating with the planet) or in an inertial frame. In the planet-fixed frame, the person standing on the scale is stationary. The sum of all apparent forces acting on the person is zero. These apparent forces include gravity, the normal force, and the centrifugal force. Since they sum to zero, the normal force is just the additive inverse of the sum of gravitational force and the centrifugal force.

    In the inertial frame, the person is rotating with the planet. Thus there is some net centripetal force acting on the person. The only forces (in Newtonian mechanics) acting on the person are gravitation and the normal force. These must sum (vectorially) to yield the centripetal force: normal force = centripetal force - gravitational force.
  10. Dec 1, 2008 #9
    Thanks for all the help, by the way. It's really appreciated.

    I'm still doing something wrong though.

    So we know the centripetal force makes a difference of 0.05mg, right?
    Because without it it's 1.05times greater.
    I solved for g, and got 8.457, so mv^2/r = 0.05g, or v^2=0.05rg
    I solve for V, then plug it into T=2piR/V, and got way too small of a number.
    T=2piR/V, and I'm getting the wrong answer.
  11. Dec 1, 2008 #10

    D H

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    You have two problems here. One is a bit nit picky. The planetary radius you calculated for (C) is wrong in light of the information given in (D). I think you might have some other small errors.

    The big error, I suspect is a units problem. The reason I suspect this is because you said you obtained "way too small of a number". Check your units.
  12. Dec 1, 2008 #11
    Ah, I finally got it. Stupid me forgot 5170 was in kilometers, not meters.

    Just out of curiosity though, why was the radius in C wrong? I mean it got me the right answer but they're pretty generous with the range. Assuming a sphere, isn't the radius the same everywhere?

    Yet again, thanks a bunch for the help and be patience. Truly appreciated.
  13. Dec 1, 2008 #12

    D H

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    The answer for C was fine. You weren't told about the rotation until D. However, with this new information in hand you should to refine your estimate of the radius. One way to look at it: With this new information, you should reinterpret question C as asking about a weight of 172.2 pounds rather than 164.
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