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Anyone good at physics? If so, I have a few questions...

Discussion in 'Fred's House of Pancakes' started by eddiehaskell, Nov 3, 2008.

  1. eddiehaskell

    eddiehaskell Member

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    I have a test tomorrow and I'm trying to study, but a few questions have me stumped. I'd be eternally grateful if anyone could explain them to me.:) Here goes...

    1. A pilot makes an outside vertical loop of radius 320m. At the top of his loop he is upright and pushing down on his seat with only one-half of his normal weight. How fas is he moving?

    and

    2. A satellite orbits the earth once every 6.0 hours in a circle. What is the radius of the orbit if the satellite has a mass of 2000 kg?
     
  2. EZW1

    EZW1 Active Member

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    I don't recall the question off the top of my head, but if you'' look up the equasion for centripetal motion, that will do it for you.

    Essentially what it means is, an object in motion will continue on a straight line until acted upon by another force. As a satellite orbits the earth, gravity tries to constantly pull it down. When its forward motion equals the force of gravity, it will orbit instead of falling back or shooting off into space. The same forces are in effect for the airplane, but you have one more variable to consider: he is pushing down at 1/2 his weight (this is a hint).
     
  3. Betelgeuse

    Betelgeuse Active Member

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    So, what are your questions about these questions? The only way that we can actually help you is if we know where your confusion lies.
     
  4. carz89

    carz89 I study nuclear science...

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    Yes, there are common physics equations for these. I don't have them handy.

    I do know that #2 doesn't depend on the mass at all. Any object in orbit at X miles radius above the earth must travel at the same speed. Think of it this way: all objects in orbit are actually falling toward earth, and we know that all objects fall at the same rate of acceleration towards earth (neglecting atmospheric drag).
     
  5. eagle33199

    eagle33199 Platinum Member

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    I know there are equations for these situations that are just plug and play, but since it's been 5 years since i've taking a physics class, i have to start at the beginning.

    F=MA, and an appropriately drawn free body diagram for each situation.

    1. You know that the Pilot's downward force on the plane seat is 1/2 what it normally is. His mass is constant. Thus at the very top of the loop, the change in vertical acceleration of the plane must be exactly 1/2 standard earth gravity - roughly 4.9 m/s^2.Since this is talking about a moment in time, it seems that some calculus would be in order... At least, that's how i'd go about setting up the problem.

    2. Again, F=MA and a free body diagram. The downward force on the object is obvious, so all you have to do is balance this with the outward force of the object as it travels around the earth. That should give you a required speed for the object - take that speed and the 6 hours to figure out the total distance traveled in one loop, and you've got the circumference of the circle - from that the radius.

    Those are how i would approach the problem... we'll see if i can come up with some answers here (mostly because i excel at these types of problems).
     
  6. eagle33199

    eagle33199 Platinum Member

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    since 2 is easier (in my mind) i'll start there...

    mass = 2000kg
    full loop time = 6 hours = 21600s
    radius of orbit = ?
    downward acceleration = 9.8 m/s^2

    For this, we can use a few equations:

    T=2*pi*R/v
    T=2*pi*v/a

    We know a and T, so we can solve for v in the second equation:
    v=T*a/(2*pi) = 21600*9.8/(2*3.14159) = 33689.95m/s

    Subbing that into the first equation, we can solve for R:
    R=T*v/(2*pi) = 21600*33689.95/(2*3.14159) = 1.158*10^8m

    or as a final answer, 115,800km (i think).

    huh... guess i didn't use the mass there. go figure!
     
  7. eagle33199

    eagle33199 Platinum Member

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    Now question 1 doesn't seem as complicated. if we use those same equations:

    T=2*pi*R/v
    T=2*pi*v/a

    R = 320m
    a = 4.9m/s^2 (1/2 of 9.8)

    set those equations equal to each other:

    2*pi*R/v=2*pi*v/a
    R/v = v/a
    v^2 = R*a
    v^2 = 320m * 4.9m/s^2 = 1568m^2/s^2

    so your final answer would be v = 39.5979797 m/s
     
  8. eagle33199

    eagle33199 Platinum Member

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    So in conclusion, it seems that i initially tried to make things a lot harder than they needed to be... once i found the equations (wikipedia ftw!) things became a lot simpler.
     
  9. qbee42

    qbee42 My other car is a boat

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    Force = V^2/R

    That's the formula for centripetal acceleration.

    Tom
     
  10. SageBrush

    SageBrush Senior Member

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    1) Weight = mass * acceleration
    If his weight is half, and he has not lost any mass, then net acceleration is half, or g/2.
    Gravity has not changed, so an upward acceleration of g/2 must exist.

    This leads us to: g/2 = v^2/r