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Homework Help: Power and Satellites

  1. Apr 7, 2006 #1
    Hello I need osme help with these two problems. I was told the ansers but do not undersatnd how the answers were gotten.

    1. The power dissipated in lifting a 2.0 kg ball vertically at 3.0m/s equals what?
    The answer is 59 W. ALl I know is that power equals Work/ delta T but there is no time for this one.

    2. Increasing the orbital radius of a satellite by a factor of 2 requires its period to change by a factor of what?

    For thi I know R^3/T2=R^4/T^2.

    I would appreciate any help.
     
  2. jcsd
  3. Apr 7, 2006 #2

    Astronuc

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    In the first problem:

    Work = Force * distance (and actually it is an itegral) but let's assume the force is constant.

    Power = d/dt (Work) or d/dt (Energy).

    Now, if a mass m, which is subjected to gravity of acceleration g, is moved up a distance, h, it experiences a change in potential energy of mgh. Now if it moves at constant velocity, v, then v = h/t, where t is the time it takes to move a distance h.

    In the second problem, be sure the formula is correct:

    See - http://en.wikipedia.org/wiki/Orbital_period#Calculation

    http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion
     
  4. Apr 7, 2006 #3

    BobG

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    Your second one is wrong.

    Kepler's Third Law: The square of the period is proportional to the cube of the mean distance.

    In other words:
    [tex]t = C a^{\frac{3}{2}}[/tex]

    If the radius is doubled, then you have:

    [tex]t = C (2a)^{\frac{3}{2}}[/tex]

    C is just some constant, since Kepler's law just explains the relationship between the period and the average radius. However, C is equal to:
    [tex]C = \frac{2 \pi}{\sqrt{G M}}[/tex]
    where G is the universal gravitational constant and M is the mass of the object the satellite is orbiting.

    For Earth:
    [tex]G M = 398600.4418 \frac{km^3}{sec^2}[/tex]

    If your satellite's radius was 42,164 km, your orbital period would be:

    [tex]t=\frac{2 \pi}{\sqrt{398600.4418 \frac{km^3}{sec^2}}} * (42164 km)^{\frac{3}{2}}[/tex]
    which, coincidentally, happens to equal the number of seconds required for the Earth to rotate 360 degrees.
     
  5. Apr 7, 2006 #4
    Hello,

    The answer to the first question is much more simple than what Astronuc said. (He is not wrong at all though, but he uses big artillery for a really simple problem in my opinion :smile: ).

    The power dissipated in lifting a 2.0 kg ball vertically at 3.0m/s equals what?
    Here is my answer:

    In this exercice, the only force that apply on the ball is the weight P=mg.
    We know that: Power = Work / Time = (Force * Distance) / Time

    With the infos provided, the simplest thing is to take 1 second for time:
    Power = (2 * 9.81 * 3) / 1 = 58.86 # 59 W

    You could have taken any amount of time, provided you calculate the associated distance.

    What astronuc said is more of the general case, but is still totally true though.

    I hope I could help,

    Cyril

    PS: Sorry if there are some english mistakes, I'm not a native english speaker ^^
     
  6. Apr 8, 2006 #5
    Or, if the force and velocity are constant, P=Fv. But, of course, that was his end result anyway. :smile:

    -Dan
     
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