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Force Needed to Propel Object to 100m on Moon

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    You have an object of mass 6.154 kg on the surface of the moon. What force is required to propel the object vertically to a height of 100m?

    acceleration due to gravity on the moon (g): 1.625 m/s^2

    2. Relevant equations
    y = (v^2 - v0^2) / (2*g)

    y: vertical displacement
    v: velocity at displacement
    v0: initial velocity
    g: acceleration due to gravity

    y = v0*t - (1/2)*g*t^2

    t: time of flight

    KE = (1/2)*m*v^2

    KE: Kinetic energy

    3. The attempt at a solution
    Weight of object on the moon is 6.154kg * 1.625 m/s^2 = 10 N.

    Using the first equation above with v = 0 (apex of flight) and solving for v0, I find that the required initial velocity is 18.03 m/s.

    The second equation gives a flight time of 22.19 seconds--11.095 seconds to apex and 11.095 seconds back to the surface.

    The third equation yields a required kinetic energy needed of 988.25 J which is the same as the work required by the propulsion system because the object is initially at rest.

    And this is where I get stuck. What equation do I use next? Power? How do I get the propulsive force required?

    Thanks
     
  2. jcsd
  3. Sep 8, 2009 #2

    berkeman

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    Staff: Mentor

    The problem statement is incomplete. It doesn't take a force to get the object moving at its initial velocity, it takes work, which is force through a distance. Is there any mention of a distance that this force is applied through to get the object up to its initial velocity?

    Welcome to the PF, BTW.
     
  4. Sep 8, 2009 #3
    Okay, I knew it wasn't so simple with what I was given. It may actually become very complicated. Let me reveal the operational concept/problem.

    I need to figure out if it is possible to propel an object vertically from the surface of the moon using only electromechanical means. The thought that crossed my mind was a linear solenoid. So:

    Ammended problem statement:
    You have an object of mass 6.154 kg on the surface of the moon. Is there a way to propel the object vertically to a height of 1m, 10m, 50m, 100m without using chemical rocketry, (e.g. by using a linear solenoid) or some other electromechanical means?

    What other information is needed to solve this problem? Let's say that I had a linear solenoid that provided 500N with a stroke of 2cm. Would this work? Is there a general equation that can be derived that is dependent on the characteristics of a linear-solenoid (force output, max stroke length, response time)?

    Thanks
     
  5. Sep 8, 2009 #4

    ideasrule

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    Homework Helper

    If you have a way of continuously applying a force to the mass, the force just needs to exceed the mass's weight for the weight to accelerate upwards until it reaches 100 m. Otherwise, the work done by the force determines the mass's initial speed, and its initial speed determines its final height. Your solenoid does W=500*0.02=10 J of work, not nearly the 988 J needed. That said, it definitely is possible to achieve 18 m/s with an electromagnetic gun. The U.S. military has made a rail gun that achieves seven times the speed of sound with a 3-kg projectile:

    http://en.wikipedia.org/wiki/Railgun
     
  6. Sep 8, 2009 #5
    So are you saying that if I just calculate the work done by an "engine" and have that work be 988 J, I will achieve the desired velocity? This is where I was getting confused. If I increase the stroke to 200cm then the work done by the solenoid would be 1000 J which would be enough? When evaluating different propulsion ideas, is this all I have to look at?

    Thanks
     
  7. Sep 8, 2009 #6

    ideasrule

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    Homework Helper

    I don't know what engine you have in mind, but all of that 988 J must be transferred to the mass as kinetic energy. Waste heat and sound can't propel anything.
     
  8. Sep 8, 2009 #7
    So what would you think to be reasonable for "extra" energy? 10% extra? 5%?
     
  9. Sep 9, 2009 #8

    berkeman

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    Staff: Mentor

    Pffftttt. No chemical rockets, eh? Use a compressed spring. What's the PE of a compressed spring? You don't even need the "electro" part of electromechanical... Done.
     
  10. Sep 9, 2009 #9
    I don't know much about springs, but it seems like I have the same problem in trying to figure out what characteristics of the spring are necessary to produce the required initial velocity of 18.03 m/s. What sort of spring constant and length are you talking about? How do you re-compress the spring? Do I just multiply the rated force of a spring by the length of the spring to get the work done by the spring and compare that with the required KE of 1000 J?

    edit: I guess I'm struggling with the concept of work and KE in practical applications. The last question above sums up the real question I have. Do I just find the work of any application and compare that to the 1000 J needed? Is it that simple? I may come back and then ask how to find the work of different applications.
     
  11. Sep 9, 2009 #10

    berkeman

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    Staff: Mentor

    The nice thing about the spring method is that you just have to equate the PE stored in the compressed spring with the PE that the object has at the top of the arc (when its KE is zero)...
     
  12. Sep 9, 2009 #11
    PE of spring: (1/2)*k*x^2
    PE of object on moon @ 100m: (6.154kg)*(1.625 m/s^2)*(100m) = 1000 J

    So 2000 J = k*x^2.

    Compressing a spring to 30cm, say, means that the spring constant would have to be:
    2000 J / (.3m)^2 = 22,222.22 N/m. Did I do that right?

    This would be a very stiff spring which would required a very heavy metal so I'm guessing that the mass would be more than the 6.154kg that was started with.
     
  13. Sep 9, 2009 #12

    berkeman

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    Staff: Mentor

    Yes, it would be a very strong spring, but at least you have one solution as a baseline. Now you can look more into the railgun concept that was mentioned in post #4 by ideas. I'd add one comment though -- don't just think of a vertical railgun. Consider a horizontal railgun with a 90 degree bend at the end to convert the horizontal velocity into a vertical velocity...
     
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