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A simple change in gravitational energy, but I want to be certain.

  1. Jan 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Welcome to the Australian Physics course! I have to analyse the effects of changes on gravitational energy and radiation from the Sun on a manned spaceflight to Mars. I doubt my abilities to solve simple problems, however, so I now turn to the vast collective of the Internet to aid me. I simply want to know if my calculation for the potential energy of my hypothetical rocket is correct. I seek a quantitative representation of the danger associated with takeoff and landing in this situation, so I have used the equation for gravitational potential energy to try and find a really big number that represents certain death.


    2. Relevant equations

    I'm hoping that I only need to use:

    [itex]E=mgh[/itex]


    3. The attempt at a solution

    Why do I say 'Welcome to the Australian physics course'? It seems that other countries have far less text in their physics than we do. As a result, I offer the TL;DR version, where I take the mass of the Saturn V rocket and assume low-Earth orbit at 300km, using normal Earth acceleration:

    [itex] E=mgh[/itex]
    [itex] E= (3039000)(9.8)(300000)=8.93 \times 10^{12} J[/itex]


    Alternatively, you may read the full article in which this calculation features.

    The changes in gravitational energy involved in a manned spaceflight to Mars are immense. Consider the Saturn V rocket, with a mass of just over three million kilograms. It is known that gravitational potential energy is given by E=mgh, so if this rocket is fired into a low-earth orbit of 300km, it will have a gravitational potential energy of 8.93 terajoules (8.93x10^12 J), only counting the height of the rocket from the surface of the earth. The total mass of the rocket after it had completed the takeoff process, however, was only 131,300kg. This new mass gives only 386 gigajoules (3.86x10^11 J) of gravitational potential energy. Subjecting a human being to this amount of energy would kill the person, so the rocket must be designed to protect any passengers onboard. Luckily, this impulse is being distributed over a long period of time, as the sheer size of the rocket poses a large inertia. To accelerate this mass immediately would be inefficient and painful for anyone close enough to see the rocket launch. This is using the definition of impulse as I=F/dt. The huge force of the rocket is spread over a large time, reducing the impulse, which reduces the potential of damage to the crew. So the changes in gravitational energy are immense, but the issue is resolved by reducing the acceleration of the rocket so that the related impulse is reduced. This reduction of the acceleration is performed both on takeoff from Earth and entry to Mars. There is also an issue of the crew’s bones weakening, due to the lack of significant gravity in space. Centrifuges are used to combat this, creating a force equal to gravity on the crew member’s body.
     
  2. jcsd
  3. Jan 3, 2012 #2

    BruceW

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    GPE=mgh is approximately correct for low earth orbit. But there is about 5% error in your answer due to this approximation. So your calculation is roughly right. For a more accurate answer, you should use Newton's law of gravitation.

    About the main article, It think you've got force and impulse mixed up. Impulse is the change in momentum and so impulse divided by time equals the average force, not the other way around.
     
  4. Jan 3, 2012 #3
    Bruce, I believe you are correct here. I just looked through my book and saw I=FΔt. So if I revised my wording to be along the lines of 'Force is reduced as time increases, as force is proportional to the product of impulse and the inverse of time', would I now be correct?
     
  5. Jan 3, 2012 #4

    BruceW

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    yep, pretty much. although technically, it is impulse divided by a period of time which gives the average force. So when you say 'force is reduced as time increases', this may cause confusion, because it sounds like you're talking about instantaneous force.
     
  6. Jan 3, 2012 #5
    Excellent, thanks. I'll modify my work and feel far more secure now.
     
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