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

AI Thread Summary
The discussion focuses on analyzing the gravitational energy changes during a manned spaceflight to Mars, specifically using the equation E=mgh to calculate potential energy. A calculation for the Saturn V rocket shows a gravitational potential energy of 8.93 terajoules at a height of 300 km, but after takeoff, the rocket's mass reduces the potential energy to 386 gigajoules. This energy level poses a risk to human passengers, necessitating design considerations to mitigate danger during takeoff and landing. The conversation also addresses the importance of impulse and force, clarifying the relationship between them and how spreading force over time can reduce potential harm. Overall, the calculations and discussions emphasize the significant changes in gravitational energy and the need for careful engineering in spaceflight.
US-RB
Messages
3
Reaction score
0

Homework Statement



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.


Homework Equations



I'm hoping that I only need to use:

E=mgh


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:

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


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.
 
Physics news on Phys.org
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.
 
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?
 
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.
 
Excellent, thanks. I'll modify my work and feel far more secure now.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top