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Star Trek Physics

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Artificial Gravity and General Relativity The Principle of equivalence states that ‘Gravity is the same thing as acceleration.’ Anything in free fall in a gravitational field would be unable to feel its own weight. Anything accelerating in space at the value of 9.81 metres per second per second acts in the same way as an object on the surface of the Earth.

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Weightlessness as same acceleration as the lift Reaction force due to the acceleration of the lift. v

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Death Star – Artificial Gravity The radius of the space station itself is about 70km. To find the mass I re-arranged the formula g = Gm/r² to give m = gr²/G with g being the value of gravity on Earth (9.81) and G being the gravitational constant (6.67x10-11). Using this formula I got the value of m to be 7.21x10² ⁰ kg. I then moved on to working out the death stars’ density. Assuming the station is spherical, the equation for volume is 4/3(pi x r³). This gave me the value of 1.44x Using the formula density = mass/volume I got the answer as 5.0x10+5 kgm-3. The density of lead is 11.34x10+3kgm-3 so this is not high enough to provide the artificial gravity. A more dense material that exists is the neutron star. This particle has a density of 3.99x10+17lgm-3. This is exactly what we need. To find out the volume of the matter we need I re-arranged the equation to give volume = mass/density. This gave me 1.8x10+3m³. Finally, the radius of the matter needed comes from the re-arranged volume of a sphere formula. The cube root of the volume divided by four thirds times pi will give us the radius of the matter needed and this came out conveniently as 7.5m. So, if we stick this in the middle, it will give us our artificial gravity.

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Fusion Generator to Provide energy for Impulse Drive The power needed to get to alpha centauri is 1e16W To get fusion to occur, the temperature must be high enough to overcome the repulsive force between the two particles. KE lost by particles=PE gained by electric field PE gained= electric potential x charge=4.79e-13J From Kinetic Theory, average KE=3/2 kT So Temp needed to get KE=2.32e10K Core Temp of Sun=15.7e6K Surface temp=5778K Reason for difference is to do with quantum tunneling Conclusion: Not feasible to get to alpha centauri at this time as current fusion generators do not generate enough power and in order for quantum tunneling to work properly, we would need a generator the size of the sun.

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Artificial Gravity: “Wheel” Space Station How fast would the station need to rotate to maintain g with 1000 passengers & crew? Modelled the ring as 1 strip 5300m long which = circumference. Then Circ. = 2πr to find r Used mv²/r = mg Therefore v = 90.97ms¯¹

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Antimatter Impulse Drive Calculation of the amount of antimatter needed to take the space station to alpha centauri

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Distance to Alpha Centuri : 4.37 light years 1 light year = 9.46x10^15m therefore 4.37x9.46x10^15m = distance to star Rearranging v = U+at Gives t=v-u/a Add in the values for each Gives 75x10^5s to get to the maximum velocity To get the distance remaining we use 0.5at^2 This gives us the distance travelled during the acceleration which is 2.04x10^6 To calculate the distance remaining we use 0.5xdistance to star – the distance travelled accelerating We then need to calculate the time spent accelerating + the time at a constant velocity to give the time to get to the decelerating point And the total time for the journey is given by the time x 2 Which came out as 18 years Using f=ma we can calculate the force needed which came out as 37x105N This will allow us to calculate the change in momentum for the flight by multiplying the velocity by the force which came to roughly 2.78x10^13kgms-1

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