course phys 241
Very good.
Physics 241
Experiment 26
Distance Velocity `d velocity
1.2 4000 burns -
1.4 5700 burns 1700 burns
1.6 6600 burns 1100 burns
1.8 7300 burns 700 burns
2.0 7700 burns 400 burns
Height reached Velocity
1.2 4000 burns
1.4 5700 burns
1.6 6600 burns
1.8 7300 burns
2.0 7700 burns
For this experiment, I had to determine the velocity required to reach a height of 1.2 earth radii using a simulation. The impulse of the burn required to reach that height is 4000 burns. To reach the height of 1.4 earth radii, it took a velocity of 5700 burns, which is significantly less than twice what it took to travel 1.2 earth radii. I continued to experiment with different radii until I found how fast the object would have to travel to reach heights of 1.6, 1.8, and 2.0 earth radii. It is very clear that it takes less and less velocity for the object to reach higher distances. In the next part of the experiment, I determined the kinetic energy required to reach the heights. With a mass M, I used the equation KE=.5*m*v^2. These energies are what it took to move the projectile to the given height. As the calculations show, it took twice as much energy to move the projectile from 1.2 earth radii to 1.4 earth radii, but it takes less than twice the energy to move the projectile from 1.4 earth radii to 1.8 earth radii. From the graph below, it appears that the graph is increasing at a decreasing rate.
The kinetic energy required to send the object to the given height produces a similar graph to the graph of velocity vs distance.
Height reached Kinetic Energy
1.2 8,000,000(M)m/s^2
1.4 16,245,000(M)m/s^2
1.6 21,780,000(M)m/s^2
1.8 26,645,000(M)m/s^2
2.0 29,645,000(M)m/s^2
Height reached `d KE
1.2 8,000,000(M)m/s^2
1.4 8,245,000(M)m/s^2
1.6 5,535,000(M)m/s^2
1.8 4,865,000(M)m/s^2
2.0 3,000,000(M)m/s^2
As the graph displays, the kinetic energy required is also increasing at a decreasing rate. The kinetic energy required to move from 1.0 to each increment is displayed in the next table. Each of these energies can be found on the KE vs distance graph by determining the rise from one distance to the next. For the projectile to go from 1.4 to 1.8, it would take 9,900,000 (M)m/s^2, which could be found on the graph by determining the rise from 1.4 to 1.8.
If an object was “dropped” from a distance 2 radii from the center of the earth, its initial gravity would be one quarter of what it would be on earth. As the object fell closer and closer to the target 1.6 mark, the gravitational pull would increase. The pull of gravity determines the velocity, an in turn the velocity squared times the mass determines the kinetic energy. An object actually has more PE at 1.6 earth radii than 2.0 earth radii since the pull of gravity is stronger as the object gets closer to the earth. This illustrates conservation of energy because it takes more energy to keep a satellite for example orbiting above the earth at 1.6 radii than it would take to keep a satellite at 2.0 earth radii, because the force acting against the satellite pulling it back towards earth is greater at 1.6 than at 2.0.
To get the projectile to maintain a circular orbit at 1 earth radii, it had to be launched at 1.57 radians, with a burn rate of 7800. The following table depicts the burn rate required to maintain circular orbits at distances of 1.2, 1.4, 1.6, 1.8, and 2.0 earth radii. To determine the rate required, the starting point is entered into the program, I then estimated what the velocity should be, decreasing at each increment as I moved away from the center of the earth. I then determined the kinetic energy required to maintain that path around the earth. The following table shows the results of the simulation.
Distance Burn rate `d burn rate Kinetic Energy `d KE
1.0 7800 - 30,420,000(M)m/s^2 -
1.2 7200 600 25,920,000(M)m/s^2 4,500,000(M)m/s^2
1.4 6700 500 22,445,000(M)m/s^2 3,475,000(M)m/s^2
1.6 6275 425 19,687,812.5(M)m/s^2 2,757,187.5(M)m/s^2
1.8 5900 375 17,405,000(M)m/s^2 2,282,812.5(M)m/s^2
2.0 5600 300 15,680,000(M)m/s^2 1,725,000(M)m/s^2
This table shows the kinetic energy required at each orbit, and the change in energy as the object moves further and further away. The changes in potential energy are all positive, though they are increasing at a decreasing rate. The reason that the potential energies are increasing is because PE=m*g*d. The d is increasing, so even though g is decreasing, the PE increases. At a certain point, the g will be so small that the PE will decrease until there is no PE.
If a satellite is going to change altitude from 1.0 to 1.2, it must fire its thrusted directly perpendicular to the gravitational force at a rate where the satellite stabilizes at the new orbit.
For the next part of the experiment, I will determine which of the following values, when graphed vs KE, is linear. This would determine the appropriate proportionality.
Distance KE 1/r 1/r^2 1/r^.5
1.0 30,420,000(M)m/s^2 1 1 1
1.2 25,920,000(M)m/s^2 .83 .69 .91
1.4 22,445,000(M)m/s^2 .71 .51 .85
1.6 19,687,812.5(M)m/s^2 .63 .39 .79
1.8 17,405,000(M)m/s^2 .56 .31 .75
2.0 15,680,000(M)m/s^2 .5 .25 .71
These three graphs depict, from left to right, KE vs 1/r, 1/r^2, and 1.r^.5. It appears to me that the graph of KE vs. 1/r^.5 is the closest to linear. Since Ke is proportional to r^2, than it makes sense that k=v/r^.5.
In an elliptical orbit, the object moves fastest just as it approaches the planet. The KE is the lowest at the point furthest from the planet. This means that the PE is the highest at this point. These observations are consistent with the theory of the conservation of energy.
In order to get a space shuttle into a circular orbit, I speculate that it uses just enough fuel to get to the point where its velocity and radius are consistent with the circular orbit in the simulation. Then, all that it would need is the gravitational pull of the earth to maintain its path. To get a space craft to the moon, you only need enough energy to get out of the gravitational pull of the earth, and let the moons gravitational pull take effect. Since the moon is so far from the earth, one would have to use the earths gravity to “slingshot” the spacecraft around the earth and then expel thrust to leave that pull towards the moon.