Period (s) | ||||
Mass (kg) | 1 | 2 | 3 | Average |
0.50 | 0.33 | 0.33 | 0.33 | 0.33 |
0.30 | 0.36 | 0.35 | 0.39 | 0.37 |
0.20 | 0.43 | 0.44 | 0.47 | 0.45 |
0.10 | 0.66 | 0.66 | 0.67 | 0.67 |
0.05 | 0.82 | 0.96 | 1.00 | 0.93 |
As I cannot get my scatter plots to paste here I will turn them in on the test day with the rest of my homework.
Calculations: We were instructed to use our data to calculate the mass of the rubber stopper (the “satellite” in this experiment) and find the percent error as compared to the measured mass of the stopper. Because we used five different radii in our data collection I am going to do five separate calculations of the mass of the stopper and our percent error.
1. Velocity=distance/time
=2´p´0.20 2 / 0.28 = 4.5 m/s 2
Centripetal acceleration=velocity2 / radius
=4.5 2 / 0.20 = 101
Centripetal force=mass ´ centripetal acceleration
(0.20 kg ´ 9.8 m/s)=mass ´101
1.96=mass´101
Mass=1.96/101
Mass=0.019 kg
% Error=(calculated-actual)/calculated
=(0.019-0.026)/0.019
= -0.37
=37% error
2. Velocity=distance/time
=2´p´0.30 2 / 0.34 = 5.5 m/s 2
Centripetal acceleration=velocity2 / radius
=5.5 2 / 0.30 = 100.8
Centripetal force=mass ´ centripetal acceleration
(0.20 kg ´ 9.8 m/s)=mass ´100.8
1.96=mass´101
Mass=1.96/101
Mass=0.019 kg
% Error=(calculated-actual)/calculated
=(0.019-0.026)/0.019
= -0.37
=37% error
3. Velocity=distance/time
=2´p´0.40 2 / 0.43 = 5.8 m/s 2
Centripetal acceleration=velocity2 / radius
=5.8 2 / 0.40 = 84.1
Centripetal force=mass ´ centripetal acceleration
(0.30 kg ´ 9.8 m/s)=mass ´84.1
1.96=mass´84.1
Mass=1.96/84.1
Mass=0.023 kg
% Error=(calculated-actual)/calculated
=(0.023-0.026)/0.023
= -0.13
=13% error
4. Velocity=distance/time
=2´p´0.5 2 / 0.46 = 3.4 m/s 2
Centripetal acceleration=velocity2 / radius
=3.4 2 / 0.50 = 46.2
Centripetal force=mass ´ centripetal acceleration
(0.30 kg ´ 9.8 m/s)=mass ´46.2
1.96=mass´46.2
Mass=1.96/46.2
Mass=0.042 kg
% Error=(calculated-actual)/calculated
=(0.042-0.026)/0.042
= 0.38
=38% error
5. Velocity=distance/time
=2´p´0.60 2 / 0.49 = 4.6 m/s 2
Centripetal acceleration=velocity2 / radius
=4.6 2 / 0.50 = 42.3
Centripetal force=mass ´ centripetal acceleration
(0.20 kg ´ 9.8 m/s)=mass ´42.3
1.96=mass´42.3
Mass=1.96/42.3
Mass=0.046 kg
% Error=(calculated-actual)/calculated
=(0.046-0.026)/0.046
= 0.43
=43% error
Conclusion: The conclusion from this lab is that the radius of a satellite’s orbit is directly related to its period and that the centripetal force is inversely related to the period of a satellite’s orbit. My aha moment was when I plotted the date we collected and saw more clearly the relationship between the two variables. I learned that the escape velocity of a satellite with a circular orbit is ≥8km/s and <11.2 m/s and that the escape velocity of a satellite with an ellipse orbit is ≥11.2 m/s. Finally, I learned that centripetal force is the force directed toward the center of a body’s orbit and centrifugal force dragging a body away from the center of rotation and is equal and opposite to the centripetal force.
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