In this coursework I am going to investigate the relationship between the orbital period and the distance of the planet from the sun. Assume that this relationship is a power law of the form: T = KR^n
Extracts from this document...
Introduction
Zahra Balal A/S Use of Maths
Planetary Motion
Introduction:
The German astronomer Johann Kepler studied the relative motion of the planets and discovered a relationship between their orbital periods and their means distance from the sun
Aim:
In this coursework I am going to investigate the relationship between the orbital period and the distance of the planet from the sun.
Assume that this relationship is a power law of the form:
T = KR^n
T = the time for full cycle around the sun.
R = Mean distance of the planet from the sun.
K and N are constant.
K = is the gradient
Middle
Saturn
1427
3.154423973
10753
4.031529646
Uranus
2870
3.457881997
30660
4.486572151
Neptune
4497
3.652922888
60150
4.779235632
Pluto
5907
3.771366971
90670
4.957463616
After calculating the all of the logs I have insert them into excel spreadsheet and displayed the equation of the line.
Now that you have the equation y = 1.5001x – 0.7003
I will transfer it to T = KR^n
1.5001 is the gradient, which is n
-0.7003 is the y intercept (c), which is log K but you should
Conclusion
2870
30680.93802
30660
4497
60179.6453
60150
5907
90599.87385
90670
Percentage error:
((Actual data -Model) /Actual Data)*100
((88-88.10860661)/88)*100
Model | T | |
T = 10^-0.7003 x (R)^1.5001 | Actual Data | Predicted Error |
88 | 88 | 0% |
224 | 225 | 0.444444444% |
366 | 365 | 0.273972603% |
687 | 687 | 0% |
4330 | 4329 | 0.023100023% |
10756 | 10753 | 0.027899191% |
30681 | 30660 | 0.068493151% |
60180 | 60150 | 0.049875312% |
90600 | 90670 | 0.077203044% |
Conclusion:
After doing the above course work I have found out that, I have achieved many things and I can list them as:
- Calculating the logs of R and T
- Drawing graph by using the result from R and T logs
- Determining an accurate linear equation of logs (both R and T)
- Rearranging log k= 0.7003 into 10^-0.7003 then putting it into the equation of T= kR so it will look like: T= 10^-0.7003* (R) ^1.5001.
- Applying the rules 1 and 3 to rearrange the equation T= KR
- Determining the correct equation for the line T= 10^-0.7003*(R ^1.5001)
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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