# 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

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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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