• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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.

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

rowspan="1">

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Core & Pure Mathematics essays

1.  5 star(s)

5 625 1250 1000 My theory was correct. I shall now try and binomially prove this... 2(x+h)4 - 2x4 = 2(x4 + h4 +4hx� + 6x�h� + 4xh�) - 2x4 = x + h - x h 2x4 + 2h4 +8hx� + 12x�h� + 8xh� - 2x4 = 2h4 +8hx�

2. As you can see, I have investigated all the positive points on the 'x' scale. Some results are not all to 1 decimal place. The pattern in this results table is that the 'x' value has been multiplied by 4.

1. ## Functions Coursework - A2 Maths

x f(x) 1.87930 -0.0006474767 1.87931 -0.0005715231 1.87932 -0.0004955684 1.87933 -0.0004196125 1.87934 -0.0003436555 1.87935 -0.0002676974 1.87936 -0.0001917381 1.87937 -0.0001157777 1.87938 -0.0000398162 1.87939 0.0000361464 1.87940 0.0001121102 The root therefore lies in the interval [1.87938,1.87939]. The next table show values of f(x) at values of x from 1.87938 to 1.87939, with intervals of 0.000001.

2. ## Mathematics Coursework - OCR A Level

I then displayed these rearrangements graphically in the form y=g(x). y=3x5-x2-x+0.31 Rearrangement that does not work One of the rearrangements of the formula did not work using this method of rearrangement. The rearrangement was x=(3x5-x+0.31)1/2 and I displayed it graphically in the form y=(3x5-x+0.31)1/2 The table below was calculated in Autograph.

1. ## Arctic Research (Maths Coursework)

I will start with my base camp in the middle. This is a good position to start as all the observation sites are at equal distances of 50 km (which is the radius) from the base camp. It will also enable me see how the journey times are affected in

2. ## GCSE Math Coursework: Triminoes

5 420 56 0123456 6 756 84 Largest Number 1 2 3 4 5 6 +1 +1 +1 +1 +1 Linear Equation FORMULA f (n) =an + b f (n) = 1n + 0 Sum of all numbers 6 30 90 210 420 756 +24 +60 +120 +210 +336 +36

1. ## Methods of Advanced Mathematics (C3) Coursework.

3 119 3.5 195.875 4 299 As you can see I have found only the first route and without a graph I wouldn't be aware of the two routes actually present between this interval. The reason for the failure is the fact of two routes being in one interval.

2. ## Experimentally calculating the wavelength of an He-Ne laser by means of diffraction gratings

m would simply be the order of the fringe one considers. However, the value for ? is not given. Still, by shining the laser through a diffraction grating to display its interference pattern on a screen, one may calculate the distance from the grating to the screen L, and the • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 