Question

PART 4: MASS OF THE SUN

Kepler’s 3rd law of planetary motion can be used to estimate MS the mass of the sun:

T2 = 4π2/GMs*r3

Go online and find the data for the mean distance and orbital period of the planets of the solar system. Enter the orbital periods of the planets (in seconds) and the mean distances (in meter) between the planets and the sun into separate columns in an Excel spreadsheet. Plot T versus r and note that the resulting plot is not linear.

Now calculate T2 and r3 in two adjacent columns (let Excel do the calculations) and plot T2 versus r3. Is the plot linear? Do a regression analysis and determine the slope of the plot. According to Kepler’s 3rd law,

slope=4π2 /GMs Ms= 4π2 /G⋅slope

Calculate MS using the above formula and compare with the expected value MS = 1.989 ∙ 10^30 kg

MS = ____________ %error = |measured-expected|/|expected| x 100 =_______

30

kg.

Answer 1

planet	T	r	T2	r3
mercury7,	7.5e6	57.9e9	5.6e13	1.94e32
venus	1.89e7	108.2e9	3.58e14	1.26e33
earth	3.15e7	149.6e9	9.94e14	3.34e33
mars	6e7	227e9	3.59e15	1.2e34
jupiter	3.75e8	778e9	1.41e17	4.7e35
saturn	9.3e8	1427e9	8.6e17	2.9e36
uranus	2.64e8	2871e9	7.017e18	2.6e37
neptune	5.2e8	4497e9	2.7e19	9.1e37

Here is the graph of T² vs r³ with T² on y axis and r³ on x axis.

As we can see, the fit was linear. The slope is given as

slope = 2.956e-19

Now, to find mass of sun

M_s = 4*pi² / G* slope

M_s = 4*pi² / 6.67e-11*2.956e-19

M_s = 2e30 Kg

Hurray, we got so close

The given mass of sun is 1.989e30

so,

% error = 2e30 - 1.989e30 / 1.989e30 * 100

% error = 0.668 %

The calculations are accurate.