Year 12 Physics -Term 1 2015- Gravity and Motion Ethan Jones 12PHC SJA TASK 1: a) b) Equation of line: y = 0.6656x + 8.4604 Therefore gradient = m = 0.6656 R^2= 1 log T^2/r^3 =c logT^2-logr^3=logc 3 log〖r=2logT-logc〗 ∴logr=2/3 logT-1/3 logc y=mx+c m=graident m= 2/3 ∴graident=0.6666 The gradient collected from the graph shows 0.6656 whereas the value that Kepler’s 3rd law shows is 0.6666. This difference can be explained due to the fact that some of the Data collected for the graph could be slightly incorrect. The orbital velocities of most celestial objects can be worked out and represented by Kepler’s 3rd Law: The Square of the period of any celestial body is proportional to the cube of the radius of its orbit. This is visualised in the equation- 〖T_a〗^2/〖r_a〗^3 =c -Mean Constant (c) 〖T_a〗^2/〖r_a〗^3 =c Io: 〖1.77〗^2/〖(4.22E^8)〗^3 =x c=4.17*〖10〗^(-26) Europa: c=4.17*〖10〗^(-26) Ganymede: c=4.18*〖10〗^(-26) Callisto: c=4.2*〖10〗^(-26) ∴mean (c) =4.18*〖10〗^(-26) c) Himalia: T: 250.23 r: ? 〖T_a〗^2/〖r_a〗^3 =c 〖250.23〗^2/〖r_a〗^3 =4.18*〖10〗^(-26) 〖250.23〗^2/(4.18*〖10〗^(-26) )=〖r_a〗^3 1.5*10^30=〖r_a〗^3 r=1.14*10^10m r=11441970km r=11461000km (NASA) The difference of 19030km between the two values can be attributed to two main factors, firstly the calculated (c) mean value could be offset due to differences in information on the four moons investigated; Io, Europa, Ganymede and Callisto. The second point where an issue could occur is the given

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