1. Two satellites are orbiting around Earth. One satellite has a period of 1.4hours and is 200km above Earths surface. The other satellite has a period of 6.0h. Use Kepler's laws and the fact the radius of Earth is 6.37 x 10^6meters to determine the height of the second satellite above Earths surface. 2. Keplers Third Law: (Ta/Tb)^2 = (Ra/Rb)^3 Fg=Gm1m2/r2 G=6.67x10^-11 Nm^2/kg^2 Msv2/r = GMeMs/r^2 g=Fg/m g=9.80N/kg g=GMe/Re^2 3. Ok first I know that Fg is equal to Fc, but I dont know the equation to find out Fc. So I just decided to attempt to find out the period but dividing 6hours by 1.4hours and using that number (4.29 rounded) multipled it with 200km and got 858km. I then used 858km as the height of the 2nd satellite above the Earths surface. I also calculated the velocity of satellite one, which is 7.79 x 10^3 m/s (or just 7.8m/s) Now, since I am doing a long distance course and not good at all at algrebra, and the lack of good example question within the textbook and coursebook to go on, I stumped. The answer I have cannot be worth 5 marks from the work Ive shown, but I just cant find any way to prove how my answer is correct with so little to go on.