In: Physics
The orbital speed (period) of an earth satellite is essentially independent of its mass. Definitely for anyartificial satellite, the mass doesn't matter, only the measurements of the orbit. -- If the 'radii' given in the question refer to the radii of the satellites themselves ... the physical cans ... then that doesn't matter either. The only thing that affects the speed in orbit is the size/shape of the orbit, not the size, shape, mass, age, or ethnicity of the satellite. -- So let's assume that these two satellites are both in circular earth orbits, and that the radii given in the question refer to the respective orbital radii ... one orbit has a radius of 7 units, and the other one has a radius of 9 units. According to one of Kepler's laws ... I think it's #3 ... the quantity ( T2 / R3 ) is a constant for all satellites of the same central body, where T = the period of the orbit, R = its radius. For the pair of satellites: TA2/RA3 = TB2/RB3 We're given that RA=7 and RB=9, so TA2/343 = TB2/729 TA2/TB2 = 343/729 ===> TA/TB = sqrt(343/729) ===> TA = TB x sqrt(343/729) This would give us the ratio of the orbital periods, but not the ratio of the speeds that the questions asks for. To get that, we also have to figure in the circumferences of the two orbits. CA = 2 pi RA = 14 pi units, CB = 2 pi RB = 18 pi units. The orbital speed of either one is (circumference / period). Speed of A = (CA/TA) = 14 pi / TA = 14 pi / [TB x sqrt(343/729)] Speed of #2 = (CB/TB) = 18 pi / TB Speed A / Speed B = (14 pi TB) / [18 pi TB x sqrt(343/729)] Speed A / Speed B = 7 / [9 sqrt(343/729)] = 7 / (9 x 0.68594) = 1.13389 (rounded) As expected (hoped), the speed of the satellite in the smaller orbit is greater than the speed of the one in the larger orbit. Kepler and Newton are both happy. I know we're not supposed to get personal with these contributions, but this questioner seems to have a number in mind and an understanding of the principles at work. It's been close to 40 years since I learned this stuff, and after sweating this one out, I would really like to know if my answer is anywhere close, so I'll ask the questioner to please do me a favor and put a comment on my message board.