Calculate the orbital period for jupiter's moon io, which orbits 4.22×105km from the planet's center (m=1.9×1027kg) .

150000 seconds or approximately 1.7 days. The equation for the orbital period is: Ď„=âš((4Ď€^2/ÎĽ)a^3) where Ď„ = Period ÎĽ = standard gravitational parameter (GM) a = semi-major axis It's generally better to use ÎĽ rather than the product of G and M since we know ÎĽ more accurately than either G or M in many cases from observations of satellites around the various planets. For instance in the case of Jupiter, we know ÎĽ is 1.26686534(9)Ă—10^17 whereas we only know it's mass at 1.8982x10^27 kg and G at 6.674x10^-11 m^3/(kg*s^2). Anyway, let's first calculate ÎĽ based upon the data given: ÎĽ = 1.9x10^27 kg * 6.67ex10^-11 m^3/(kg*s^2) ÎĽ = 1.26806x10^17 m^3/s^2 Now let's substitute the known values into the equation for the orbital period. Ď„=âš((4Ď€^2/ÎĽ)a^3) Ď„=âš((4Ď€^2/1.26806x10^17)(4.22x10^8)^3) Ď„=âš((4*9.869604401/1.26806x10^17)(7.5151448x10^25) Ď„=âš((39.4784176/1.26806x10^17)(7.5151448x10^25) Ď„=âš((3.11329256x10^-16)(7.5151448x10^25) Ď„=âš(2.3396844x10^10) Ď„=152960.2706 Rounding to 2 significant figures since that's the precision of the least accurate datum, we get 150000 seconds, or approximately 1.7 days.

The answer is d) by requesting advice on science internet sites