Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of $4.22 X 10^5$ km. From these data, how would you determine the mass of Jupiter?

The orbital period can be calculated using Kepler's 3rd law, which states that the square of the orbital period is proportional to the cube of the average distance from ....

Kepler's third law relates the orbit period T in years to the semi-major axis distance a in AU using the equation $T^2=a^3$. If $T=76$ then $a=17.94$. Given that the comet's...

The equatorial diameter of Jupiter is 142800 kilometer. Where as Earth's diameter is only 12756. Volume is 4/3 Pi x R ^3. .

Io 471800 kilometers Europa 671100KM Ganymmede 1070400KM Callisto 1882200KM Use this link for all satellites

Observing their position on a star map and wait till they return to that position again. and counting days elapsed..

$a$ - Semi-major distance; $\quad b$ - Semi-minor distance; $c$ - Centre-to-Focal Point distance; $\quad e$ - Eccentricity of the ellipse; $d_p$ - Perihelion distance; $\quad d_a $ - Aphelion...

The problem with deciding at what point an orbit starts is that all planets' orbits precess. According to Johannes Kepler, the start of an orbit is the perihelion when the...

The orbital period $T$ in years of all bodies in the solar system is directly related to the semi-major axis distance $a$ is AU by Kepler's third law $T^2=a^3$. So,...

Earth's orbital period is called a year. Rotation period is called a day. SOLar day is 24 hours but Earth is moving around Sun one degree every day..

To do this problem, you need the $R = 6371 km$ For the satellite to be in a stable orbit at a height, h, its centripetal acceleration $V^2/(R + h)$...