The moon Europa, of the planet Jupiter, has an orbital period of 3.55 days and an average distance from the center of the planet equal to 671,000 km. If the magnitude of the gravitational acceleration at the surface of Jupiter is 2.36 times greater than that on the surface of the Earth, what is the radius of Jupiter?:confused:

what is the radius of Jupiter?


It has a mass of 1.9 x 1027 kg and is 142,800 kilometers (88,736 miles) across the equator.

Jupiter Statistics
Mass (kg) 1.900e+27
Mass (Earth = 1) 3.1794e+02
Equatorial radius (km) 71,492
Equatorial radius (Earth = 1) 1.1209e+01
Mean density (gm/cm^3) 1.33
Mean distance from the Sun (km) 778,330,000
Mean distance from the Sun (Earth = 1) 5.2028
Rotational period (days) 0.41354
Orbital period (days) 4332.71
Mean orbital velocity (km/sec) 13.07
Orbital eccentricity 0.0483
Tilt of axis (degrees) 3.13
Orbital inclination (degrees) 1.308
Equatorial surface gravity (m/sec^2) 22.88
Equatorial escape velocity (km/sec) 59.56
Visual geometric albedo 0.52
Magnitude (Vo) -2.70
Mean cloud temperature -121°C
Atmospheric pressure (bars) 0.7
Atmospheric composition
Hydrogen
Helium
90%
10%

Still confused... :( how to find it?

Sorry for the late reply. I had to make sure I got it right.

1. The centripetal force the Europa experiences is equal to the gravitational it experiences. So,

F_g = F_c

F_g = \frac{GMm}{r^2}

F_c = mr \omega ^2

\frac{GMm}{r^2} = mr \omega ^2

->\omega = \frac{2\pi}{T}

\frac{GMm}{r^2} = mr (\frac{2\pi}{T})^2

\frac{GM\cancel{m}}{r^2} = \cancel{m}r (\frac{4\pi^2}{T^2})

\frac{GM}{r^3} = (\frac{4\pi^2}{T^2})

T is the periodic time in seconds.
G is the universal gravitational constant
M is the mass of Jupiter (unknown)
r is the radius from the centre of Jupiter to Europa.

Find M, the mass of Jupiter.

Then, you have F_J = \frac{GM}{r^2}

Since you have the mass of Jupiter, you can find r here, which is the radius of Jupiter. (I got 73799885.53 m, which is pretty close to the real radius of Jupiter :))