Question

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?
(Hint: begin by calculating the rotation speed.)

Answer 1

radius of orbit of Europa, rm = 671000km
rotational period of Europa, T = 3.55 days
Rotational velocity of Europa, ω
= 2π/(3.55*86400) radians/second
Centripetal acceleration, a
=rmω²


radius of Jupiter, r = to be determined
Acceleration due to gravity on Jupiter
= GM/r² = 2.36g
Gravitational acceleration on Europa
=GM/(rm)²
=(GM/r²)*r²/(rm)²
=(2.36g)(r/rm)²

Equating gravitational acceleration with centripetal acceleration,
(2.36g)(r/rm)² = rmω²
r=(rm)³ω²/(2.36g)
=74,038 km

According to Google, r(Jupiter) = 71,492 km

Check my numbers and my thinking please.