# Gravitational Acceleration to calculate radius of Jupiter

1. May 12, 2010

### mlostrac

1. The problem statement, all variables and given/known data
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.)

radius of orbit of Europa, rm = 671000km
rotational period of Europa, T = 3.55 days
Acceleration due to gravity on Jupiter = GM/r² = 2.36g
radius of Jupiter, r = ?

2. Relevant equations
Rotational velocity of Europa, ω = 2π/(3.55*86400) radians/second
Centripetal acceleration, a =rmω²

3. The attempt at a solution

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. Did I miss something?

2. May 12, 2010

### diazona

I don't think so, that's pretty close. You can't expect your calculation to be perfectly exact because, for example, you're implicitly assuming that Europa has a circular orbit.

3. May 13, 2010

### mlostrac

Ok. So I did all my calculations correctly? Do you get the same answer?