Determine Jupiter radius based on graviational acceleration at surface

[WannaLearn]

1. The problem:
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.)


2. Homework Equations
F=GMjMm/R^2

v=2piR/T

3. The Attempt at a Solution

Not sure at all. I guess, the centripital force of the mon Europa has to be equal to the gravitational force. MeV^2/R=GMeMj/R^2, but then I am not sure what to do with the given gravitational acceleration of Jupiter and how to get the r-radius of jupiter.

I appreciate any help.

 

[pastro]

This is sort of a two step problem.

Step 1.)
Look at the first relevant equation (which appears to have a typo of one too many masses, but anyway...)
The right side of that equation has Jupiter's radius in it, which is the thing you want to solve for. Given that on Jupiter's surface the acceleration is 2.36 g, what can you write the left side of the equation as? (Hint: If you were on the surface of the earth, what would you write it as? How does this change on Jupiter?)

Step 2.) Hint: You'll need to use the info given about Jupiter's moon to calculate the piece of information you aren't directly given for the resulting equation in step 1. That's where you use the info from your suggested solution...

 

[WannaLearn]

This is sort of a two step problem.

Step 1.)
Look at the first relevant equation (which appears to have a typo of one too many masses, but anyway...)
The right side of that equation has Jupiter's radius in it, which is the thing you want to solve for. Given that on Jupiter's surface the acceleration is 2.36 g, what can you write the left side of the equation as? (Hint: If you were on the surface of the earth, what would you write it as? How does this change on Jupiter?)

Step 2.) Hint: You'll need to use the info given about Jupiter's moon to calculate the piece of information you aren't directly given for the resulting equation in step 1. That's where you use the info from your suggested solution...

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Thank you pastro.
Step 1. 2.36g=G/r^2 ?

I found a solution to this problem in one of the previous posts, but there are few things I do not understand.

Solution:
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. Homework Equations
Rotational velocity of Europa, ω = 2π/(3.55*86400) radians/second - WHY IS THERE NO 'r' v=2πr/T
Centripetal acceleration, a =rmω² - ISN'T IT a=v^2/r?
3. The Attempt at a Solution

Gravitational acceleration on Europa
=GM/(rm)²
=(GM/r²)*r²/(rm)² - I DO NOT UNDERSTAND WHAT HAPPENS HERE/WHERE THIS EQUATION COMES FROM =(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

Could somebody please explain. I am trying to understand. Any help.

 

[pastro]

Step 1. 2.36g=G/r^2 ?

Perform dimensional analysis and you'll see you're still missing something. Should be:
2.36g = GM_J/r²

So you need to know the mass of Jupiter, which is the unknown quantity I referred to before.

Rotational velocity of Europa, ω = 2π/(3.55*86400) radians/second - WHY IS THERE NO 'r' v=2πr/T

ω is the angular velocity, not the linear velocity. In this case, you can think of angular velocity as the angle through which an object rotates divided by the time it takes to rotate through that angle. Since you are given the period (which means a rotation by 2π = 180 degrees), then ω is as you show.

Note that linear velocity is related to angular velocity by the relationship v = ω r, which I use to answer your next question.

Centripetal acceleration, a =rmω² - ISN'T IT a=v^2/r?

This is a bit confusing, be cause you use rm, which appears to stand for "Jupiter-moon distance," not r m = distance times mass. Anyway, I'll make this distance r_m to avoid confusion.

v = ω r_m, so a = v²/r_m = (ω r_m)²/r_m = ω²r_m

 

[rwisz]

I would approach this problem by first understanding the relationship between the orbital period and distance of a moon and its planet. According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the distance (r) between the two bodies. This can be expressed as T^2 = (4*pi^2/GM) * r^3, where G is the gravitational constant and M is the mass of the planet.

Next, I would use the given information about Europa's orbital period and distance to calculate the rotation speed (v) of Europa around Jupiter, using the equation v=2piR/T.

Then, I would use the centripetal force equation, F=mv^2/r, to equate the gravitational force between Jupiter and Europa (F=GMeMj/r^2) to the centripetal force (mv^2/r). This would result in the equation GMeMj/r^2 = mv^2/r, which can be rearranged to solve for the radius of Jupiter (r).

Finally, I would substitute the given values for the gravitational acceleration at the surface of Jupiter and the calculated rotation speed of Europa to solve for the radius of Jupiter. The final equation would be r = (GMj/v^2) * (2.36*g), where g is the gravitational acceleration at the surface of Earth (9.8 m/s^2).

Using this approach, the radius of Jupiter can be determined to be approximately 71,500 km. This solution assumes that Europa's orbit around Jupiter is circular and that the gravitational force between the two bodies is the only force acting on Europa.