Gravitational Force on a Particle in Jupiter's Core

In summary, the gravitational force on an object of mass located a distance r from Jupiter's center is given by:F=4mgR^2pi
  • #1
Tubefox
9
0

Homework Statement


Jupiter has a core of liquid metallic hydrogen, with uniform density $\rho_c$, with radius $R_c$. This is surrounded by a gaseous cloud $R_g$, where $R_g>R_c$. Assume the cloud is of uniform density $\rho_g$.

The problem also specifies that we are to assume both regions of Jupiter are spherical (so it's two spheres, one inside the other)

What is the gravitational force on an object of mass $m$ located a distance $r$ from Jupiter's center? Consider the following cases:

A) $r<R_c$ (i.e., within the core)
B) $R_c<r<R_g$
C) $r>R_g$

Homework Equations



$$F=\frac{Gm dM}{|d-d'|^2}\hat{d-d'}$$

(With the hat denoting unit vector)

The Attempt at a Solution



I think what we're supposed to do here, since the density is not given as a continuous function, is to calculate the gravitational force due to the core, and the gravitational force due to the surrounding gas, and then superpose them. Since the density is uniform, we know that in a small chunk of mass dM (I think, this is one of the main thing's I'm not sure about):

$$dM=\rho P^2 \sin\theta d\theta d\phi dP$$

In spherical coordinates, $(P, \phi, \theta)$, that is. The $\rho$ is either $\rho_c$ or $\rho_g$, depending on which integral we're trying to do. The distance term will be given by:

$$\vec{d-d'}=(P \cos \phi \sin \theta - r_1) \hat{i} + (P \sin \theta \sin \phi-r_2) \hat{j} + (P \cos \theta - r_3)\hat{j}$$

Where $(r_1,r_2,r_3)$ are the coordinates of the mass located a distance $r$ from the center. This gives:

$$|d-d'|=\sqrt{P^2 - 2r_1P\cos(\phi)\sin(\theta) - 2r_2 P \sin \theta \sin \phi - 2r_3 P cos\theta + r^2}$$

(Note that $r_1^2 + r_2^2 + r_3^2 = r^2$ from the way we defined them)

I'm not sure this is right, especially because doing this produces the obscenely complicated integral for the force from the core region:

$$F=\int_0^{2\pi}\int_0^\pi \int_0^{R_c} \frac{mG\rho_c P^2 sin \theta}{P^2 - 2r_1P\cos(\phi)\sin(\theta) - 2r_2 P \sin \theta \sin \phi - 2r_3 P cos\theta + r^2} dP d\theta d\phi$$

Which looks too complicated to be right, especially since it contains three constants that weren't given. Could somebody either confirm I'm on the right track, or give me some guidance as to where I went wrong/how to get back on the right track? Thanks.
 
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  • #2
Way too complicated! Start by answering this problem: A mass m sits on the surface of a spherically symmetric planet with mass density ##\rho## and radius R. What gravitational force does the planet exert on the mass?

No calculus required! Just Newton's law of gravity.
 
  • #3
Hint: Take advantage of spherical symmetry.
 
  • #4
Thanks for the reply. I think I see what you're saying. Since we have uniform density, we just have:

$$M=4\pi R^2 \rho$$

And since it's on the surface, the distance is just R

So $$F=4mg\rho \pi$$

(Note I'm not sure that's exactly right but I woke up a few minutes ago so I can't quite get my brain to work well enough to check)

How do I deal with the outer shell, though? That seems more complicated.
 
  • #5
Tubefox said:
Thanks for the reply. I think I see what you're saying. Since we have uniform density, we just have:

$$M=4\pi R^2 \rho$$
Careful. You want the volume of the sphere, not the surface area.

Tubefox said:
(Note I'm not sure that's exactly right but I woke up a few minutes ago so I can't quite get my brain to work well enough to check)
Understood. Give it another try. :wink:

Tubefox said:
How do I deal with the outer shell, though? That seems more complicated.
Actually, it's just as easy---even easier. First take a guess as to what the field inside a uniform spherical shell would be. Then look up Newton's Shell Theorems.
 
  • #6
Aha! I suspected t was something like that (I was thinking Gauss' Law), but I couldn't quite make the connection. I will attempt that solution when I get home, thank you!

And yeah I realize that I should've plugged (4/3)pi R^3 in there, like I said, I had just woken up and was not in a state of full coherence, haha.
 

Related to Gravitational Force on a Particle in Jupiter's Core

1. What is the gravitational force on a particle in Jupiter's core?

The gravitational force on a particle in Jupiter's core is approximately 2.5 times stronger than the gravitational force on Earth. This is due to Jupiter's larger mass and higher surface gravity.

2. How does the gravitational force on a particle in Jupiter's core compare to Earth's?

As mentioned earlier, the gravitational force on a particle in Jupiter's core is 2.5 times stronger than Earth's. This means that objects on Jupiter's surface will feel a stronger pull towards the planet's center compared to objects on Earth's surface.

3. Does the gravitational force on a particle in Jupiter's core vary throughout the planet?

Yes, the gravitational force on a particle in Jupiter's core varies throughout the planet. This is due to the varying distribution of mass within Jupiter's atmosphere and core.

4. How does the gravitational force on a particle in Jupiter's core affect the planet's structure?

The strong gravitational force on particles in Jupiter's core causes the planet to have a dense and compact core, with layers of gas and liquid surrounding it. This is in contrast to Earth's structure, which has a solid inner core and a thicker, more spread out outer core.

5. Can the gravitational force on a particle in Jupiter's core be measured?

Yes, the gravitational force on a particle in Jupiter's core can be measured using mathematical equations and scientific instruments. This allows scientists to study the planet's structure and better understand its formation and evolution.

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