There is not enough gravity at the center to start nuclear fusion, but it seems that there would be plenty enough to collapse the planet.
[Physics] What keeps a gas giant from falling in on itself
astrophysicsnewtonian-gravityplanetspressurethermodynamics
Related Solutions
Gravitational Field from a Ring of Mass
...the force on a unit mass at $P$ from the two masses $M$ is $$F=-\frac{2GMx}{(x^2+a^2)^{3/2}}$$
Now, as long as we look only on the $x$-axis, this identical formula works for a ring of mass $2M$ in the $y$,$z$ plane! It’s just a three-dimensional version of the argument above, and can be visualized by rotating the two-mass diagram above around the $x$ -axis, to give a ring perpendicular to the paper, or by imagining the ring as made up of many beads, and taking the beads in pairs opposite each other.
Bottom line: the field from a ring of total mass $M$, radius $a$, at a point $P$ on the axis of the ring distance $x$ from the center of the ring is $$F=-\frac{GMx}{(x^2+a^2)^{3/2}}$$
Gravity of a Torus
People sometimes think that, perhaps for reasons of symmetry, an object in the interior of a ring of matter would be drawn toward the center, but this is not the case – at least not for objects in the plane of the ring. To see why, consider a very thin ring of mass treated as a circle of radius $R$ in the plane, and a particle inside this ring at a distance $r$ from the center. Construct an arbitrary line passing through this particle, striking the ring in two opposite directions at distances $L_1$ and $L_2$. If we rotate this line about the particle through an incremental angle $\mathrm{d}q$, it will sweep out sections of the ring proportional to $L_1\cos(a)\mathrm{d}q$ and $L_2\cos(a)\mathrm{d}q$, where $a$ is the angle the chord makes with the normals to the circle at the points of intersection. The net gravitational force exerted by these two opposing sections of the ring is proportional to the masses in these small sections divided by the squares of the distances, i.e., the force is proportional to $\mathrm{d}q \cos(a) (\frac{1}{L_1} - \frac{1}{L_2}$) in the direction of the $L_1$ intersection point. Hence the net force is in the direction of the closest point on the ring, directly away from the center.
Gravitation field due to rigid bodies
Fun additional reading: Ringworld :)
This might not fully answer your question, but maybe it will be a good start.
Things to consider
- Thermal energy received by Jupiter from the sun
- Thermal energy radiated by Jupiter (hence, net thermal energy)
- Jupiter's composition
- Jupiter's temperature
- Jupiter's gravitation
- Hydrogen's thermal properties (among other properties)
For the first 2 items to consider, we actually don't have much of a problem. We could go about figuring out the sun's radiation power (well...on wiki it says $3.846 \times 10^{26}$ if you're interested), estimate the energy received by Jupiter by making a surface area argument and then figuring out Jupiter's radiation power. From this site http://www.tritonfun.com/custom.em?pid=594668, it actually says that Jupiter is radiating 1.9 times the amount of heat it receives from the sun. But this is because it is also creating it's own energy from a variety of methods (including radionuclides). We can ignore all of that though because I'll bet that Jupiter is close to thermal equilibrium in the short term (ie, it's not constantly changing temperature as a whole extremely drastically). If you weren't given the information as to how hot a planet was, this is the kind of analysis you would have to perform in order to get the average temperature of the body. Luckily we can use data from other sources for the temperature of Jupiter.
That addresses points 1 and 2. From a number of online resources, Jupiter has multiple layers, namely, the outer, gaseous atmosphere, a transition area between gas and liquid, a liquid/metallic section and the a mostly hydrogen core. Unfortunately I can't find exact distances of all of these...which would make answering this much easier. But, what we need to know is that the layers differ drastically in terms of pressure and temperature. In fact many layers that are farther from the center of Jupiter are actually hotter, but with less pressure. So the pressure and the temperature combined will affect what phase hydrogen is in. It's extremely complicated and probably well beyond the scope of most physicists to understand why some layers have more pressure than others, but we could just make a vast simplification and say the pressure and temperature increase/decrease as a simple function of distance from the center of Jupiter.
Jupiter's temperature and pressure increase steadily as we get closer and closer to the center only at the core. Near the phase transition region between gas and liquid, the temperature is about 10,000 K and the pressure is 200 Gpa. The temperature at the core's edge is around 36,000 K and 3000-4500 Gpa. So it seems like both temperature and pressure increase throughout the liquid layer/transition layer too. I'm not necessarily sure how to calculate this from scratch, given the elements, the mass of the planet, it's radius and density. Of course it will involve balancing the forces of gravity with electrostatic forces (I just tried a quick calculation using those 2 forces only, but it was WAY off). The fact that my calculation was pretty far off tells me there are a lot more things you have to consider that are extremely important for accuracy. For example, as the hydrogen molecules heat up, they will be bouncing around faster and faster, causing a greater repulsion. Details like this make the calculation very difficult.
I know this doesn't help you a whole lot, but hopefully I've illustrated some important points. Firstly, this kind of calculation (if you want it to be accurate) is something people with PhD's who specialize is this kind of stuff calculate. You can make huge simplifications and maybe get an okay answer. But I feel like it's a wasted effort to actually go through the calculations if they don't give you a very accurate answer. But like I said, if you want to get a rough answer, think about all of the forces going on. Gravity, electrostatic repulsion, average thermal energy, and then make the net force = 0. Electrostatic repulsion of hydrogen can be found on the wiki page and hopefully you know how to take into account gravity as a function of distance from the center =p.
Best Answer
Pulsar's answer is indeed correct, but let me expand a bit more.
What happens when a gas giant shrinks?
A uniform mass will have a self gravitational potential of $-\frac{3GM^2}{5R}$. If we decrease its radius, its potential will decrease as well and the difference will be turned into thermal energy. Although gas giants and stars are not uniform mass balls, their gravitational binding energy is still proportional to $\frac{GM^2}{R}$, Thus if the radius decreases it will release energy, which will raise the temperature in return.
What happens when the temperature increases?
Assuming the gas in those planets obey the ideal gas law $$PV=nRT$$ (where $R$ is not the radius but the molar gas constant $R=8.314\,\text{J K}^{−1}\text{mol}^{-1}$), it's obvious that when $T$ increases and $V$ decreases (due to the shrink in the previous section) $P$ must increase. Note that most real gases behave qualitatively like an ideal gas, so this is not a crazy assumption.
So what is the big picture?
The planet shrinks a little bit, the potential difference turns into thermal energy and its temperature rises. The rise in temperature will cause the pressure to rise and prevent the planet from shrinking further (holding the planet in hydrostatic equilibrium). However, the planet also loses energy due to EM radiation as well, so it will continuously shrink and radiate. The process is called Kelvin–Helmholtz mechanism.
For instance, Jupiter is shrinking the tiny bit of $2\,\text{cm}$ each year. Although you might think this is really nothing, the amount of heat produced is similar to the total solar radiation it receives.
Addendum (Nov. 2020)
As Rob Jeffries has correctly pointed out, what ultimately keeps a gas giant from collapsing indefinitely is the electron degeneracy pressure. Eventually because of high pressure the hydrogen and other elements in the deep interior of the gas giant will undergo a phase transition to a metallic phase and will not compress any further.