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Your (gardener's intuition) is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre. The nuclear reactions supply just enough energy to equal that radiated from the surface and thus prevent the need for further contraction.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$

The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$

Your (gardener's) intuition is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre. The nuclear reactions supply just enough energy to equal that radiated from the surface and thus prevent the need for further contraction.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$

The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$

Your (gardener's intuition) is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$

The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$

Your (gardener's intuition) is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre. The nuclear reactions supply just enough energy to equal that radiated from the surface and thus prevent the need for further contraction.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$

The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$

Your intuition is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

Your (gardener's intuition) is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.

To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.

But why this particular temperature/density combination? Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre.

The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.

$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$

The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$