When a planet forms, it gains mass by accumulating smaller bodies like planetesimals. These bodies fall into the planet’s gravitational field and release energy, which is converted into heat. If we assume that this heat is retained, we can estimate the initial temperature of the planet.

Assume the planet has:

• Radius $R_p$
• Constant density $\rho$
• Mass $M$

The volume of a sphere is $V = \frac{4}{3} \pi R_p^3$, so the total mass of the planet is:

$$ M = \frac{4}{3} \pi R_p^3 \rho \tag{1} $$

As mass is added gradually, the energy released during the accretion of a shell of mass $\delta M$ at radius $r$ is due to gravitational potential energy. The gravitational potential energy per unit mass at radius $r$ is:

$$ U(r) = -\frac{G M(r)}{r} $$

where $M(r)$ is the mass already assembled within radius $r$. Since the density is constant:

$$ M(r) = \frac{4}{3} \pi r^3 \rho \tag{2} $$

An infinitesimal shell at radius $r$ with thickness $dr$ has mass:

$$ dM = 4 \pi r^2 \rho \, dr \tag{3} $$

So the gravitational energy released when this shell falls onto the growing planet is:

$$ dE = \frac{G M(r) \, dM}{r} = \frac{G \left( \frac{4}{3} \pi r^3 \rho \right) \left( 4 \pi r^2 \rho \, dr \right)}{r} \tag{4} $$

Simplifying:

$$ dE = \frac{G \cdot \frac{4}{3} \pi r^3 \rho \cdot 4 \pi r^2 \rho}{r} \, dr = \frac{16}{3} \pi^2 G \rho^2 r^4 \, dr \tag{5} $$

Now integrate this from $r = 0$ to $r = R_p$ to get the total gravitational energy released during the planet’s formation:

$$ E = \int_0^{R_p} \frac{16}{3} \pi^2 G \rho^2 r^4 \, dr \tag{6} $$

This is a straightforward integral:

$$ \int_0^{R_p} r^4 \, dr = \frac{R_p^5}{5} $$

Therefore:

$$ E = \frac{16}{3} \pi^2 G \rho^2 \cdot \frac{R_p^5}{5} = \frac{16}{15} \pi^2 G \rho^2 R_p^5 \tag{7} $$

This is the total energy released and assumed to be converted entirely into heat.

The total thermal energy needed to raise the temperature of the planet from 0 K to $T$ is:

$$ E = C_p M T \tag{8} $$

Where:

• $C_p$ is the specific heat capacity of the material (assumed uniform and constant)
• $M$ is the total mass from Equation (1)

Substitute Equation (1) into Equation (8):

$$ E = C_p \cdot \left( \frac{4}{3} \pi R_p^3 \rho \right) \cdot T = \frac{4}{3} \pi C_p \rho R_p^3 T \tag{9} $$

Set the gravitational energy (Equation 7) equal to the thermal energy (Equation 9):

$$ \frac{16}{15} \pi^2 G \rho^2 R_p^5 = \frac{4}{3} \pi C_p \rho R_p^3 T \tag{10} $$

Cancel out common factors:

• $\pi$ from both sides
• One power of $\rho$
• $R_p^3$

This leaves:

$$ \frac{16}{15} \pi G \rho R_p^2 = \frac{4}{3} C_p T \tag{11} $$

Multiply both sides by 3:

$$ \frac{48}{15} \pi G \rho R_p^2 = 4 C_p T \tag{12} $$

Simplify:

$$ \frac{16}{5} \pi G \rho R_p^2 = 4 C_p T \tag{13} $$

Now solve for $T$:

$$ T = \frac{16}{5} \cdot \frac{\pi G \rho R_p^2}{4 C_p} = \frac{4 \pi G \rho R_p^2}{5 C_p} \tag{14} $$

$$ T = \frac{4 \pi G \rho R_p^2}{5 C_p} \tag{15} $$

This equation estimates the average internal temperature a planet would reach if all gravitational energy during its formation were converted into heat and retained.

• The temperature increases with the square of the radius: larger planets get much hotter.
• It also increases with density and decreases with specific heat.
• In reality, some energy would be lost via radiation during accretion, so this gives an upper bound.