Two Jupiter-size planets are released from rest 1.0 x 10¹¹m apart. What are their speeds as they crash together?

The distance between the two planets at rest is 1.0 x 10¹¹ m.

The mass of the planet is 1.90 x 10²⁷ kg

the mean radius of the planet is 6.99 x 10⁷ m

From the law of conservation of energy,

K₁ + U₁ = K₂ + U₂ ..............................................(1)

K₁ is the kinetic energy of the planets before collision.
U₁ is the potential energy of planets before collision.
K₂ is the kinetic energy of the planets after collision.
U₂ is the potential energy of the planets after collision.

Calculate the initial kinetic energy of the planets using
$K_1=\frac{1}{2}m{v}_1^2$
m = mass of planets.
v₁ = initial velocity of the planets.
As planets are initially at rest so their initial kinetic energy is zero.
K₁ =0

Initial potential energy (due to gravitational force) is

$U_1=\frac{-G{m}^2}{r_1}$

Similarly find the final kinetic energy

$K_2=\frac{1}{2}m{v}_2^2$

and final potential energy

$U_2=\frac{-G{m}^2}{r_2}$

Substitute these in the equation (1) and solve for velocity

$$\begin{gathered} 0+\frac{-G{m}^2}{r_1}=\frac{1}{2}m{v}_2^2+\frac{-G{m}^2}{r_2} \\ \frac{v_2^2}{2}=\frac{G{m}^2}{r_2}-\frac{G{m}^2}{r_1} \\ {v}_2=\sqrt{2Gm\left(\frac{1}{r_2}-\frac{1}{r_1}\right)} \end{gathered}$$

Now substitute values and calculate velocity

$$\begin{gathered} =\sqrt{2\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(1.90\times {10}^{27}\mathrm{kg}\right)\left[\left(\frac{1}{\left(13.38\times {10}^7\mathrm{m}\right)}-\frac{1}{\left(1.0\times {10}^{11}\mathrm{m}\right)}\right)\right]} \\ =\sqrt{\left(2.53\times {10}^{17}\mathrm{N}·{\mathrm{m}}^2/\mathrm{kg}\right)\left(7.463\times {10}^{-9}{\mathrm{m}}^{-1}\right)} \\ =\sqrt{\left(1888.14\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2\right)} \\ =4.35\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$