TOI-560c

For the calculations we used the following formula: $$\rho = \frac {M}{V}$$ where $M$ and $V$ are the mass and volume of the planet respectively. After converting the units grams and $cm^3$ we have calculated the density of TOI-560c: $\mathbf{3.9469 g / cm^3}$

Based on an adjusted version of the Kepler’s third law, we determined the distance between TOI-560c and its host star using the following formula: $$T^2 = (\frac {4 \pi^2}{G M_s}) d^3$$ where $T$ is the orbital period and $M_s$ is the mass of the star, from which we get $$d = \sqrt[3]{\frac {T^2 G M_S}{4 \pi^2}}$$ which returns a distance of $\mathbf{1.873 \cdot 10^7 km^2}$

In fact, this radiant flux would be 10 times stronger than it is on Earth. If we were to theoretically send a space probe or some other spacecraft, we can rely on an easily accessible power source using solar energy.

To better understand TOI-560c’s behaviour we have decided to calculate the power radiated by its host star. Using the Stefan-Boltzmann’s law we can form the following formula: $$P_{rad} = 4 \pi r^2 \sigma T^4$$ where $T$ is the host star’s temperature and $\sigma$ is Stefan-Boltzmann's constant. This resulted in the power of $\mathbf{6.074×10^{25} W}$, which is about 5 times smaller than that of the Sun.

With that information we calculated the power which reaches TOI-560c's orbit using the following equation: $$P = \frac {P_{rad}}{a}$$ where $P$ is the radiant flux on planet’s orbit, $P_{rad}$ is power radiated by the star and $a$ is the area of an imaginary sphere, radius of which equals to the distance between the star and the planet. With that we got the result of $\mathbf{13,779 W / m^2}$, which is about 10 times greater than the radiant flux that reaches Earth.

To calculate rough orbital speed of TOI-560c we used the following formula: $$v_o \approx \sqrt{\frac {G M}{r}}$$ where $M$ is the planet’s mass, $r$ is the planet’s radius and $G$ is the gravitational constant. This resulted in the speed of $\mathbf{15,949.92 m/s}$. By using the equation $$v_e = \sqrt{2} \cdot v_o$$ we got the escape velocity of $\mathbf{22,556 m/s}$.

TOI-560b is one of 4 Mini-Neptunes observed by NASA that show slow transformation to Super-Earths. Due to solar winds, these exoplanets slowly lose their atmospheres, leaving only rocky cores. These remnants are then classified as Super-Earths. After this discovery a theory was proposed that all or most Mini-Neptunes have a Super-Earth like core. We have decided to try to find out if such composition could be possible. For this we calculated the density our theoretical model would have and compared it to TOI-560c’s density. The model consists of a typical Super-Earth like rocky core of a radius of $1 R_{EARTH}$ and the density of $5.51 g / cm^3$. and a layer with the density of a gas, which consists of 80% hydrogen and 20% helium. The latter was estimated to have a density of about $0.10772 kg / m^3$. The next step was calculating the volume of the model’s core and atmosphere. For this we’ve decided to take the formula $$V = \frac{4}{3} \pi r^3$$ For convenience, let $r^3 = a^3 \cdot R_{EARTH}^3$ and let’s replace $\frac {4}{3} r^3$ with a variable $u$. With this we can rewrite the formula as $$V = u \cdot a^3$$ from which we can get the volume of the core $V_C = u$ and, through the volume of TOI-560c, the volume of the atmosphere $V_{ATM} = 12.5494u$. Through this we can calculate the individual masses and then the mass of the entire model: $m_C = 5510.505u$, $m_{ATM} = 1.3518u$ and $m = 5511.8568u$. There we can apply the formula $$\rho = \frac {M}{V}$$ using the model’s mass and TOI’s volume, which using our variable $u$ would be $13.5494u$. With this we get a theoretical density of $\mathbf{4.068 g / cm^3}$, which is very similar to the real density of TOI-560c. Therefore, we believe this theory to be possible.