If the Earth were to suddenly stop its orbit around the Sun and fall into the Sun, how long would it take for the Earth to reach the Sun?
I. solution. If the Earth were to suddenly stop its orbit around the Sun, it would fall straight into the Sun due to gravity. We can consider its path as a degenerate ellipse, with the major axis being the Earth-Sun distance, the minor axis being zero, and one of its foci being the Sun. According to Kepler's Third Law, the square of the orbital period of the planets is proportional to the cube of their average distance from the Sun. Let's write down Kepler's Third Law for the Earth in two states of its orbit, using the appropriate data for the two elliptical orbits:

$$
\frac{T^{2}}{T_{e}^{2}}=\frac{R^{3}}{R_{e}^{3}}
$$

where the $e$ index denotes the data corresponding to the degenerate ellipse. The major axis of the degenerate ellipse is $R$, so here the average Earth-Sun distance is $R_{e}=R / 2$. Using this,

$$
T_{e}=\sqrt{\frac{T^{2}(R / 2)^{3}}{R^{3}}}=T \sqrt{\frac{1}{8}}=\sqrt{\frac{1}{8}} \text { year }
$$

since the Earth's orbital period is 1 year. The Earth would reach the Sun along the degenerate elliptical orbit in half of the orbital period there, i.e., $(1 / 2) \sqrt{1 / 8}$ year $\approx 64.5$ days.

Lengyel György (Bp., Apáczai Csere J. Gyak. Gimn., II. o. t.)

II. solution. The Earth gains kinetic energy from the decrease in the potential energy of the Earth-Sun system:

$$
\gamma m M\left(\frac{1}{R-x}-\frac{1}{R}\right)=\frac{1}{2} m v^{2}(x)
$$

where $\gamma$ is the gravitational constant $\left(\gamma=6.67 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$, $M$ is the mass of the Sun $\left(M=2 \cdot 10^{30} \mathrm{~kg}\right)$, $m$ is the mass of the Earth, and $v(x)$ is the Earth's speed after it has traveled a distance $x$ towards the Sun.

Substituting $\mathrm{d} x / \mathrm{d} t$ for $v(x)$, we get the following differential equation:

$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{\frac{2 \gamma M}{R}} \sqrt{\frac{x}{R-x}}
$$

or equivalently:

$$
\sqrt{\frac{R-x}{x}} \frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{\frac{2 \gamma M}{R}}
$$

Integrate both sides from 0 to $t$:

$$
\int_{0}^{t} \sqrt{\frac{R-x}{x}} \frac{\mathrm{d} x}{\mathrm{~d} t} \mathrm{~d} t=\sqrt{\frac{2 \gamma M}{R}} t
$$

The integral on the left side can be calculated by substitution:

$$
\int_{0}^{t} \sqrt{\frac{R-x}{x}} \frac{\mathrm{d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{0}^{R} \sqrt{\frac{R-x}{x}} \mathrm{~d} x=\left[\sqrt{x(R-x)}-R \operatorname{arctg} \sqrt{\frac{R-x}{x}}\right]_{0}^{R}=\frac{\pi}{2} R
$$

Thus,

$$
t=\frac{\pi}{2} \sqrt{\frac{R^{3}}{2 \gamma M}}
$$

This essentially gives us the form of Kepler's Third Law used in the first solution. Numerically, $t=$ 64.5 days, in agreement with the value obtained in the previous solution.

Marth Gábor (Bp., Apáczai Csere J. Gyak. Gimn., IV. o. t.)