newtonian mechanics - What is the stable range for orbit of the Earth?

Suppose a force pushes/pulls Earth straightly toward Sun and make earth x kilometers closer/farther from Sun. For what x the Earth remains in a stable orbit, rather than spirally fall to the Sun or going farther away? What math is involved?

Answer

For the Sun-Earth system alone, the important quantity here is the effective gravitational potential $$ U_{eff}=\frac{\ell^2}{2mr^2}-\frac{GmM}{r} $$ with $m$ the mass of Earth and $M$ the mass of the Sun, and $\ell$ is the angular momentum of the Earth about the Sun. Since the orbit of the Earth is basically circular we can take $r$ to be $r=R_0=149600\times 10^6m$ to be the average distance to the Sun, and estimate $$ \ell=m v r= m \times 29.78\times 10^3 \times R_0 \approx 2.66\times 10^{37} $$ The energy scale $U_0$ can be set using $U_0=\vert U_{eff}(r=R_0)\vert $, so that plotting $U_{eff}/U_0$ as a function of $\rho=r/R_0$ gives the graph below.

The minimum of this potential, which is the radius where the orbit is circular, of course occurs at $\rho=1$ and $U_{eff}/U_0=-1$. If you "push" the Earth a bit towards the Sun, the Earth will oscillate about two radial turning points that will define the minimum and maximum radius of the elliptical orbits. For instance, if you push the Earth (keeping $\ell$ constant) so that $U_{eff}/U_0=-3/4$, the radius would oscillate between $\rho_{min}\approx 0.67$ and $\rho_{max}\approx 2.0$.

The orbit remains bound as long at the total energy remains negative. You can reach the threshold $\rho_{c}$ by solving for $\rho$ when you set $U_{eff}=0$. This gives $\rho_{c}\approx 1/2$. Thus, if you pushed the Earth to a radius smaller than one half of it's current average radius (managing $v$ so that $\ell$ would remain constant), the orbit would become unbound.