91Ó°ÊÓ

The kinetic energy is given by$\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$ where mis mass of the object and vis the velocity of the object, the kinetic energy of spring is given by$\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$ where k is spring constant and xis the distance from the equilibrium point and the potential energy at the planet’s surface is$U=-\frac{GMm}{R}$ where Gis universal gravitational constant, Mis the mass of planet, Ris radius of planet and mis the mass of other object.

According to physics, energy is the quantity that must be supplied to a body or physical system in order to perform work on it or heat it.

According to the principle of conservation of energy, energy can be transformed in form but not created or destroyed.

Let the system is free of non-conservative forces, therefore energy is conserved between the starting and end states.

Elastic potential energy and gravitational potential energy will present at the beginning state but only kinetic energy will present at the final state as the object is so far away.

Write equation of conservation of energy and use the formula of energy.

$$\begin{gathered} E_i=E_f \\ \frac{1}{2}{k}_s{s}^2+\left(-\frac{GMm}{R}\right)=\frac{1}{2}m{v}_f^2 \end{gathered}$$

Solve the equation for s.

$$\begin{gathered} k_s{s}^2=m{v}_f^2+\frac{2GMm}{R} \\ s=\sqrt{\frac{m}{ks}\left(v_f^2+\frac{2GM}{R}\right)} \end{gathered}$$

Substitute m = 9 kg, $k_s=2.8×{10}^{5}N/m$, $v_f=226m/s$, $G=6.67×{10}^{-11}{m}^3/kg·s^2$, $M=8×{10}^{20}kg$and $R=8.7×{10}^{5}m$into the obtained result.

$$\begin{gathered} s=\sqrt{\frac{9}{2.8×{10}^5}\left[{\left(226\right)}^2+\frac{2(6.67×{10}^{-11})(8.0×{10}^{20})}{8.7×{10}^5}\right]} \\ =2.36\mathrm{m} \end{gathered}$$

Therefore, the string must be compressed up to 2.36 m.