Apackage of mass 5 kg sits at the equator of an airless asteroid of mass 7.0 1020 kg and radius 3.5 105 m. we want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 165 m/s. we have a large and powerful spring whose stiffness is 1.6 105 n/m. how much must we compress the spring?

3.03m

Explanation:

Assuming that energy is conserved between the initial and the final states.

In the initiat state, from the question, we have both gravitational potential and elastic potential energy, while in the final state, we have only kinetic energy.

Hence:

1/2ks * s^2 + (-GMm/R) = 1/2mvf^2

ks * s^2 = mvf^2 + 2GMm/R

s = √m/ks(vf^2 + 2GM/R)

= √5/1.6*10^5(165^2 + 2(6.67*10^-11)(7.0*10^20)/3.5*10^5

= √(0.00003125 * (27225 + 266800)

= √(0.00003125 * 294025

= √ (9.18828125)

= 3.03m

568.98

explanation:

its math

neither accurate nor precise.