Question

A package 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?

Answer 1

Answer:

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

Answer 2

The elongation of the spring during compression will be $s=1.82m$

What will be the compression of the spring?

It is given that

Mass of package, m = 5 kg

Mass of asteroid, $M=3\times10^{20} kg$

The radius of the asteroid, $R=4.3\times 10^5m$  

Velocity is an asteroid, $V_f=163\ \frac{m}{s}$

Spring constant $K=1.8\times10^5\frac{N}{m}$  

Now from the conservation of energy

Initial Energy = Final Energy

$\dfrac{1}{2} ks^2+\dfrac{GMm}{R} = \dfrac{1}{2} mV_f^2$

$Ks^2=\dfrac{2GMm}{R} +mV_f^2$

$Ks^2=m(\dfrac{2GM}{R} +V_f^2)$

$s^2=\dfrac{m}{k} (\dfrac{2GM}{R} +V_f^2)$

$s^2= \dfrac{5}{1.8\times10^5} [(\dfrac{2\times 6.67\times10^{-11}\times 3\times10^{20}}{4.3\times10^5} )+163^2]$

$s^2=3.314$  

$s=\sqrt{3.314}$

$s=1.82m$

Thus the elongation of the spring during compression will be $s=1.82m$

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