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A package of mass 5 kg sits at the equator of an airless asteroid of mass 3.0 1020 kg and radius 4.3 105 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 163 m/s. We have a large and powerful spring whose stiffness is 1.8 105 N/m. How much must we compress the spring?

Answer :

Answer: 1.82 m

Explanation:

Given

Mass of package, m = 5 kg

Mass of asteroid, M = 3*10^20 kg

Radius of asteroid, R = 4.3*10^5 m

Velocity is asteroid, v(f) = 163 m/s

Force constant of spring, k = 1.8*10^5 N/m

Using conservation of energy

E(i) = E(f)

1/2ks² + (GMm/R) = 1/2mv(f)²

ks² = 2GMm/R + mv(f)²

ks² = m[2GM/R + v(f)²]

s² = m/k * [2GM/R + v(f)²]

s² = 5/1.8*10^5 * [((2 * 6.67*10^-11 * 3*10^20)/4.3*10^5) + 163²]

s² = 2.77*10^-5 * [(93070 + 26569)]

s² = 2.77*10^-5 * 119639

s² = 3.314

s = √3.314

s = 1.82 m

The distance for compressing the speed is 1.82 m.

Conversation of energy:

Since

Mass of package, m = 5 kg

Mass of asteroid, M = 3*10^20 kg

Radius of asteroid, R = 4.3*10^5 m

Velocity is asteroid, v(f) = 163 m/s

Force constant of spring, k = 1.8*10^5 N/m

Now we know that

E(i) = E(f)

1/2ks² + (GMm/R) = 1/2mv(f)²

ks² = 2GMm/R + mv(f)²

ks² = m[2GM/R + v(f)²]

s² = m/k * [2GM/R + v(f)²]

s² = 5/1.8*10^5 * [((2 * 6.67*10^-11 * 3*10^20)/4.3*10^5) + 163²]

s² = 2.77*10^-5 * [(93070 + 26569)]

s² = 2.77*10^-5 * 119639

s² = 3.314

s = √3.314

s = 1.82 m

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