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

Question

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?

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bonexptip 4 years 2021-07-20T14:43:29+00:00 2 Answers 63 views 0

Answers ( )

  1. Xavia
    0
    2021-07-20T14:44:30+00:00
    Reply

    Answer:

     The Spring must be compressed by a length  x= 2.0m

    Explanation:

    From the question we are told that

             The mass is M_p = 5kg

            The mass of the  asteroid M_a= 3.0* 10^{20} \ kg

             The radius of the asteroid is R_a = 4.3 * 10^5 \ m

              The speed of the asteroid is v = 163 \ m/s  

             The stiffness is k = 1.8*10^{5}N/m

    From this question we can see that it the spring that provide the required energy to lunch this package

        Hence according to the law of conservation of energy

    Energy Provided by the spring =  kinetic energy of the package +  the gravitational potential energy of the package

       

      Energy Provided by the spring   = \frac{1}{2} kx^2

    Where x is the length the spring is to be compressed to obtain the desired result

      kinetic energy of the package  =\frac{1}{2} M_p v^2

    the gravitational potential energy of the package   =\frac{GM_a M_p}{R_a}

    Expressing the equation mathematically

                 

    \frac{1}{2} kx^2 = \frac{1}{2} M_p v^2 +\frac{GM_a M_p}{R_a}

    Making x the subject  

        x =\sqrt{ \frac{ M_p v^2 +\frac{GM_a M_p}{R_a} }{k} }

    Substituting values

        x = \sqrt{\frac{5 * 163^2 + \frac{(6.67*10^{-11}) (3.0*10^{20}) (5)}{4.3*10^5} }{1.8*10^5} }

           x= 2.0m

       

     

  2. Giakhanh
    0
    2021-07-20T14:44:47+00:00
    Reply

    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

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