Several companies (Planetary Resources, Deep Space Industries, etc.) and even NA

Question

Several companies (Planetary Resources, Deep Space Industries, etc.) and even NASA are considering asteroid exploration missions that could involve moving an asteroid closer to Earth (i.e. to a lunar orbit, etc.) for study. Most asteroids are located at about 2.5 AU, and the Earth is at 1 AU (1 AU = 1.49598 times 10^6 km). So... how much energy would it take to move an asteroid to Earth? In this problem, we will step through how to answer that question by applying the vis viva equation which expresses the kinetic, potential, and total energy of a closed orbit: kinetic + potential = total m nu^2/2 - GMm/r = -GMm/2a where M is the mass of the parent body, m is the mass of the orbiting body, nu is the velocity of the orbiting body, G is the universal gravitational constant (6.673 times 10^-11 m^3 kg^-1 s^-2), r is the distance between the orbiting body and the parent body, and a is the semimajor axis of the orbit. The mass of the Sun, for reference 1.9891 times 10^30 kg. Also recall that the orbital velocity for a circular orbit (one where the radius of the orbit is the same as the semimajor axis) is given by: v = Squareroot GM/a (a) Assume the asteroid starts with a circular orbit around the Sun with semimajor axis of 2.5 AU. What is its kinetic energy per mass of the asteroid (in units of J/kg)? (b) What is the potential energy per mass of the asteroid (in units of J/kg) at 2.5 AU? (c) If the kinetic energy of this asteroid were magically removed (it stopped moving on its orbital path somehow), what would the total energy per mass be? What is the semimajor axis of an orbit with that much total energy? What do you suppose would

Explanation / Answer

a) G = 6.67e-11

mass of SUN M = 1.989e+30 kg

orbit radius a = 2.5 AU = 2.5*1.496e+6 km = 3.74e+9 m

velocity of the asteriod v = sqrt (GM/a)

KE of the ateriod = mv2/2

KE per unit mass = v2/2 = GM/2a = 6.77e-11*1.989e+30/2*3.74e+9 = 1.8e+10 J/kg

2) potential energy U = GMm/a

     PE per unit mass = U/m = GM/a = 6.77e-11*1.989e+30/3.74e+9 = -0.9e+10 J/kg

3) Total energy remains const. if the KE is removed then

Total energy per unit mass = - 0.9 e+10

a = GM/2 = 3.74e+9 m

v =0 and the only force acting on the asteriod is Gravitional attraction force and it will fall into SUN.

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