6. Three perfectly spherical masses, A, B and C, each of radius 6.00 x 10^3 km a

Question

6. Three perfectly spherical masses, A, B and C, each of radius 6.00 x 10^3 km and mass 5.00 x 10^24 kg and uniform composition, are arranged as shown in the
diagram. Their coordinates are A (0, 10^6 km), B (-10^6 km, -10^6 km), C (+10^6 km, -10^6 km). A space ship of mass m = 10,000 kg is at the origin of the coordinate system, the point (0,0).

B) The spaceship is to be given an initial speed that will allow it to escape from the vicinity of the three planets (i.e.,instead of being pulled back by their gravitational pull). Calculate the minimum possible value of this initial speed. (Hint:use gravitational potential energy and the work energy theorem. If the rocket escapes to infinity, that is, as place where there are no gravitating masses apart from the rocket itself, what is its final gravitational potential energy?)

Explanation / Answer

AT infinity the energy of system is 0
SO,initial kinetic energy should be enough to escape the initial gravitational potential energy that are holding them together

Kinetic energy = gravitational potential energy
0.5*m*V^2 = G*m* (mA/rA + mB/rB + mC/rC)
0.5*V^2 = G* (mA/rA + mB/rB + mC/rC)

here mA =mB =mC = 5*10^24 Kg
rA = 10^9 m
rB = sqrt ( (10^9)^2 + (10^9)^2)
= 1.4142 * 10^9 m
rC = sqrt ( (10^9)^2 + (10^9)^2)
= 1.4142 * 10^9 m

0.5*V^2 = G* (mA/rA + mB/rB + mC/rC)
0.5*V^2 = (6.673*10^-11)*m (1/rA + 1/rB + 1/rC)
0.5*V^2 = (6.673*10^-11)*(5*10^24)*(1/(10^9) + 1/(1.4142*10^9) + 1/(1.4142*10^9))
0.5*V^2 = 3.3365*10^5 * [1 + 1/1.4142 + 1/1.4142]
0.5*V^2 = 3.3365*10^5 *2.4142
V= 1269 m/s

Answer: 1269 m/s

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