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This question is best solved using the law of conservation of energy. The orbital potential energy equation is best derived through the definition of potential energy as negative work. Work equals the integral of force with respect to x. If you integrate GMm/r2 dr from infinity to the orbital radius, you will find that orbital potential energy is -GMm/R. If we plug in the universal gravitational constant for G- 6.6742*10-11 m3/(kg*s2), the Mass of the Sun for M- 1.9891*1030 kg, the mass of Pluto for m- 1.305*1022 kg, and plutos average orbital radius for R- 5.906*1012 m, we will find the potential energy of Pluto to be -2.933*1029 J. This is also the value of kinetic energy, just positive (because KE is a scalar value). If KE=2.933*1029 =mv2/2, v=SQRT(2KE/m) = (m being mass of Pluto) 6,715 m/s or15,002 mph.

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