At t = 509 s after midnight, a spacecraft of mass 1240 kg is located at position

Question

At t = 509 s after midnight, a spacecraft of mass 1240 kg is located at position ‹ 7 × 105, 6 × 105, 7 × 105 › m, and at that time an asteroid whose mass is 4 × 1015 kg is located at position ‹ 6 × 105, 6 × 105, 14 × 105 › m. There are no other objects nearby.

(a) Calculate the (vector) force acting on the spacecraft.

F with arrownet =

(b) At t = 509 s the spacecraft's momentum was p with arrowi, and at the later time t = 518 s its momentum was p with arrowf. Calculate the (vector) change of momentum p with arrowf p with arrowi. p with arrowf p with arrowi = kg·m/s

Explanation / Answer

Since all of the position coordinates have the same power (10^5), we can ignore it and consider the two positions <7,6,-7> and <6,-6,-14>. The distance between them, then, is
d = ( (7 - 6)² + (6 - -6)² + (-7 - -14)²) = (1 + 144 + 49) = 253 = 13.928 x 10^5 m
(we can't ignore the 10^5 forever; it just helps with the math!)

F = GmM/d² = 6.674e-11N·m²/kg² * 1240kg * 4e15kg / (13.928e5m)² = 1.705e-4 N
If we take the asteroid to be at the "origin", then the spacecraft is at
P = (7 - 6) i + (6 - -6) j + (-7 - -14) k = 1 i + 12 j + 7 k
Then the unit vector p = P / d = 0.072 i + 0.86 j + 0.50 k
I simply divided each component by 13.928; the sum of these squares = 1.
But I am very rusty with unit 3D vectors and may be missing the point of the question.
If I'm right so far, then the vector force is F times p.

(b) The unit vector for the change in momentum must be the same as above.
momentum = Ft = 1.705e-4N * 9s = 15.345e-4 N·s
so the vector change in momentum should be 1.53e-3N·s * p

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