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A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.8 × 104, 0⟩ m/s, and its position with respect to the center of the earth is ⟨8 × 106, 9 × 106, 0⟩ m. (The center of the earth is at the origin of the coordinate system.) What is the (approximate) new momentum of the asteroid 1.5 × 103 seconds later?

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Rau7star

Answer:

[tex]P_{f}[/tex] =(5.7 x [tex]10^{7[/tex] i - 2.24 x [tex]10^{7[/tex] j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

[tex]F= \frac{Gm_{1}m_{2} }{r^{2} }[/tex] r^

Plugging in the values, we have

F= [(6.67x[tex]10^{-11}[/tex])(1500)(5.97 x [tex]10^{24}[/tex])(8x[tex]10^{6}[/tex]i + 9x[tex]10^{6[/tex] j)] / ((8x[tex]10^{6}[/tex])² + (9x[tex]10^{6[/tex] )²[tex])^{1.5}[/tex]

F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

[tex]P_{f}[/tex] = [tex]P_{i[/tex] + FΔt

[tex]P_{f}[/tex] = 1500(3.5 x [tex]10^{4[/tex] i - 1.8x[tex]10^{4[/tex] j) + (2736i + 3078j)(1.5x[tex]10^{3[/tex])

[tex]P_{f}[/tex] =(5.7 x [tex]10^{7[/tex]  i- 2.24 x [tex]10^{7[/tex] j)kgm/s

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