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?

=(5.7 x i - 2.24 x j) kgm/s

Explanation:

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

r^

Plugging in the values, we have

F= [(6.67x)(1500)(5.97 x )(8xi + 9x j)] / ((8x)² + (9x )²

F= 2736 i^ + 3078 j^

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

= + FΔt

= 1500(3.5 x i - 1.8x j) + (2736i + 3078j)(1.5x)

=(5.7 x  i- 2.24 x j)kgm/s

volts times amps 6 x 0.5 3w

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