Question

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?

Answer 1

Answer:

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

Explanation:

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

$F= \frac{Gm_{1}m_{2} }{r^{2} }$ r^

Plugging in the values, we have

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

F= 2736 i^ + 3078 j^

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

$P_{f}$ = $P_{i$ + FΔt

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

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