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

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

Step-by-step explanation:

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

`F= (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