Chapter 1: Q35P (page 40)

The crew of a stationary spacecraft observe an asteroid whose mass is $4 \times 10^{17}$ kg. Taking the location of the spacecraft as the origin, the asteroid is observed to be at location $< - 3 \times 10^{3}, - 4 \times 10^{3}, 8 \times 10^{3} >$ m at a time $18.4$ s after lunchtime. At a time $21.4$ s after lunchtime, the asteroid is observed to be at location $< - 1.4 \times 10^{3}, - 6.2 \times 10^{3}, 9.7 \times 10^{3} >$ m. Assuming that the velocity of the asteroid does not change during this time interval, calculate the vector velocityrole="math" localid="1656672441674" $\vec{v}$ of the asteroid.

Short Answer

Expert verified

The velocity of the asteroid is: $\overset{⇀}{V} = <5.33 \times 10^{2}, - 7.33 \times 10^{2}, 5.67 \times 10^{2}>$ m/s.

Step by step solution

Identification of given data

Mass- $4 \times 10^{17}$ kg

Location at a time $18.4$ s after lunchtime- $<- 3 \times 10^{3}, - 4 \times 10^{3}, 8 \times 10^{3}>$ m

Location at a time s after lunchtime- $<- 1.4 \times 10^{3}, - 6.2 \times 10^{3}, 9.7 \times 10^{3}>$ m

Calculating velocity

The velocity of the asteroid is the change in position vector over the change in time, that is

$\overset{⇀}{v} = \frac{∆ \overset{⇀}{r}}{∆ t} = \frac{<- 1.4 \times 10^{3}, - 6.2 \times 10^{3}, 9.7 \times 10^{3}> - <- 3 \times 10^{3}, - 4 \times 10^{3}, 8 \times 10^{3}>}{21.4 - 18.4} \overset{⇀}{v} = \frac{<1.6 \times 10^{3}, - 2.2 \times 10^{3}, 1.7 \times 10^{3}>}{3} \overset{⇀}{v} = <5.33 \times 10^{2}, - 7.33 \times 10^{2}, 5.67 \times 10^{2}> m / s$