Question

Problem 3.27 Your answer is incorrect. Try again At t = 484 s after midnight, a spacecraft of mass 1200 kg is located at position <3 x 105, 4 × 105-3 × 105> m, and at that time an asteroid whose mass is 7 x 1015 kg is located at position <9 x 105, -3 x 105, -17 x 105>m. There are no other objects nearby. (a) Calculate the (vector) force acting on the spacecraft. F net0.036 8.29e5 1.66e4 > N (b) At t = 484 s the spacecraft's momentum was pi, and at the later time ț = 491 s its momentum was pf. Calculate the (vector) change of momentum p-P pf-pi = || 0.0000546 58.03e5 11.62e4 > kg.m/s

Answer 1

Distance between the asteroid and spacecraft which will be given as -

| $\vec{r}$| = _$\sqrt{}$(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

| $\vec{r}$| = _$\sqrt{}$[(9 x 10⁵ m) - (3 x 10⁵ m)]² + [(-3 x 10⁵ m) - (4 x 10⁵ m)]² + [(-17 x 10⁵ m) - (-3 x 10⁵ m)]²

| $\vec{r}$| = _$\sqrt{}$2.81 x 10¹² m²

| $\vec{r}$| = 1.67 x 10⁶ m

The unit vector can be calculated by -

$\hat{r}$ = $\vec{r}$ / | $\vec{r}$|

$\hat{r}$ = (x₂ - x₁) $\hat{i}$ + (y₂ - y₁) _$\hat{j}$ + (z₂ - z₁) $\hat{k}$ / _$\sqrt{}$(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

$\hat{r}$ = [(9 x 10⁵ m) - (3 x 10⁵ m)] $\hat{i}$ + [(-3 x 10⁵ m) - (4 x 10⁵ m)] _$\hat{j}$+ [(-17 x 10⁵ m) - (-3 x 10⁵ m)] $\hat{k}$ / (1.67 x 10⁶ m)

$\hat{r}$ = (6 x 10⁵ m) $\hat{i}$ + (-7 x 10⁵ m) _$\hat{j}$ + (-14 x 10⁵ m) $\hat{k}$ / (1.67 x 10⁶ m)

$\hat{r}$ = (0.359) $\hat{i}$ + (-0.419) _$\hat{j}$ + (-0.838) $\hat{k}$

(a) The force acting on the spacecraft which will be given by -

$\vec{F}$_net = {G M m / | $\vec{r}$|² } $\hat{r}$

where, M = mass of a spacecraft = 1200 kg

m = mass of an asteroid = 7 x 10¹⁵ kg

G = gravitational constant = 6.67 x 10^-11 Nm²/kg²

then, we get

$\vec{F}$_net = { [(6.67 x 10^-11 Nm²/kg²) (1200 kg) (7 x 10¹⁵ kg)] / (2.81 x 10¹² m²) } [(0.359) $\hat{i}$ + (-0.419) _$\hat{j}$ + (-0.838) $\hat{k}$]

$\vec{F}$_net = (1.99 x 10^-4 N) [(0.359) $\hat{i}$ + (-0.419) _$\hat{j}$ + (-0.838) $\hat{k}$]

$\vec{F}$_net = (7.14 x 10^-5 N) $\hat{i}$ + (-8.33 x 10^-5 N) _$\hat{j}$ + (-1.66 x 10^-4 N) $\hat{k}$

Then, we have

$\vec{F}$_net = < 7.14 x 10^-5 , -8.33 x 10^-5 , -1.66 x 10^-4 > N