Question

At t = 532.0s after midnight, a spacecraft of mass 1400 kg is located at position (3x10⁵, 7x10⁵,-4x10⁵ ) m, and at that time an asteroid whose mass is 7 x 10 ¹⁵ kg is located at position (9 x 10 ^ 5, - 3 x 10 ^ 5, - 12 x 10 ^ 5) m. There are no other objects nearby. (a) Calculate the (vector) force acting on the spacecraft. (b) At t=532.0: the spacecraft's momentum was ${\overrightarrow{\mathbf{p}}}_{\mathbf{i}}$and at the later time t = 538.0s its momentum was ${\overrightarrow{\mathbf{p}}}_{\mathbf{f}}$.Calculate the (vector) change of momentum ${\overrightarrow{\mathbf{p}}}_{\mathbf{f}}\mathbf{-}{\overrightarrow{\mathbf{p}}}_{\mathbf{i}}$.

Short answer

1. The value of vector force
2. Change in vector momentum

Step-by-step solution

Identification of given data

• $t=532.0s$
• Mass of spacecraft$m=1400kg$
• Positions of spacecraft$(3\times {10}^5,7\times {10}^5,-4\times {10}^5)$
• Mass of asteroid${\overrightarrow{p}}_f-{\overrightarrow{p}}_i=40\times {10}^3\mathrm{N}·\mathrm{m}$

Calculation of the vector force

(a) According to the Newton’s law of gravitation

$F=\frac{G{m}_1{m}_2}{r^2}$

Where

G – Gravitational constant

$m_1$- Mass of 1 object

$m_2-$Mass of 2 objects

$r-$Distance between two object

$$\begin{gathered} F=\frac{6.67\times {10}^{-11}\times 1400\times 7\times {10}^{15}}{{\left(3\times {10}^5\right)}^2} \\ F=7.26\times {10}^{3}N \\ F=⟨7.26\times {10}^3,0,0⟩N \end{gathered}$$

Calculation of change in vector momentum

(b) The spacecraft momentum

When t=532 s

$$\begin{gathered} {\overrightarrow{p}}_i=F·t \\ {\overrightarrow{p}}_i=7.26\times {10}^3\times 532 \\ {\overrightarrow{p}}_i=⟨3.86\times {10}^6,0,0⟩Kg\cdot m/s \end{gathered}$$

When t=538 s

$$\begin{gathered} {\overrightarrow{p}}_f=F·t \\ {\overrightarrow{p}}_f=7.26\times {10}^3\times 538 \\ {\overrightarrow{p}}_f=⟨3.90\times {10}^6,0,0⟩Kg\cdot m/s \end{gathered}$$

Then the value of role="math" localid="1658079511166" ${\overrightarrow{p}}_f-{\overrightarrow{p}}_i$

$$\begin{gathered} {\overrightarrow{p}}_f-{\overrightarrow{p}}_i=3.86\times {10}^6-3.90\times {10}^6 \\ {\overrightarrow{p}}_f-{\overrightarrow{p}}_i=⟨40\times {10}^3,0,0⟩Kg\cdot m/s \end{gathered}$$