A planet of mass $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{\text{\hspace{0.17em}kg}}$ is at location$\langle 5e11,-2e11,0\rangle \mathbf{\text{\hspace{0.33em}m}}$. A star of mass $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{50}}\mathbf{\text{\hspace{0.33em}kg}}$ is at location$\langle -2e11,3e11,0\rangle \mathbf{\text{\hspace{0.33em}m}}$. It will be useful to draw a diagram of the situation, including the relevant vectors.

(a) What is the relative position vector pointing from the planet to the star?

(b) What is the distance between the planet and the star?

(c) What is the unit vector role="math" localid="1657099553132" $\widehat{\mathbf{r}}$in the direction of $\widehat{\mathbf{r}}$?

(d) What is the magnitude of the force exerted on the planet by the star?

(e) What is the magnitude of the force exerted on the star by the planet?

(f) What is the (vector) force exerted on the planet by the star?

(g) What is the (vector) force exerted on the star by the planet?

• Mass of the planet,$m_P=4\times {10}^{24}\text{\hspace{0.33em}kg}$.
• Mass of the star,$m_S=5\times {10}^{30}\text{\hspace{0.33em}kg}$.
• Planet location,${\overrightarrow{R}}_P=\left(5\times {10}^{11},-2\times {10}^{11},0\right)\text{\hspace{0.33em}m}$.
• Star location,${\overrightarrow{R}}_S=\left(-2\times {10}^{11},3\times {10}^{11},0\right)\text{\hspace{0.33em}m}$ .

The force exerted by an object which has a mass on an object which also has a mass can be written as:

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

Here $m_1$and$m_2$are the masses, and r is the distance between them.

Force exerted by object-1 on object-2 equals the force exerted by object-2 on object-1.

The relative position vector pointing from the planet to the star can be calculated as:

${\overrightarrow{R}}_{SP}={\overrightarrow{R}}_S-{\overrightarrow{R}}_P$

Substituting the values in the above expression, and we get,

$\begin{array}{c}{\overrightarrow{R}}_{SP}=\left(-2\times {10}^{11},3\times {10}^{11},0\right)\text{\hspace{0.33em}m}-\left(5\times {10}^{11},-2\times {10}^{11},0\right)\text{\hspace{0.33em}m}\\ =\left(-7\times {10}^{11},5\times {10}^{11},0\right)\text{\hspace{0.33em}m}\end{array}$

The relative position vector pointing from the planet to the star is $\left(-7\times {10}^{11},5\times {10}^{11},0\right)\text{\hspace{0.33em}m}$

Step 3: Distance between star and planet

b)

${\overrightarrow{R}}_{SP}=\left(-7\times {10}^{11},5\times {10}^{11},0\right)\text{\hspace{0.33em}m}$

Distance between star and planet can be calculated as:

$\begin{array}{c}\left|\overrightarrow{R_{SP}}\right|=\sqrt{{\left(-7\times {10}^{11}\right)}^2+{\left(5\times {10}^{11}\right)}^2+0}\\ =8.605\times {10}^{11}\text{\hspace{0.33em}m}\end{array}$

Distance between star and planet is $8.605\times {10}^{11}\text{\hspace{0.33em}m}$ .

$\overrightarrow{R_{SP}}=\left(-7\times {10}^{11},5\times {10}^{11},0\right)\text{\hspace{0.33em}m}$

$\left|\overrightarrow{R_{SP}}\right|=8.605\times {10}^{11}\text{\hspace{0.33em}m}$

Unit vector can be calculated as:

${\widehat{R}}_{SP}=\frac{{\overrightarrow{R}}_{SP}}{\left|{\overrightarrow{R}}_{SP}\right|}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}{\widehat{R}}_{SP}=\frac{\left(-7\times {10}^{11},5\times {10}^{11},0\right)\text{\hspace{0.33em}m}}{8.605\times {10}^{11}\text{\hspace{0.33em}m}}\\ =\left(-0.813,0.5810,0\right)\end{array}$

A unit vector is$\left(-0.813,0.5810,0\right)$.

Gravitational force can be calculated as:

$F=G\frac{m_P\times m_S}{{\left|{\overrightarrow{R}}_{SP}\right|}^2}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}F=6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times \frac{4\times {10}^{24}\text{\hspace{0.33em}kg}\times 5\times {10}^{30}\text{\hspace{0.33em}kg}}{{\left(8.605\times {10}^{11}\text{\hspace{0.33em}m}\right)}^2}\\ =1.80\times {10}^{21}\text{\hspace{0.33em}N}\end{array}$

Thus the value of force is$1.80\times {10}^{21}\text{\hspace{0.33em}N}$

Force exerted by object-1 on object-2 equals the force exerted by object-2 on object-1. The gravitational force exerted by the planet on the start is the same force exerted on the planet by the star.

Thus the value of force will be$1.80\times {10}^{21}\text{\hspace{0.33em}N}$ .

Gravitational force can be calculated as:

$\overrightarrow{F}=G\frac{m_P\times m_S}{{\left|{\overrightarrow{R}}_{SP}\right|}^2}{\widehat{R}}_{SP}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}\overrightarrow{F}=1.80\times {10}^{21}\text{\hspace{0.33em}N}\times \left(-0.813,0.5810,0\right)\\ =\left(-1.495\times {10}^{21},1.0458\times {10}^{21},0\right)\text{\hspace{0.33em}N}\end{array}$

Thus the value of force(vector) will be $\left(-1.495\times {10}^{21},1.0458\times {10}^{21},0\right)\text{\hspace{0.33em}N}$

${\overrightarrow{R}}_{PS}={\overrightarrow{R}}_P-{\overrightarrow{R}}_S$

Substituting the values in the above expression, and we get,

$\begin{array}{c}{\overrightarrow{R}}_{PS}=\left(5\times {10}^{11},-2\times {10}^{11},0\right)\text{\hspace{0.33em}m}-\left(-2\times {10}^{11},3\times {10}^{11},0\right)\text{\hspace{0.33em}m}\\ =\left(7\times {10}^{11},-5\times {10}^{11},0\right)\text{\hspace{0.33em}m}\end{array}$

$\begin{array}{c}{\overrightarrow{R}}_{PS}=\left(7\times {10}^{11},-5\times {10}^{11},0\right)\text{\hspace{0.33em}m}\\ \left|\overrightarrow{R_{PS}}\right|=\sqrt{{\left(7\times {10}^{11}\right)}^2+{\left(-5\times {10}^{11}\right)}^2+0}\\ =8.605\times {10}^{11}\text{\hspace{0.33em}m}\end{array}$

${\widehat{R}}_{PS}=\frac{{\overrightarrow{R}}_{PS}}{\left|{\overrightarrow{R}}_{PS}\right|}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}{\widehat{R}}_{PS}=\frac{\left(7\times {10}^{11},-5\times {10}^{11},0\right)\text{\hspace{0.33em}m}}{8.605\times {10}^{11}\text{\hspace{0.33em}m}}\\ =\left(0.813,-0.5810,0\right)\end{array}$

$\overrightarrow{F}=G\frac{m_S\times m_P}{{\left|{\overrightarrow{R}}_{PS}\right|}^2}{\widehat{R}}_{PS}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}\overrightarrow{F}=1.80\times {10}^{21}\text{\hspace{0.33em}N}\times \left(0.813,-0.5810,0\right)\\ =\left(1.495\times {10}^{21},-1.0458\times {10}^{21},0\right)\text{\hspace{0.33em}N}\end{array}$

Thus the value of force (vector) will be$\left(1.495\times {10}^{21},-1.0458\times {10}^{21},0\right)\text{\hspace{0.33em}N}$.