91Ó°ÊÓ

• Mass of the planet,$m_P=4×{10}^{24}\text{ k²µ}$.
• Mass of the star,$m_S=5×{10}^{30}\text{ k²µ}$.
• Planet location,${\stackrel{→}{R}}_P=\left(5×{10}^{11},−2×{10}^{11},0\right)\text{ m}$.
• Star location,${\stackrel{→}{R}}_S=\left(−2×{10}^{11},3×{10}^{11},0\right)\text{ m}$ .

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:

${\stackrel{→}{R}}_{SP}={\stackrel{→}{R}}_S−{\stackrel{→}{R}}_P$

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

$\begin{array}{c}{\stackrel{→}{R}}_{SP}=\left(−2×{10}^{11},3×{10}^{11},0\right)\text{ m}−\left(5×{10}^{11},−2×{10}^{11},0\right)\text{ m}\\ =\left(−7×{10}^{11},5×{10}^{11},0\right)\text{ m}\end{array}$

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

Step 3: Distance between star and planet

b)

${\stackrel{→}{R}}_{SP}=\left(−7×{10}^{11},5×{10}^{11},0\right)\text{ m}$

Distance between star and planet can be calculated as:

$\begin{array}{c}\left|\stackrel{→}{R_{SP}}\right|=\sqrt{{\left(−7×{10}^{11}\right)}^2+{\left(5×{10}^{11}\right)}^2+0}\\ =8.605×{10}^{11}\text{ m}\end{array}$

Distance between star and planet is $8.605×{10}^{11}\text{ m}$ .

$\stackrel{→}{R_{SP}}=\left(−7×{10}^{11},5×{10}^{11},0\right)\text{ m}$

$\left|\stackrel{→}{R_{SP}}\right|=8.605×{10}^{11}\text{ m}$

Unit vector can be calculated as:

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

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

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

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

Gravitational force can be calculated as:

$F=G\frac{m_P×m_S}{{\left|{\stackrel{→}{R}}_{SP}\right|}^2}$

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

$\begin{array}{c}F=6.67×{10}^{−11}\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}×\frac{4×{10}^{24}\text{ k²µ}×5×{10}^{30}\text{ k²µ}}{{\left(8.605×{10}^{11}\text{ m}\right)}^2}\\ =1.80×{10}^{21}\text{ N}\end{array}$

Thus the value of force is$1.80×{10}^{21}\text{ N}$

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$1.80×{10}^{21}\text{ N}$ .

Gravitational force can be calculated as:

$\stackrel{→}{F}=G\frac{m_P×m_S}{{\left|{\stackrel{→}{R}}_{SP}\right|}^2}{\widehat{R}}_{SP}$

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

$\begin{array}{c}\stackrel{→}{F}=1.80×{10}^{21}\text{ N}×\left(−0.813,0.5810,0\right)\\ =\left(−1.495×{10}^{21},1.0458×{10}^{21},0\right)\text{ N}\end{array}$

Thus the value of force(vector) will be $\left(−1.495×{10}^{21},1.0458×{10}^{21},0\right)\text{ N}$

${\stackrel{→}{R}}_{PS}={\stackrel{→}{R}}_P−{\stackrel{→}{R}}_S$

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

$\begin{array}{c}{\stackrel{→}{R}}_{PS}=\left(5×{10}^{11},−2×{10}^{11},0\right)\text{ m}−\left(−2×{10}^{11},3×{10}^{11},0\right)\text{ m}\\ =\left(7×{10}^{11},−5×{10}^{11},0\right)\text{ m}\end{array}$

$\begin{array}{c}{\stackrel{→}{R}}_{PS}=\left(7×{10}^{11},−5×{10}^{11},0\right)\text{ m}\\ \left|\stackrel{→}{R_{PS}}\right|=\sqrt{{\left(7×{10}^{11}\right)}^2+{\left(−5×{10}^{11}\right)}^2+0}\\ =8.605×{10}^{11}\text{ m}\end{array}$

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

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

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

$\stackrel{→}{F}=G\frac{m_S×m_P}{{\left|{\stackrel{→}{R}}_{PS}\right|}^2}{\widehat{R}}_{PS}$

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

$\begin{array}{c}\stackrel{→}{F}=1.80×{10}^{21}\text{ N}×\left(0.813,−0.5810,0\right)\\ =\left(1.495×{10}^{21},−1.0458×{10}^{21},0\right)\text{ N}\end{array}$

Thus the value of force (vector) will be$\left(1.495×{10}^{21},−1.0458×{10}^{21},0\right)\text{ N}$.