Question

Question: The masses and coordinates of three spheres are as follows: 20 kg ,x = 0. 50 m , y =1.0 m ; 40 kg , x = - 1. 0 , y = - 1 . 0 m ; 60 kg , x = 0 m ,y = - 0.50 m .What is the magnitude of the gravitational force on 20 kg sphere located at the origin due to these three spheres?

Short answer

Answer:

The magnitude of the gravitational force on the sphere is $\left|\overrightarrow{F}\right|=3.2\times {10}^{-7}N$

Step-by-step solution

Listing the given quantities

The mass of the first sphere is${\text{m}}_{\text{1}}=20\mathrm{kg}$

The mass of the second sphere is${\text{m}}_{\text{2}}=40\mathrm{kg}$

The mass of the third sphere is ${\text{m}}_{\text{3}}=60\mathrm{kg}$

The mass of the central sphere is $\text{m}=20\mathrm{kg}$

The coordinates of the first sphere are ${\text{m}}_{\text{1}}(0.50\mathrm{m},1.0\mathrm{m})$

The coordinates of the second sphere are ${\text{m}}_{\text{2}}(-1.0\mathrm{m},-1.0\mathrm{m})$

The coordinates of the third sphere are ${\text{m}}_{\text{3}}(0\mathrm{m},-0.50\mathrm{m})$

The coordinates of the central sphere are $\text{m}(0,0)$

Understanding the concept of gravitation and the principle of superposition

We can use the concept of gravitation and superposition principle.

Formula:

$$\begin{gathered} \overrightarrow{\mathrm{F}}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}\hat{\mathrm{r}} \\ \overrightarrow{\mathrm{F}}=\sum _{}\stackrel{\rightharpoonup }{\mathrm{F}} \end{gathered}$$

Explanation

From the figure, the magnitude of r₁ is

$$\begin{gathered} {\text{r}}_{\text{1}}=\sqrt{{\left(0.50\text{m}\right)}^2+{\left(1.0\text{m}\right)}^2} \\ =1.118m \end{gathered}$$

The magnitude of r₂ is

$$\begin{gathered} {\text{r}}_{\text{2}}=\sqrt{{\left(1.0\text{m}\right)}^2+{\left(1.0\text{m}\right)}^2} \\ =1.414m \end{gathered}$$

The magnitude of r₃ is

${\text{r}}_{\text{2}}=0.50m$

According to Newton’s law of gravitation, the force of attraction is

$\stackrel{\rightharpoonup }{F}=\frac{\text{GMm}}{{\text{r}}^{\text{2}}}\hat{r}$

According to the superposition law,

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3$

$\begin{array}{c}\overrightarrow{{\mathrm{F}}_{\mathrm{net}}}=\mathrm{Gm}\left[\left(\underset{\mathrm{n}=1}{\overset{3}{}}\frac{{\mathrm{m}}_{\mathrm{n}}{\mathrm{x}}_{\mathrm{n}}}{{\mathrm{r}}_{\mathrm{n}}^3}\right)\widehat{\mathrm{i}}+\left(\underset{\mathrm{n}=1}{\overset{3}{}}\frac{{\mathrm{m}}_{\mathrm{n}}{\mathrm{y}}_{\mathrm{n}}}{{\mathrm{r}}_{\mathrm{n}}^3}\right)\widehat{\mathrm{j}}\right]\\ =(6.67\times {10}^{-11}\frac{\mathrm{N}\cdot {\mathrm{m}}^2}{{\mathrm{kg}}^2})\times (20\mathrm{kg})\left[\left(\frac{20\mathrm{kg}\times 0.50\mathrm{m}}{{(1.118\mathrm{m})}^3}\right)+\left(\frac{40\mathrm{kg}\times (-1.0\mathrm{m})}{{(1.414\mathrm{m})}^3}\right)+\left(\frac{20\mathrm{kg}\times (0\mathrm{m})}{{(-0.50\mathrm{m})}^3}\right)\right]\widehat{\mathrm{i}}+\\ (6.67\times {10}^{-11}\frac{\mathrm{N}\cdot {\mathrm{m}}^2}{{\mathrm{kg}}^2})\times (20\mathrm{kg})\left[\left(\frac{20\mathrm{kg}\times 1.0\mathrm{m}}{{(1.118\mathrm{m})}^3}\right)+\left(\frac{40\mathrm{kg}\times (-1.0\mathrm{m})}{{(1.414\mathrm{m})}^3}\right)+\left(\frac{20\mathrm{kg}\times (-0.50\mathrm{m})}{{(-0.50\mathrm{m})}^3}\right)\right]\widehat{\mathrm{j}}\\ =(-9.3\times {10}^{-9}\mathrm{N})\widehat{\mathrm{i}}+(-3.2\times {10}^{-7})\widehat{\mathrm{j}}\end{array}$

$$\begin{gathered} F=\sqrt{{\left(-9.3\times {10}^{-9}N\right)}^2+{\left(-3.2\times {10}^{-7}N\right)}^2} \\ =3.2\times {10}^{-7}N \end{gathered}$$

The magnitude of the gravitational force on the sphere m is $\left|\overrightarrow{F}\right|==3.2\times {10}^{-7}N$