A solid uniform sphere has amass of$1.0\times {10}^4\mathbf{kg}$and a radius of$1.0\mathbf{m}$. What is the magnitude of the gravitational force due to the sphere on a particle of massmlocated at a distance of(a) 1.5 m and (b) 0.50 m from the center of the sphere? (c) Write a general expression for the magnitude ofthe gravitational force on the particle at a distance$r\le 1.0\mathbf{m}$from the center of the sphere.

i) Mass of sphere is$1\times {10}^4\text{\hspace{0.33em}kg}$

ii) Radius of sphere is$1.0\text{\hspace{0.33em}m}$

To find the gravitational force, we have to use Newton’s law of gravitation in which force is directly proportional to the product of masses and inversely proportional to the distance between the objects.

Formula:

Volume of sphere, …(i)

Density of a material, …(ii)

Gravitational force, …(iii)

Magnitude of the force at a distance of 1.5 m using equation (iii), we get

$\begin{array}{c}F=\frac{\left(6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\times \left(1\times {10}^4\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\times \left(m\right)}{{\left(1.5\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2}\\ =(2.96\times {10}^{-7}m)\text{\hspace{0.33em}N/kg}\end{array}$

The gravitational force at a distance 1.5m from the centre of the sphere is$\left(2.96\times {10}^{-7}m\right)\text{N/kg}$

Using equation (iii), the net gravitational force at distance 1 m is

$\begin{array}{c}F=\frac{\left(6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\times \left(1\times {10}^4\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\times \left(m\right)}{{\left(1\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2}\\ =\left(3.0\times {10}^{-7}m\right)\text{N/kg}\end{array}$

To find gravitational force at distance 0.5 m, we need to find the mass at that distance using density of sphere as:

Density of sphere

$\begin{array}{c}\rho =\frac{M}{\frac{4}{3}\pi r^3}\mathrm{.............................}\text{(fromequation(ii))}\\ =\frac{1\times {10}^4\text{\hspace{0.17em}\hspace{0.17em}kg}}{\frac{4\pi }{3}\times {\left(1\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3}\\ =2.4\times {10}^3{\text{kg/m}}^{\text{3}}\end{array}$

So, mass at distance of$r=0.5m$

$\begin{array}{c}M=\rho \times \left(\frac{4}{3}\right)\pi r^3\\ =2.4\times {10}^3{\text{\hspace{0.17em}\hspace{0.17em}kg/m}}^{\text{3}}\times \frac{4}{3}\pi \times {\left(0.5\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3\\ =1.3\times {10}^3\text{\hspace{0.33em}kg}\end{array}$

So now gravitational force is:

$\begin{array}{c}F=\frac{\left(6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\times \left(1.3\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\times m}{{\left(0.5\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2}\\ =\left(3.3\times {10}^{-7}m\right)\text{N/kg}\end{array}$

The gravitational force at a distance 0.5m from the centre of the sphere is$\left(3.3\times {10}^{-7}m\right)\text{N/kg}$

Mass of sphere in terms of radius from equation (ii), can be written as:

$\begin{array}{c}M=p\times \left(\frac{4}{3}\right)\pi r^3\\ =2.4\times {10}^3\times \left(\frac{4}{3}\right)\pi r^3\\ =10053.09r^3\end{array}$

$\begin{array}{c}F=\frac{\left(6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\times \left(m\right)\times \left(10053.09r^3\text{\hspace{0.17em}\hspace{0.17em}kg}\right)}{r^2}\\ =\left(6.7\times {10}^{-7}mr\right)\frac{\text{N}}{\text{kg.m}}\end{array}$

The general expression for force is

$\left(6.7\times {10}^{-7}mr\right)\frac{\text{N}}{\text{kg.m}}$