Question

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.

Short answer

a) Gravitational force due to sphere on particle of mass m at$1.5\text{\hspace{0.33em}m}$ is$\left(3\times {10}^{-7}m\right)\text{N/kg}$

b) Gravitational force due to sphere on particle of mass m at $0.5\text{\hspace{0.33em}m}$is$\left(3.3\times {10}^{-7}m\right)\text{N/kg}$

c) General expression for force on particle at distance of$r\le 1.0m$ from center of sphere is$\left(6.7\times {10}^{-7}mr\right)\text{N/kgm}$

Step-by-step solution

The given data

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

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

Understanding the gravitational force

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)

(a) Calculation of net force at distance 1.5 m from center of sphere

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$\left(2.96\times {10}^{-7}m\right)\text{N/kg}$

(b) Calculation of net force at distance 0.5 m from center of sphere

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$\left(3.3\times {10}^{-7}m\right)\text{N/kg}$

(c) General expression for force at from center of sphere

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

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

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

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

The general expression for force is

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