An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its center at a distance x to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (see Fig. E13.35). What is the gravitational force that the sphere exerts on the ring-shaped object? Show that your result reduces to the expected result when x is much larger than a.

• The radius of the object is a.
• The mass of the object is M.
• The mass of the uniform sphere is m.
• The radius of the uniform sphere is R.

This law states that every particle attracts another particle which is directly proportional to the product of the masses and inversely proportional to the square of the distances amongst them.

The gravitational constant multiplied by the mass and divided by the square of their distances gives the gravitational force on the sphere.

From Newton’s law of gravitation, the force exerted by the sphere is expressed as:

$F=G\frac{m_1{m}_2}{r^2}$

Here, F is the force exerted by the sphere, G is the gravitational constant,$m_1\hspace{0.33em}and\hspace{0.33em}{m}_2$ the mass of the object and the sphere, and r is the distance between the objects.

The free body diagram of these objects can be expressed as-

Analyzing the free body diagram, the force on the mass of the thin ring due to m mass acting at a distance L is described as:

As the ring is symmetrical, therefore, the “vertical component” of this particular force mainly cancels out which leaves only the horizontal component.

Hence, the horizontal component mainly adds from the two “symmetrically opposite mass elements” , the horizontal component of the force is expressed as-

$d{F}^{\text{'}}=dFcos\varphi$…………………….(1)

Here, is the angle between the x-axis and the L distance. Hence, evaluating the equation (1), we get,

$d{F}^{\text{'}}=G\frac{m\times dM}{L^2}cos\varphi$………………(2)

Apart from that, we can calculate the length L from the above figure using the Pythagorean theorem, such that

$$\begin{gathered} L^2=a^2+x^2 \\ L={\sqrt{a^2+x}}^2 \end{gathered}$$………………………..(3)

Where a is the radius of the ring and x is the distance of the axis from the sphere to the axis.

Then, we can define the cosine function, such that,

$\cos \phi =\frac{x}{L}$……………………………..(4)

Hence, by using the equation (2) and equation (3), we get,

$$\begin{gathered} \mathrm{dF}\text{'}=\mathrm{G}\frac{\mathrm{m}\times \mathrm{dM}}{{\mathrm{L}}^2}\frac{\mathrm{x}}{\mathrm{L}} \\ =\mathrm{G}\frac{\mathrm{x}\times \mathrm{m}\times \mathrm{dM}}{{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}}^{\frac{\mathbf{3}}{\mathbf{2}}}} \end{gathered}$$………………….(5)

Hence, integrating both the sides of equation (5), we get,

localid="1658899304839" $$\begin{gathered} =\int \left(G\frac{x\times m\times dM}{{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}}^{\frac{\mathbf{3}}{\mathbf{2}}}}\right) \\ \int dF\text{'}=\left(Gm\frac{x}{{(a^2+x^2)}^{{\displaystyle \frac{3}{2}}}}\right)\int dM \end{gathered}$$

Using $\int dM=M$, we get,

localid="1658899754124" $F\text{'}=G\frac{mMx}{{\left(x^2\left(1+{\left({\displaystyle \frac{a}{x}}\right)}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}}$ …………………..(6)

Hence, the gravitational force of the ring-shaped object islocalid="1658899055092" $G\frac{mMx}{{(a^2+x^2)}^{{\displaystyle \frac{3}{2}}}}$.

We can write equation (6) as,

$F\text{'}=G\frac{mMx}{{\left(x^2\left(1+{\left({\displaystyle \frac{a}{x}}\right)}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}}$

In the condition x>>a, we can write that ${\left(\frac{a}{x}\right)}^2$will be very small and then we can write the above equation as,

$$\begin{gathered} F\text{'}=G\frac{mMx}{{\left(x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =G\frac{mM}{x^2} \end{gathered}$$

It is verified that the result is expected when ‘x’ is much larger than ‘a’.