Question

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.
What is the gravitational force that the sphere exerts on the ring-shaped object?
Express your answer in terms of the variables a, M, m, R, x, and appropriate constants.
F = 2GMmx/(x2+ R2)3/2

Answer 1

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

Answer 2

The gravitational force the sphere exerts on the ring-shaped object will be

$F=\dfrac{ GMmx}{\sqrt{a^2+x^2}^3 }$

What is gravitational force?

From the gravitational law of the forces, every object present in the universe applies force on each other this force depends upon the mass of the body and the distance between them, and this force is called the gravitational force.

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

$dF=\dfrac{GmdM}{L^2}$

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

$dF'=dFcos\phi$

where Ф is the angle between L and the x-axis

$dF'=\dfrac{GmdMcos\phi}{L^2}$

$L^2=a^2+x^2$

where a = radius of ring and x = distance of the axis of the ring from the sphere.

$L=\sqrt{a^2+x^2}$

$cos\phi=\dfrac{x}{L}$

$dF'=\dfrac{GmdMcos\phi}{L^2}$

$dF'=\dfrac{GmMd x}{L^3}$

$dF'=\dfrac{GmdM }{(\sqrt{a^2+x^2} )^3}$

Integrating both sides we have

$\int dF'=\int \dfrac{GmdMcos\phi}{L^2}$

$\int dF'=\dfrac{Gm \int dMx}{\dsqrt{(a^2+x^2)^3} } \ \ Here \ \int dM=M$

$F=\dfrac{GmMx}{\dsqrt{(a^2+x^2)^3} }$

Thus the gravitational force the sphere exerts on the ring-shaped object will be

$F=\dfrac{GmMx}{\dsqrt{(a^2+x^2)^3} }$  

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