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Question: Find the gravitational force acting on a particle A inside a uniform spherical layer of matter....

Find the gravitational force acting on a particle A inside a uniform spherical layer of matter.

Explanation

Solution

Divide the sphere into thin spherical shells and find the force acting on the particle lying inside the sphere by a thin spherical shell and integrate within the suitable limits.

Complete step by step solution:
Let M be the mass of the uniform spherical layer of matter and m be the mass of the particle lying inside it at a distance of r from the centre. Let the sphere be divided into thin spherical shells of density σ\sigma per unit area. Force acting on the particle by a thin spherical shell of radius R is given by


dF=GmdMs2cos⁡αdF = \dfrac{{GmdM}}{{{s^2}}}\cos \alpha
Mass =dM=σ2πRsin⁡θRdθdM = \sigma 2\pi R\sin \theta Rd\theta
The force from the entire spherical shell is given by
F=2πGσmR2∫θ=0πcos⁡αsin⁡θdθs2F = 2\pi G\sigma m{R^2}\int\limits_{\theta = 0}^\pi {\dfrac{{\cos \alpha \sin \theta d\theta }}{{{s^2}}}} -----(1)
In order to find the value of the integral we need to express in terms of. To do so we use cosine formula. Using the cosine law, we have s2=R2+r2θ−2Rrcos⁡θ{s^2} = {R^2} + {r^2}\theta - 2Rr\cos \theta

Differentiating both sides, we get
2sds=2Rrsin⁡θdθ2sds = 2Rr\sin \theta d\theta
⇒sin⁡θdθ=sdsRr\Rightarrow \sin \theta d\theta = \dfrac{{sds}}{{Rr}} ------------(2)
Now using the cosine formula for the external angle
R2=r2+s2−2rscos⁡α{R^2} = {r^2} + {s^2} - 2rs\cos \alpha
cos⁡α=r2+s2−R22rs\cos \alpha = \dfrac{{{r^2} + {s^2} - {R^2}}}{{2rs}} ------------(3)
cos⁡α=r2+s2−R22rs\cos \alpha = \dfrac{{{r^2} + {s^2} - {R^2}}}{{2rs}} ------------(3)
Now, put the values of equations (2) and (3) in equation (1). Also, forθ=0,s=R−r\theta = 0,s = R - rand for θ=π,s=r+R\theta = \pi ,s = r + R
Now (1) reduces to
F=−2πGσmR2∫s=R−rs=R+r(r2+s2−R2)sdss2.2rs.RrF = - 2\pi G\sigma m{R^2}\int\limits_{s = R - r}^{s = R + r} {\dfrac{{\left( {{r^2} + {s^2} - {R^2}} \right)sds}}{{{s^2}.2rs.Rr}}}
⇒F=−πGσmR2r2∫s=R−rs=R+r(1+r2−R2s2)ds\Rightarrow F = \dfrac{{ - \pi G\sigma m{R^2}}}{{{r^2}}}\int\limits_{s = R - r}^{s = R + r} {\left( {1 + \dfrac{{{r^2} - {R^2}}}{{{s^2}}}} \right)ds}
Using the area density expression ⇒σ=M4πR2 \Rightarrow \sigma = \dfrac{M}{{4\pi {R^2}}} the above equation reduces to
F=−GmM4Rr2∫s=R−rs=R+r(1+r2−R2s2)dsF = \dfrac{{ - GmM}}{{4R{r^2}}}\int\limits_{s = R - r}^{s = R + r} {\left( {1 + \dfrac{{{r^2} - {R^2}}}{{{s^2}}}} \right)ds}
⇒F=−GmM4Rr2[s−(r2−R2)1s]s=R−rs=R+r\Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {s - \left( {{r^2} - {R^2}} \right)\dfrac{1}{s}} \right]_{s = R - r}^{s = R + r}
⇒F=−GmM4Rr2[(R+r)−(R−r)−(r2−R2)(1R+r−1R−r)]\Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {\left( {R + r} \right) - \left( {R - r} \right) - \left( {{r^2} - {R^2}} \right)\left( {\dfrac{1}{{R + r}} - \dfrac{1}{{R - r}}} \right)} \right]
⇒F=−GmM4Rr2[(2r)+(−2r)]⇒F=0\Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {\left( {2r} \right) + \left( { - 2r} \right)} \right] \Rightarrow F = 0

Thus the algebraic sum of the forces acting on the particle lying inside a sphere is zero.

Note: The gravitational force acting on a particle lying inside a spherical layer of matter is always zero. This is true for all particles lying inside a sphere.