Question

Two spherical bodies of masses M

and are 4M kept at separation of d

as shown in figure. Where should we

place a third particle of mass m so

that it will experience a zero net

force? ​

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

• Two masses m and 4m .
• They are kept distance d m apart .

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

• The distance at which the third particle of mass m should be placed so that the net force is zero.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } FormulA\:UseD:-}}}}}$

We will use universal law of gravitation that is stated as ,

$\boxed{\red{\bf{\green{\dag}\:Force=G\dfrac{m_1m_2}{r^2}}}}$

where ,

• r is distance of separation.
• m1 & m2 are two masses.
• G is universal gravitational constant.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

Let us take that the third mass m should be placed at a distance of x from mass m .

Now , the mass m will experience force from both the masses of 4 m and m.

$\underline{\blue{\tt{\longmapsto Case\:One: Attraction\: between\:4m\:\&\:m}}}$

$\tt Here$

• $\sf Mass_1 = 4m$
• $\sf Mass_2 = m$
• $\sf Distance = d - x$

$\bf \qquad\qquad Force\: exerted:$

$\tt \implies Force_1=G\dfrac{m_1m_2}{r^2}$

$\tt \implies Force_1 = G\dfrac{4m\times m}{(d-x)^2}$

$\boxed{\bf \red{\hookrightarrow} Force_1 = G\dfrac{4m^2}{(d-x)^2}}$

$\underline{\blue{\tt{\longmapsto Case\:Two: Attraction\: between\:m\:\&\:m}}}$

$\tt Here$

• $\sf Mass_1 = m$
• $\sf Mass_2 = m$
• $\sf Distance = x$

$\bf \qquad\qquad Force\: exerted:$

$\tt \implies Force_2=G\dfrac{m_1m_2}{r^2}$

$\tt \implies Force_2= G\dfrac{m\times m}{x^2}$

$\boxed{\bf \red{\hookrightarrow}Force_2 = G\dfrac{m^2}{(x)^2}}$

__________________________________

$\pink{\sf \purple{\longrightarrow}Now\:to\:net\:force\:to\:be\:0,}$

$\tt:\implies Force_1=Force_2$

$\tt:\implies G\dfrac{m^2}{(x)^2}=G\dfrac{4m^2}{(d-x)^2}$

$\tt:\implies \dfrac{4\cancel{m^2G}}{\cancel{m^2G}}=\dfrac{(d-x)^2}{x^2}$

$\tt:\implies 4=\dfrac{(d-x)^2}{x^2}$

$\tt:\implies \bigg\lgroup \dfrac{d-x}{x}\bigg\rgroup^2 =2^2$

$\tt:\implies \dfrac{d-x}{x}=2$

$\tt:\implies d-x=2x$

$\tt:\implies d = 2x + x$

$\tt:\implies 3x = d$

$\underline{\boxed{\red{\tt{\longmapsto\:\:\:\:x\:\:=\:\:\dfrac{d}{3}}}}}$

$\boxed{\green{\bf\pink{\dag}\:Hence\:it\:should\:be\:placed\:at\:a\: distance\:of\:\bf{\dfrac{d}{3}}}}$

Answer 2

Explanation:

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