Question

$\huge\underline{\fbox{Challenge}}\ \textless \ br /\ \textgreater$

The masses of two different bodies are 6000 kg and 24000 kg respectively. The distance between both the bodies is 3 km. Find the position where an object should be kept along the line joining the centres of both the bodies, such that the net gravitational force on the object is zero.
(Take G = 6.7 x 10-¹¹ N m² kg-²)

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Answer 1

⇝ Given :-

• The masses of two different bodies are 6000 kg and 24000 kg respectively.

• The distance between both the bodies is 3 km.

⇝ To Find :-

• The position where an object should be kept along the line joining the centres of both the bodies, such that the net gravitational force on the object is zero.

⇝ Solution :-

We Have,

• Mass of 1st body = m₁ = 6000 kg

• Mass of 2nd body = m₂ = 24000 kg

• Distance b/w two bodies = r = 3000 m

We know Gravitational Force between two masses m₁ and m₂ separated by a distance r is given by the formula :

$\boxed{\red{ \boxed{\bf F_G=\dfrac{Gm_1m_2}{r^2}}}}$

Now,

Let a object of Mass m is placed at a distance x meter from m₁ between on the line joining the two masses m₁ and m₂ as shown in attached figure.

❒ As net Gravitational Force on the object should be zero :

So,

$\implies$Force on m due to m₁ and m₂ should be equal.

Hence,

$\bf\dfrac{Gm_1m}{x^2}=\dfrac{Gm_2m}{(3000-x)^2}$

$:\longmapsto \dfrac{ \cancel{6000}}{ \text{x}^{2} } = \dfrac{ \cancel{24000}}{(3000 - \text x)^{2} }$

$:\longmapsto \dfrac{1}{ { \text x}^{2} } = \frac{4}{ {(3000 - \text x)}^{2} }$

⭐ Taking Square Root Both Side :

$:\longmapsto \dfrac{1}{ \text x} = \frac{2}{3000 - \text x}$

$:\longmapsto2 \text x = 3000 - \text{x}$

$:\longmapsto3 \text{x} = 3000$

$:\longmapsto \text{x} = \cancel\dfrac{3000}{3}$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf x=1000} }}}$

So, the Mass should be placed at the distance of 1000 m near the m₁.

That is ;

The Mass should be placed at the distance of 1 km from m₁ .

Answer 2

$\texttt{\textsf{\large{\underline{Question}:}}}$

The masses of two different bodies are 6000 kg and 24000 kg respectively. The distance between both the bodies is 3 km. Find the position where an object should be kept along the line joining the centres of both the bodies, such that the net gravitational force on the object is zero.

$\texttt{\textsf{\large{\underline{Solution}:}}}$

m₁ = 6000 kg

m₂ = 24000 kg

r = 3000 m

Where,

m₁ = Mass of 1st body

m₂ = Mass of 2nd body

r = Distance between two bodies

Formula :-

$FG = \frac{ Gm₁m₂ }{r²}$

As given net Gravitational Force on the object should be zero

So,Force on m due to m₁ and m₂ should be

So,Force on m due to m₁ and m₂ should beequal

Hence,

$\frac{6000}{ {x}^{2} } = \frac{24000}{(3000 - x) ^{2} }$

$\frac{1}{ {x}^{2} } = \frac{4}{(3000 - x) ^{2} }$

★Take square root both the side

$\frac{1}{x} = \frac{2}{3000 - x }$

$2x = 3000 - x$

$3x = 3000$

★Now cancel

$x = \frac{3000}{3}$

$x = 1000$

$\huge\tt\pink{WHICH \: IS \: THE \: REQUIRED \: ANSWER}$