Question

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Answer 1

Question :-

Infinite number of bodies each of mass 2 kg are situated on x - axis at distance 1 m ,2 m , 4 m and 8 m respectively from the origin the resulting gravitational potential due to this system at the origin will be -

Answer :-

$\mapsto \sf \: \boxed{V_{r} = - 4G \:} \:$

Option (d) is correct .

To Find :-

→ Gravitational potential at origin.

Used Formula :-

Sum of infinite term if first term is "a" and common ratio is "r" -

$\mapsto \: \boxed{ \sf \: s_{\infty} \: = \frac{a}{1 - r} }$

Solution :-

According to the question ,

→ All bodies have same mass of 2 kg , with increasing distance 1 m , 2 m ,4 m ,8 m ....

We know that

The gravitational potential due to a mass "m" at a distance "d" is -

$</p><p> \to \: \boxed{ \sf \: V = \frac{ - Gm}{d} \: \: }$

Because there is infinite number of bodies ,so total gravitational potential is -

$\to \sf \: - \frac{Gm}{1} - \frac{Gm}{2} - \frac{Gm}{4} - ...... \infty \: \\ \\ \to \sf - Gm \bigg \{ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ..... \infty \bigg\} \\ \\ \to \sf \: - Gm \bigg \{1 + \frac{1}{2} + \frac{1}{ {2}^{2} } + \frac{1}{ {2}^{3} } + .... \infty \bigg \}$

Now ,apply sum of infinite terms of a geometric progression ,here

• first term (a) = 1

• common ration $\frac{1}{2}$

Hence ,

$\mapsto \sf - Gm \bigg( \frac{1}{1 - \frac{1}{2} } \bigg) \\ \\ \mapsto \sf \: - Gm \: \times \frac{1}{ \frac{1}{2} } \\ \\ \mapsto \sf \: - Gm \: \times 2 \\ \\ \mapsto \sf \: - 2 \: Gm \\ \\ \sf because \: each \: of \: mass \: m = 2 \: kg$

Hence ,resultant gravitational potential is -

$\mapsto \sf \: \boxed{V_{r} = - 4G \:}$