Question

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The force of attraction (A) between two objects, whose masses are fixed, varies inversely with the square of their distance of separation (d). The force of attraction between them is 2 units when the dis- tance is 5 cm. What is the distance between the two objects, if the attraction between them is 8 units?
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Answer 1

Answer:

$\boxed{\sf distance \: between \: the \: objects \: \: is \: 2.5 \: cm}$

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

$A = G \cdot \frac{m_1.m_2}{ {d}^{2} }$

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

Given:

A1 = 2 units

d1= 5 cm

A2 = 8 units

To find:

Distance of separation when the force of attraction is 8 units d2 = ?

Solution:

Substituting the given values in above at each point,

$A_1 = G \cdot \frac{m_1.m_2}{ {d_1}^{2} } \\ 2 = G \cdot \frac{m_1.m_2}{ {5}^{2} } \\ \sf similarly \: at \: second \: point \: of \: attraction \\ A_2 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ 8 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ \sf \: \: dividing \: both \: of \: the \: equation \\ \frac{2}{8} = \frac{G \cdot \frac{m_1.m_2}{ {5}^{2} }}{G \cdot \frac{m_1.m_2}{ {d_2}^{2} } } \\ \sf \: G \: m_1 and \: m_2 \: are \: \: constant \: hence \\ \sf they \: can \: be \: cancelled \: out \\ \frac{2}{8} = \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } } \\ \sf rearranging \: above \: equation \\ \frac{1}{4} = \frac{{d_2}^{2} }{ {5}^{2} } \\ {d_2}^{2} = \frac{ {5}^{2} }{4 } \\ {d_2}^{2} = \frac{25}{4} \\ {d_2}^{2} = 6.25 \\ \sqrt{{d_2}^{2} } = \sqrt{6.25} \\ \boxed{ \sf{d_2} = 2.5 \: cm}$

Answer: the distance between the two objects is 2.5 cm, if the attraction between them is 8 units.

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