Question

at what distance apart would to equal masses of 150 kg need to be placed for the force between them to be 2×
${10}^{ - 5}$

Answer 1

$\: as \: you \: know \: the \: force \: of \: attraction \: \\ between \: two \: bodies \: according \: \\ to \: inverse \: square \: law \: is \: given \: as \: = \\ \\ \\ \: \: \: \: \: \: \: \: f \: = \: \alpha \times \frac{m \times m}{ {r}^{2} } \\ \\ \: \: \: \: \: \: where \: \: \alpha \: = \: gravitational \: constant \\ \\ but \: \: \: \: \: \: f \: \: = \: 2 \times {10}^{ - 5} \: newton \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: m \: = \: 150 \: kg \\ \\hence \: \: \: \: \: 2 \times {10}^{ - 5} = \: 6.67 \times \: {10}^{ - 11} \times \: \frac{150 \times 150}{ {r}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: {r}^{2} = \: 6.67 \times {10}^{ - 11} \times \frac{150 \times 150}{2 \times {10}^{ - 5} } \\ \\ \: \: \: \: \: \: \: {r}^{2} \: = \: 0.075037 \: \: {m}^{2} \\ \\ \: \: \: \: \: \: \: \: \: r \: = \: 0.2739 \: m \: = \: 0.28 \: m \\ \\ hence \: the \: distance \: between \: \\ them \: should \: be \: 0.28 \: m.$
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