Question

if the gravitational force between a teacher weighting 68 kg and student 52 kg is 1.48 x 10^-8 Nm^2 / kg at what distance are they standing​

Answer 1

GIVEN :–

$\\ \: \bf {\huge{.}} \: \: M_{1} = 68 \: Kg \: \: \: \: M_{2} = 52 \: Kg$

$\\ \: \bf {\huge{.}} \: \: Gravitational \: \: Force(F)= 1.48 \times {10}^{ - 8} \: N$

TO FIND :–

• Distance between them = ?

SOLUTION :–

• We know that –

$\\ \large\implies{\boxed{ \bf F= \dfrac{GM_{1}M_{2}}{ {r}^{2} }}}$

• Here –

$\\ \implies \bf G = 6.67 \times {10}^{ - 11}{ \dfrac{N {m}^{2} }{ {Kg}^{2} } }$

• Now put the values –

$\\ \implies \bf 1.48 \times {10}^{ - 8} = \dfrac{(6.67 \times {10}^{ - 11})(68)(52)}{ {r}^{2} }$

$\\ \implies \bf {r}^{2} = \dfrac{(6.67 \times {10}^{ - 11})(68)(52)}{1.48 \times {10}^{ - 8} }$

$\\ \implies \bf {r}^{2} = \dfrac{6.67 \times {10}^{ - 11} \times 68 \times 52}{1.48 \times {10}^{ - 8} }$

$\\ \implies \bf {r}^{2} = \dfrac{6.67 \times {10}^{ - 11 + 8} \times 68 \times 52}{1.48}$

$\\ \implies \bf {r}^{2} = \dfrac{6.67 \times {10}^{ -3} \times 68 \times 52}{1.48}$

$\\ \implies \bf {r}^{2} = \dfrac{23585.12 \times {10}^{ -3} }{1.48}$

$\\ \implies \bf {r}^{2} = 15935.89 \times {10}^{ -3}$

$\\ \implies \bf {r}^{2} = 15.93$

$\\ \implies \bf r= \sqrt{15.93}$

$\\ \implies \bf r=3.99$

$\\ \implies \large{ \boxed{ \bf r=4 \: m \: (Approx .)}}$