Question

calculate the force of gravitation between two protons which are placed at a distance of 10*15m the mass of a proton is 1.67×10^-27​

Answer 1

Answer:

gravitational force = (G.m1.m²)/r² = 6.67×10^-11×(1.67 ×10^-27)²/(10^-15)²

Force = 6.67*1.67*1.67 × 10^-31 = 1.86 × 10^-30N

Answer 2

Answer

Given -

$\longrightarrow$mass of each proton - $\bf 1.67 \times 10^{-27}$

$\longrightarrow$Distance between them - $\bf 10^{15}$

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To find -

$\longrightarrow$Force of gravitation.

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Formula used -

$\bf \boxed{\bf \dfrac{GM_1M_2}{ {r}^{2} }}$

where -

$\longrightarrow$G is the universal gravitation constant.

$\longrightarrow$$\bf M_1$ is mass of one proton.

$\longrightarrow$$\bf M_2$ is mass of another proton.

$\longrightarrow$r is the distance between them.

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Solution -

$\implies$$\bf G = 6.67 \times 10^{-11}$

$\implies$$\bf M_1 = 1.67 \times 10^{-27}$

$\implies$$\bf M_2 = 1.67 \times 10^{-27}$

$\implies$$\bf r = 10^{15}$

Substituting the value in formula -

$\bf\dfrac{GM_1M_2}{{r}^{2}}$

$\implies$$\bf = \dfrac{6.67 \times {10}^{ - 11} \times 1.67 \times {10}^{ - 27} \times 1.67 \times {10}^{ - 27}}{ { ({10}^{15}) }^{2} }$

$\implies$$\bf = \dfrac{6.67 \times 1.67 \times 1.67 \times {10}^{ - 11} \times{10}^{ - 27} \times {10}^{ - 27} }{ {10}^{30} }$

$\implies$$\bf = \dfrac{18.6 \times {10}^{ - 11 - 27 - 27} }{ {10}^{30} }$

$\implies$$\bf = \dfrac{18.6 \times {10}^{ - 65} }{ {10}^{30} }$

$\implies$$\bf = 18.6 \times {10}^{ - 95}$

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