Find the gravitational force between two protons kept at a separation of 1 femtometre ( 10^-15 m ) . THe mass of the proton is 1.67 × 10 ^ - 24 kg

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Solution

$S i n c e, F = \frac{G m_{1} m_{2}}{r^{2}} G i v e n, m_{1} = m_{2} = 1.67 \times 10^{- 24} k g r = 10^{- 15} m S i n c e, G = 6.67 X 10^{- 11} N m^{2} k g^{- 2} F = \frac{6.67 X 10^{- 11} N m^{2} k g^{- 2} X (1.67 \times 10^{- 24})^{2} k g^{2}}{(10^{- 15})^{2} m^{2}} F = 18.60 X 10^{- 29} N$

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Similar questions

What is the ratio of electrostatic force and gravitational force between a proton and electron at 1 mm distance?

Mass of proton – $1.67 \times 10^{- 27} k g$

Mass of electron – $9.11 \times 10^{- 31} k g$

Charge of electron – $1.6 \times 10^{- 19} C$

What is the ratio of electrostatic force and gravitational force between a proton and electron at 1mm distance?

Mass of proton – $1.67 \times 10^{- 27}$ kg

Mass of electron – $9.11 \times 10^{- 31}$ kg

Charge of electron – $1.6 \times 10^{- 19}$ kg

1

1

= 10^{- 15} m

9.11 \times 10^{- 31}

1.67 \times 10^{- 27}

F_{e} / F_{g}

m_{p} = 1.67 \times 10^{- 27} k g

m_{e} = 9.1 \times 10^{- 31} k g

q_{p} = 1.6 \times 10^{- 19} C

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