State the universal law of gravitation. The mass of the electron and proton in a Hydrogen atom is given by 9×10^-31 Kg and 1.9×10^-27 Kg respectively which is separated by a distance 6×10^-11m. Calculate the gravitational force of attraction between them? [3 MARKS]

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Solution

Definition: 1 Mark
Problem: 2 Mark

1.The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

2. The given parameters are:
M_e=9×10^-31 Kg and M_p=1.9×10^-27 Kg, G=6.673×10^-11Nm²/Kg², r=6×10^-11m
$F = \frac{G m_{1} m_{2}}{r^{2}}$

F=6.673×10−11×9×10−31×1.9×10−27(6×10−11)2

F=3.16×10^-47N

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Similar questions

5 \times 10^{- 11} m

= 1.6 \times 10^{- 19} C

= 9.1 \times 10^{- 31}

= 1.6 \times 10^{- 27} k g, G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})

5.3 \times 10^{- 11} m

= 9.1 \times 10^{- 31} k g, m p = 1.67 \times 10^{- 27} k g, e = 1.6 \times 10^{- 19} C

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

5 \times 10^{- 11} m

= 1.6 \times 10^{- 19} C

= 9.1 \times 10^{- 31}

= 1.6 \times 10^{- 27} k g, G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})

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

9.1 \times 10^{- 31}

10 A^{0}

1 A^{0} = 10^{- 10} m .

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