Answer to Question #115863 in Astronomy | Astrophysics for Julia french

Question #115863
When the gravitational force of attraction between 2 protons is compared to the electrostatic force of repulsion between the 2 protons, the electrostatic force of repulsion will be what?
1
Expert's answer
2020-05-15T08:43:52-0400

Compare the magnitudes of forces of interaction between two protons at the same distance. Gravitational force:


Fg=Gmp2r2.F_g=G\frac{m_p^2}{r^2}.

Electrical force:


Fe=ke2r2.F_e=k\frac{e^2}{r^2}.

Divide the electrical by the gravitational:


FeFg=ke2Gmp2= =(1/(4πϵ0))(1.6021019)2(6.6731011)(1.6731027)2=1.2361036.\frac{F_e}{F_g}=\frac{ke^2}{Gm_p^2}=\\ \space\\=\frac{(1/(4\pi\epsilon_0))(1.602\cdot10^{-19})^2}{(6.673\cdot10^{-11})(1.673\cdot10^{-27})^2}=1.236\cdot10^{36}.

The electrical force is 1036 times stronger than the gravitational one.


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