Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Explanation

The given ratio is `(ke^(2))/(Gm_(e) m_(p))`
Where,
G = Gravitational constant
Its unit is `N m^(2) kg^(−2)`.
`m_(e)` and `m_(p)` = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
k = A constant
`=(1)/(4piepsilon_(0))`
`epsilon_(0)` = Permittivity of free space
Its unit is `N m^(2) C^(−2)`.
Therefore, unit of the given ratio `(ke^(2))/(Gm_(e)m_(p))=([Nm^(2) C^(-2)][C^(-2)])/([Nm^(2) kg^(-2)][kg][kg])`
`" " =M^(0)L^(0) T^(0)`
Hence, the given ratio is dimensionless.
`e = 1.6 × 10^(−19) C`
`G = 6.67 × 10^(−11) N m_(2) kg^(-2)`
`me = 9.1 × 10^(−31) kg`
`mp = 1.66 × 10^(−27) kg`
Hence, the numerical value of the given ratio is .
`(ke^(2))/(Gm_(e)m_(p)) = (9 xx 10^(9)xx(1.6 xx 10^(-19))^(2))/(6.67 xx 10^(-11) xx 9.1 xx 10^(-3)xx 1.67 xx 10^(-22))~~2.3 xx 10^(39)`
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.