Two lead spheres of `20 cm` and `2 cm` diametre respectively are planet with centres `100 cm` apart. Calculate the attraction between them, given the radius of the Earth as `6.37 xx 10^(8) cm` and its mean density as `5.53 xx 10^(3) kg m^(-3)`. Speciffic gravity of lead `= 11.5`. If the lead spheres are replaced by bress sphere of the same radii, would the force of attraction be the same?

Explanation

Here, `r_(1) = 0.20//2 = 0.1 m`,
`r_(2) = 0.02//2 = 0.01 m, r = 1.0 m`,
`rho = 5.53 xx 10^(3) kg m^(-3)`,
`rho' = 11.5 xx 10^(3) kg//m^(3)`
Mass of first lead sphere, `m_(1) = (4)/(3) pi r_(1)^(3) rho'`
`= (4)/(3) xx 3.14 xx (0.1)^(3) xx 11.5 xx 10^(3)`
`= 4.815 xx 10^(-2) kg`.
We know that, `g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho`
or `G = (3g)/(4pi R rho)`
Force of attraction between lead spheres will
`F = (Gm_(1) m_(2))/(r^(2)) = (3g)/(4pi R rho) xx (m_(1) m_(2))/(r^(2))`
`= (3 xx 9.8 xx 48.15 xx 4.815 xx 10^(2))/(4 xx 3.14 xx(6.37 xx 10^(6)) xx 5.53 xx 10^(3) xx (1)^(2))`
`= 15.4 xx 10^(-11) N`