What is magnitude of gravitational force between the earth and a 1 kg object on its surface ? Take mass of earth to be `6 xx 10^24 kg` and radius of the earth is` 6.4 xx 10^6 m.G =6.67 xx10^-11nm^2kg^-2`.

Explanation

Gravitational constant,
`G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`
`G = 6.67 xx 10^(-11) Nm^(2)kg^(-2)`
Mass of the earth, `m_(1) = 6 xx 10^(24) kg`
Mass of the body, `m_(2) = 1 kg`
Radius of the earth, `R = 6.4 xx 10^(6) m`
By universal law of gravitation,
`F = (G xx m_(1) xx m_(2))/(R^(2))`
`F = (6.67 xx 10^(-11) Nm^(-2) kg^(-2) xx 6 xx 10^(24) kg xx 1 kg)/((6.4 xx 10^(6)m)^(2))`
F = 9.8 N