What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface?

According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by 

= / ²  

Where, 

Mass of Earth, M = 6 × 10²⁴ kg 

Mass of object, m = 1 kg 

Universal gravitational constant, G = 6.7 × 10⁻¹¹ Nm² kg⁻² 

Since the object is on the surface of the Earth, 

r = radius of the Earth (R) 

r = R = 6.4 × 10⁶ m 

Therefore, the gravitational force 

= / 2 = 6.7 × 10⁻¹¹ × 6 × 10²⁴ × 1 / (6.4 × 106) 2 = 9.8  

Gravitational force between the earth and an F = \(\frac{GMm}{R^2}\)
Where G = Gravitational constant = 6.67 × 10^-11 Nm²kg^-2.
M = Mass of earth = 6 × 10²⁴ kg.
R = Radius of earth = 6.4 × 10⁶m
m = Mass of object = 1 kg.
Hence F = \(\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1}{(6.4\times 10^6)^2}\)
= 9.8N.