Question

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × \(10^{24}\) kg and radius of the earth is 6.4 × \(10^6\) m.)

Answer 1

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

F = \(\frac{GMm}{r^2}\)
Where, 
Mass of Earth, M = 6 × \(10^{24}\) kg 
Mass of object, m = 1 kg 
Universal gravitational constant, G = 6.7 × \(10^{−11} Nm^2 kg^{−2 }\)
Since the object is on the surface of the Earth,
\(r\) = radius of the Earth (\(R\))
\(r\) = \(R\) = 6.4 × \(10^6\) m 

Therefore, the gravitational force 
\(F\) = \(\frac{GMm}{r^2}\) 

\(F\) = \(\frac{6.7×10^{−11}× 6×10^{24}×1}{ (6.4×10^6)^2}\) 

\(F\) = 9.8 𝑁