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.)
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.)
Solution and Explanation
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
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