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What is the magnitude of the gravitational force between the earth and a $1 \,kg$ object on its surface? (Mass of the earth is $6 \times 10^{24}\, kg$ and radius of the earth is $6.4 \times 10^6\, m$.)
$8.9\, N$
$9.8\, N$
$5.3\, N$
$2.3\, N$
Solution
According to the universal law of gravitation, gravitational force exerted on an object of mass $m$ is given by
$F=\frac{G M m}{r^{2}}$
Where,
Mass of Earth, $M = 6 \times 10^{24}\, kg$
Mass of object, $m = 1 \,kg$
Universal gravitational constant, $G = 6.7 \times 10^{-11}\, Nm^2 \,kg^{-2}$
Since the object is on the surface of the Earth,
$r =$ radius of the Earth $(R)$
$r = R = 6.4 \times 10^6\, m$
Therefore, the gravitational force
$F=\frac{G M m}{r^{2}}=\frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}=9.8\, N$
