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

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Solution

Given that,
Mass of the body, m = 1 kg
Mass of the Earth, $M = 6 \times 10^{24} k g$
Radius of the Earth, $R = 6.4 \times 10^{6} m$
Hence, magnitude of the gravitational force (F) between the earth and the body is given by,
$F = G \frac{M \times m}{r^{2}} = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^{6})^{2}}$ = 9.77 N = 9.8 N (approx)

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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 6×1024 kg and radius of the earth is 6.4×106 m)

Given that, Mass of the body, m = 1 kg Mass of the Earth, M=6×1024 kg Radius of the Earth, R=6.4×106 m Hence, magnitude of the gravitational force (F) between the earth and the body is given by, F=GM×mr2=6.67×10−11×6×1024×1(6.4×106)2 = 9.77 N = 9.8 N (approx)

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