The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data : The universal constant of gravitation $G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2},$ mass of the moon = $7.36 \times 10^{22} k g,$ mass of the earth $= 6 \times 10^{24} k g$ and the distance between the earth and the moon $= 3, 8 \times 10^{5} k m .$

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Solution

$F = \frac{G M m}{R^{2}}$

$= \frac{6.67 \times 10^{- 11} \times 7.36 \times 10^{22} \times 6 \times 10^{24}}{3.8 \times 3.8 \times 10^{16}}$

$= 2 \times 10^{20} N$

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The mass of the earth is $6.00 \times 10^{24} k g$ and that of the moon is $7.40 \times 10^{22} k g$ . The constant of gravitation $G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2}$ . The potential energy of the system is $- 7.79 \times 10^{28} j o u l e$ . The mean distance between the earth and moon is

The masses of the earth and moon are $6 \times 10^{24} k g$ and $7.4 \times 10^{22} k g$ , respectively. The distance between them is $3.84 \times 10^{5} m$ . Calculate the gravitational force of attraction between the two. Use $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$

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F=GMmR2 =6.67×10−11×7.36×1022×6×10243.8×3.8×1016 =2×1020 N

$F = \frac{G M m}{R^{2}}$

$= \frac{6.67 \times 10^{- 11} \times 7.36 \times 10^{22} \times 6 \times 10^{24}}{3.8 \times 3.8 \times 10^{16}}$

$= 2 \times 10^{20} N$