Calculate the force of gravitation due to earth on a ball of $1 k g$ mass lying on the ground.(Mass of earth $= 6 \times 10^{3} k m$ ; and $G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})$

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Solution

Gravitational force between earth and ball,
$= \frac{G M m}{R^{2}}$ [Neglecting the radius of ball as it is small]
$= \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 1}{(6400 \times 10^{3})^{2}}$
$= \frac{9.77 \times 10^{- 3} \times 10^{- 11} \times 10^{29}}{10^{10}} = 9.77 N$

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Calculate the force of gravitation due to earth on a ball of 1 kg mass lying on the ground.(Mass of earth =6×103 km; and G=6.7×10−11Nm2/kg2)

Gravitational force between earth and ball,=GMmR2 [Neglecting the radius of ball as it is small]=6.67×10−11×6×1024×1(6400×103)2=9.77×10−3×10−11×10291010=9.77N

Calculate the force of gravitation due to earth on a ball of $1 k g$ mass lying on the ground.(Mass of earth $= 6 \times 10^{3} k m$ ; and $G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})$

Gravitational force between earth and ball,
$= \frac{G M m}{R^{2}}$ [Neglecting the radius of ball as it is small]
$= \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 1}{(6400 \times 10^{3})^{2}}$
$= \frac{9.77 \times 10^{- 3} \times 10^{- 11} \times 10^{29}}{10^{10}} = 9.77 N$