Question.
What is the magnitude of the gravitational force between the earth and a $1 \mathrm{~kg}$ object on its surface? (Mass of the earth is $6 \times 10^{24} \mathrm{~kg}$ and radius of the earth is $6.4 \times 10^{6} \mathrm{~m}$.)
What is the magnitude of the gravitational force between the earth and a $1 \mathrm{~kg}$ object on its surface? (Mass of the earth is $6 \times 10^{24} \mathrm{~kg}$ and radius of the earth is $6.4 \times 10^{6} \mathrm{~m}$.)
Solution:
$\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{G} \times \mathrm{M} \times \mathrm{m}}{\mathrm{R}^{2}}$
Where, M = mass of the earth ; m = mass of object; R = radius of earth.
$\mathrm{F}_{8}=\frac{6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right) \times(1)}{\left(6.4 \times 10^{6}\right)^{2}}$
$=9.770 \mathrm{~N} \approx 9.8 \mathrm{~N}$
$\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{G} \times \mathrm{M} \times \mathrm{m}}{\mathrm{R}^{2}}$
Where, M = mass of the earth ; m = mass of object; R = radius of earth.
$\mathrm{F}_{8}=\frac{6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right) \times(1)}{\left(6.4 \times 10^{6}\right)^{2}}$
$=9.770 \mathrm{~N} \approx 9.8 \mathrm{~N}$
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