What is the magnitude of the gravitational fo | vedclass

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(B) According to the universal law of gravitation,the gravitational force $F$ exerted on an object of mass $m$ is given by the formula:$F = \frac{G M

Vedclass · Accepted Answer

$9.8$

Vedclass · Answer

(B) According to the universal law of gravitation,the gravitational force $F$ exerted on an object of mass $m$ is given by the formula:$F = \frac{G M m}{r^2}$Where:Mass of Earth,$M = 6 	imes 10^{24} \,kg$Mass of object,$m = 1 \,kg$Universal gravitational constant,$G = 6.67 	imes 10^{-11} \,Nm^2 \,kg^{-2}$Radius of the Earth,$r = R = 6.4 	imes 10^6 \,m$Substituting the values into the formula:$F = \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 1}{(6.4 	imes 10^6)^2}$$F = \frac{40.02 	imes 10^{13}}{40.96 	imes 10^{12}}$$F \approx 9.8 \,N$Thus,the magnitude of the gravitational force is approximately $9.8 \,N$.

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

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