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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 $6\times10^{24}\ kg$ and radius of the earth is $6.4\times 10^6\ m$.
Given:
Mass of the earth is $6\times10^{24}\ kg$ and radius of the earth is $6.4\times 10^6\ m$.
To do:
To find the magnitude of the gravitational force between the earth and a $1\ kg$ object on its surface.
Solution:
We know the formula that is used to calculate the force of gravitation between two objects:
$F=G\frac{Mm}{R^2}$
Here, $F\rightarrow$ Force of gravitation
$M\rightarrow$ Mass of the object 1st
$m\rightarrow$ mass of the object 2nd
$R\rightarrow$ distance between the two objects
Let us calculate the magnitude of the gravitational force between the earth and a $1\ kg$ object on its surface by using the above formula:
Calculation of gravitational force:
Here given, the mass of the body $m=1\ kg$
Mass of the earth $M=6\times10^{24}\ kg$
Radius of earth $R=6.4\times10^{6}\ m$
Gravitational constant $G=6.7\times10^{-11}\ Nm^2kg^{-2}$
Therefore, the force of gravity on the body $F=G\frac{Mm}{R^2}$
Or $F=6.7\times10^{-11}\times\frac{(6\times10^{24})\times1}{(6.4\times10^{6})^2}$
Or $F=9.82\ N$
Therefore, the force of gravity on a body of mass $1\ kg$ lying on the surface of the earth is $9.82\ N$.
Updated on: 10-Oct-2022
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