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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: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 $9.82\ N$.

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