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What is the force of gravity acting on a point of mass of $1$ kg released from the height of $1$ m from the surface of the earth; the radius of earth is equal to \[\mathbf{6}.\mathbf{4}\text{ * }\mathbf{10}\] to the power $6$ metre of $8$ is equal to \[\mathbf{6}\text{ *}\mathbf{10}\] to the power $24$ kg?

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint: First of all, we are supposed to know the formula to calculate the force. After that we will take the required values and will put in the formula according to the requirement. We have to put extra care on the units as well.

Formula used:
Here we are supposed to use the formula for force of gravity:
where $F$= force, $G$= gravitational constant, ${{m}_{1}}$=mass of object $1$, ${{m}_{2}}$=mass of object 2 and $r$= distance between centres of the masses.

Complete step by step answer:
The law of Gravitation says that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centres-of-mass of both points, and here, this force is basically proportional to the masses of the objects and also it is inversely proportional to their separation (distance). Basically, this force of attraction always points inward from one point to another point.

This Law is applied on all the objects with masses either big or small. We can consider two big objects as a point like masses if the distance between them is very huge as compared to their sizes or even if they are symmetric spherically. For such types of cases the mass of each object will be represented as a point mass that is located at its center of mass.

To calculate the force acting on the object, it will be given by:
Mass of earth (${{m}_{1}}$) =\[6\times 1{{0}^{24}}\,kg\]
Mass of the object (${{m}_{2}}$)=\[1\text{ }kg\]
Radius of the earth = \[6.4\times {{10}^{6}}\,m\]
Here, we have to find the force of gravity acting on the object and from the law of gravitation we know that
\[\text{Force}=\text{Gravitational constant} \times \dfrac{(\text{mass of object1}\times \text{mass of object2)}}{{{(\text{distance between centres of the masses)}}^{2}}}\]

So, $F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
The value of universal gravitational constant (G) is known to us. It is:
\[G = 6.67\times {{10}^{-11}}\,N{{m}^{2}}k{{g}^{-2}}_{^{{}}}\]
Putting the values in formula:
$\Rightarrow F=6.67 \times {{10}^{-11}}\dfrac{(6\times {{10}^{24}} \times 1)}{{{(6.4\text{ }\times \text{ }{{10}^{6}})}^{2}}}\,N$
$\therefore F=9.8\,N$

Therefore, the force acting on an object can be given as $F=9.8\,N$.

Note: It is very interesting to know that an apple fell over Sir Isaac Newton and from there he started working on his theories and eventually ended up discovering gravitational force. So we can say the inspiration for the Law of Universal Gravitation was actually from an apple which dropped from a tree.
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