What is the magnitude of the gravitational force between the earth and 1 kg object on its surface Mass of the earth is 6 into 24?

AcademicPhysicsNCERTClass 9

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 $9.82\ N$.

What is the magnitude of the gravitational force between the earth and 1 kg object on its surface Mass of the earth is 6 into 24?

Updated on 10-Oct-2022 13:22:37

Answer

What is the magnitude of the gravitational force between the earth and 1 kg object on its surface Mass of the earth is 6 into 24?
Verified

Hint To solve this problem, use the universal law of gravitation, using the data given in the question. The formula for the universal law of gravitation gives us the magnitude of the force of attraction (or the gravitational force) between the two objects. $\Rightarrow F = \dfrac{{GMm}}{{{r^2}}}$Where $F,G,M,m,r$ represent gravitational force, universal gravitation constant, the mass of the body (1), the mass of the body (2), and the distance between the two bodies respectively.

Complete step by step answer

Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. From the given data in the question.$\Rightarrow M = 6 \times {10^{24}}kg =$ Mass of earth$\Rightarrow m = 1kg =$ Mass of object$\Rightarrow G = 6.67 \times {10^{ - 11}}\dfrac{{N{m^2}}}{{k{g^2}}} =$ Universal gravitational constant$\Rightarrow r = 6.4 \times {10^6}m =$ Radius of earthThe distance between the object and the earth is taken as the radius of the earth because by definition the distance is taken from the center of the two bodies. Since the object is placed on the surface of the earth, the distance from the center of the earth to the surface is the same as the radius of the earth.Substituting the above values in the formula for gravitational force,$\Rightarrow F = \dfrac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}} \times 1}}{{{{(6.4 \times {{10}^6})}^2}}} $$\therefore F = 9.77 \approx 9.8N $So, the magnitude of the gravitational force between the earth and a $1kg$ object is found to be $9.8N$.

Additional Information

Newton has not only given the universal law of gravitation but also gave three other laws of motion. These are considered as the most important laws of physics and they are,(1) Every object moves in a straight line unless acted upon by a force. (2) The acceleration of an object is directly proportional to the net force exerted and inversely proportional to the object's mass. (3) For every action, there is an equal and opposite reaction.

Note

Usually, the distance between the earth and any object placed on earth’s surface is taken equal to the radius of the earth because the size of the object is negligible compared to the size of the earth and it is practically impossible to have an object whose size can be compared to the size of the earth. So, even if the radius of the object is provided to us, it is neglected as it is very less than the radius of the earth.

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