
If the distance between two masses be doubled then the force between them will become
$\begin{align}
& \text{A}\text{. }\dfrac{\text{1}}{\text{4}}\text{ times} \\
& \text{B}\text{. 4 times} \\
& \text{C}\text{. }\dfrac{\text{1}}{\text{2}}\text{ times} \\
& \text{D}\text{. 2 times} \\
\end{align}$
Answer
509.4k+ views
Hint: Newton gave the law of gravitation. According to Newton’s law of gravitation the force between two masses is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. First take the distance between the masses be $r$ and calculate the gravitational force. Then take the distance $2r$ and calculate the force. Compare both the forces to get the relationship between them.
Formula used:
The gravitational force of attraction between two masses ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $R$ is given by
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{R}^{2}}}$
Where $G$ is the universal gravitational constant and its value in S.I unit is given by
$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Complete step-by-step answer:
Consider the two masses be ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $r$
Then according to Newton’s law of gravitation the force of attraction between them is given by
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
If the distance between them is doubled then the new distance is $r'=2r$
Now the gravitational force of attraction between them is given by
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{r{{'}^{2}}}=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{(2r)}^{2}}}=G\dfrac{{{m}_{1}}{{m}_{2}}}{4{{r}^{2}}}$
Now comparing it with the initial force $F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$ we get
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{4{{r}^{2}}}=\dfrac{F}{4}$ i.e if the distance is doubled then the force between the masses will become ${{\dfrac{1}{4}}^{th}}$of the original force.
So the correct option is $\text{A}\text{. }\dfrac{\text{1}}{\text{4}}\text{ times}$
So, the correct answer is “Option A”.
Additional Information: Gravitation: Everybody in the universe attracts every other body with a force called force of gravitation which is directly proportional to the product of the masses and inversely proportional to the square of distance between them. It was given by Newton thus called Newton’s law of gravitation.
If the two masses are ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $r$. Then according to the law the force between them is proportional to the product of the masse and inversely proportional to the square of distance between them. i.e.
$F\propto {{m}_{1}}{{m}_{2}}\text{ and }F\propto \dfrac{1}{{{r}^{2}}}$
Combining these two we get
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\text{ or, }F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where G is a constant called universal gravitational constant and its value is $G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Note: These are some important property of gravitational force
(i) The gravitational force between two masses is independent upon intervening medium.
(ii)The mutual gravitational force between
the two bodies are equal and opposite. i.e they form an action reaction pair. Hence the gravitational force obeys Newton’s third law.
(iii) The law of gravitation strictly holds for point masses(masses are considered to reside at the centre of a body ).
(iv)The gravitational force is always attractive and the central force i.e. acts along the line joining the centre of the masses.
(v)The gravitational force is conservative in nature. I.e. The work one around a closed path is zero.
Formula used:
The gravitational force of attraction between two masses ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $R$ is given by
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{R}^{2}}}$
Where $G$ is the universal gravitational constant and its value in S.I unit is given by
$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Complete step-by-step answer:
Consider the two masses be ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $r$
Then according to Newton’s law of gravitation the force of attraction between them is given by
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
If the distance between them is doubled then the new distance is $r'=2r$
Now the gravitational force of attraction between them is given by
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{r{{'}^{2}}}=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{(2r)}^{2}}}=G\dfrac{{{m}_{1}}{{m}_{2}}}{4{{r}^{2}}}$
Now comparing it with the initial force $F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$ we get
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{4{{r}^{2}}}=\dfrac{F}{4}$ i.e if the distance is doubled then the force between the masses will become ${{\dfrac{1}{4}}^{th}}$of the original force.
So the correct option is $\text{A}\text{. }\dfrac{\text{1}}{\text{4}}\text{ times}$
So, the correct answer is “Option A”.
Additional Information: Gravitation: Everybody in the universe attracts every other body with a force called force of gravitation which is directly proportional to the product of the masses and inversely proportional to the square of distance between them. It was given by Newton thus called Newton’s law of gravitation.
If the two masses are ${{m}_{1}}\text{ and }{{m}_{2}}$ separated by a distance $r$. Then according to the law the force between them is proportional to the product of the masse and inversely proportional to the square of distance between them. i.e.
$F\propto {{m}_{1}}{{m}_{2}}\text{ and }F\propto \dfrac{1}{{{r}^{2}}}$
Combining these two we get
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\text{ or, }F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where G is a constant called universal gravitational constant and its value is $G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
Note: These are some important property of gravitational force
(i) The gravitational force between two masses is independent upon intervening medium.
(ii)The mutual gravitational force between
the two bodies are equal and opposite. i.e they form an action reaction pair. Hence the gravitational force obeys Newton’s third law.
(iii) The law of gravitation strictly holds for point masses(masses are considered to reside at the centre of a body ).
(iv)The gravitational force is always attractive and the central force i.e. acts along the line joining the centre of the masses.
(v)The gravitational force is conservative in nature. I.e. The work one around a closed path is zero.
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