The gravitational force between two equal masses at a distance $r$ is $F$. When one of the masses becomes half, the distance at which it should be kept so that the gravitational force remains the same
(A) $\sqrt 2 r$
(B) $\dfrac{r}{{\sqrt 2 }}$
(C) $\sqrt 3 r$
(D) $2r$
Answer
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Hint: According to Newton’s universal law of gravitation, everybody in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the product of their masses and inversely proportional to the square of the distance between them.
Formula used:
$F = \dfrac{{GMm}}{{{r^2}}}$
Where $F$ stands for the gravitational force, $G$ stands for the gravitational constant, $M$ stands for the mass of the earth, $m$stands for the mass of the other object, and $r$ stands for the distance between the two bodies.
Complete step by step answer:
The gravitational force can be written as,
$F = \dfrac{{GMm}}{{{r^2}}}$
When one of the mass is reduced to half
$F = \dfrac{{GM\left( {\dfrac{m}{2}} \right)}}{{{x^2}}}$
We consider the distance between the two bodies to be $x$ because, when the mass changes the distance also changes so the changed distance is given as $x$.
We need to equate the two equations for gravitational force,
$F = \dfrac{{GMm}}{{{r^2}}} = \dfrac{{GM\left( {\dfrac{m}{2}} \right)}}{{{x^2}}}$
Canceling the common terms
$\dfrac{1}{{{r^2}}} = \dfrac{1}{{2{x^2}}}$
Taking the reciprocal
${r^2} = 2{x^2}$
From this,
${x^2} = \dfrac{{{r^2}}}{2}$
Taking the square root,
$x = \dfrac{r}{{\sqrt 2 }}$
The answer is: Option (B): $\dfrac{r}{{\sqrt 2 }}$
Additional Information:
Acceleration due to gravity is the acceleration experienced by a body falling freely towards the earth. The value of $g$ is taken to be $9.8m/{s^2}$. There is a slight variation for $g$ from place to place. It varies as we go to the poles due to the nonspherical shape of the earth and the rotation of the earth.
Note:
The value of the gravitational constant is given by $G = 6.67 \times {10^{ - 11}}N{m^2}K{g^{ - 2}}$. The gravitational force is negative because it is an attractive force. Forces on the two masses are equal and opposite. When the distance between the two bodies is reduced to half the gravitational force doubles.
Formula used:
$F = \dfrac{{GMm}}{{{r^2}}}$
Where $F$ stands for the gravitational force, $G$ stands for the gravitational constant, $M$ stands for the mass of the earth, $m$stands for the mass of the other object, and $r$ stands for the distance between the two bodies.
Complete step by step answer:
The gravitational force can be written as,
$F = \dfrac{{GMm}}{{{r^2}}}$
When one of the mass is reduced to half
$F = \dfrac{{GM\left( {\dfrac{m}{2}} \right)}}{{{x^2}}}$
We consider the distance between the two bodies to be $x$ because, when the mass changes the distance also changes so the changed distance is given as $x$.
We need to equate the two equations for gravitational force,
$F = \dfrac{{GMm}}{{{r^2}}} = \dfrac{{GM\left( {\dfrac{m}{2}} \right)}}{{{x^2}}}$
Canceling the common terms
$\dfrac{1}{{{r^2}}} = \dfrac{1}{{2{x^2}}}$
Taking the reciprocal
${r^2} = 2{x^2}$
From this,
${x^2} = \dfrac{{{r^2}}}{2}$
Taking the square root,
$x = \dfrac{r}{{\sqrt 2 }}$
The answer is: Option (B): $\dfrac{r}{{\sqrt 2 }}$
Additional Information:
Acceleration due to gravity is the acceleration experienced by a body falling freely towards the earth. The value of $g$ is taken to be $9.8m/{s^2}$. There is a slight variation for $g$ from place to place. It varies as we go to the poles due to the nonspherical shape of the earth and the rotation of the earth.
Note:
The value of the gravitational constant is given by $G = 6.67 \times {10^{ - 11}}N{m^2}K{g^{ - 2}}$. The gravitational force is negative because it is an attractive force. Forces on the two masses are equal and opposite. When the distance between the two bodies is reduced to half the gravitational force doubles.
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