Two objects have the same mass and are located near each other at a distance $r$. If the mass of one of the objects is doubled and the mass of the other object is tripped, find out the change in the gravitational attraction between them.
A. Decrease by 1/6
B. Decrease by 2/3
C. Increased by 3/2
D. Increase by 6
Answer
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Hint: Each body in the universe attracts another body and the force of attraction between the two bodies is directly proportional to the product of their masses.Gravitation or just gravity is the force of attraction between any two bodies. All the objects in the universe attract each other with a certain amount of force, but in most of the cases, the force is too weak to be observed due to the very large distance of separation.
Formula used:
The force of attraction between the two objects is given by:
$F = G\dfrac{{{m_1} \cdot {m_2}}}{{{r^2}}}$
where, $G$ is gravitational constant, ${m_1}$ mass of body 1 and ${m_2}$ mass of body 2 and $r$ is the radius between the two objects.
Complete step by step answer:
From the question, we know that the mass of the two objects is the same let us consider as $M$ and the distance between the objects is $r$.From the law of gravitation, the force of attraction between the two objects is expressed as,
$F = G\dfrac{{m \cdot m}}{{{r^2}}}$
Here, $G$ is gravitational constant.
Now when the mass of one of the objects is doubled and the mass of the other object is tripped, the force of attraction between the two objects is expressed as,
$
F' = G\dfrac{{\left( {2m} \right) \cdot \left( {3m} \right)}}{{{r^2}}}\\
\Rightarrow F' = 6G\dfrac{{m \cdot m}}{{{r^2}}}
$
Now we rewrite the above equation,
$\therefore F' = 6F$
Thus, the change in the gravitational attraction between them increases by 6.
Hence, option D is the correct answer.
Note:The universal constant G may be defined as the force of attraction between the two bodies per unit mass and unit separation between the two bodies. Its value is $6.67 \times {10^{ - 11}}\;{\rm{N}}{{\rm{m}}^2}{\rm{/k}}{{\rm{g}}^2}$. Do not get confused between G and g (gravitational acceleration of earth) G is the acceleration produced in an object because of gravitational acceleration of earth.
Formula used:
The force of attraction between the two objects is given by:
$F = G\dfrac{{{m_1} \cdot {m_2}}}{{{r^2}}}$
where, $G$ is gravitational constant, ${m_1}$ mass of body 1 and ${m_2}$ mass of body 2 and $r$ is the radius between the two objects.
Complete step by step answer:
From the question, we know that the mass of the two objects is the same let us consider as $M$ and the distance between the objects is $r$.From the law of gravitation, the force of attraction between the two objects is expressed as,
$F = G\dfrac{{m \cdot m}}{{{r^2}}}$
Here, $G$ is gravitational constant.
Now when the mass of one of the objects is doubled and the mass of the other object is tripped, the force of attraction between the two objects is expressed as,
$
F' = G\dfrac{{\left( {2m} \right) \cdot \left( {3m} \right)}}{{{r^2}}}\\
\Rightarrow F' = 6G\dfrac{{m \cdot m}}{{{r^2}}}
$
Now we rewrite the above equation,
$\therefore F' = 6F$
Thus, the change in the gravitational attraction between them increases by 6.
Hence, option D is the correct answer.
Note:The universal constant G may be defined as the force of attraction between the two bodies per unit mass and unit separation between the two bodies. Its value is $6.67 \times {10^{ - 11}}\;{\rm{N}}{{\rm{m}}^2}{\rm{/k}}{{\rm{g}}^2}$. Do not get confused between G and g (gravitational acceleration of earth) G is the acceleration produced in an object because of gravitational acceleration of earth.
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