Question

The force of attraction between two bodies at a certain separation is $10N$. What will be the force of attraction between them if the separation between them is reduced to half?

• A. $2.5$ $N$
• B. $5$ $N$
• C. $20$ $N$
• D. $40$ $N$

Answer 1

The correct option is D $40$ $N$

$F_g=G.\frac{m_1\times m_2}{r^2}=10N$
where, $F=$ Gravitational force
$m_1,m_2=$ Masses of two bodies
$r=$ Distance
$G=$Gravitaional constant $=6.673\times {10}^{-11}N{m}^{2}k{g}^{-2}$
From the above equation we can conclude that if distance between the object is reduceed to half then force would become four times.
$F_g^{\prime }=G.\frac{m_1\times m_2}{r^2/4}$
Therefore, $F_g^{\prime }=4\times F_g$
So, new force $=40N$