Question

Mass particles of $1\text{kg}$ each are placed along $x-$ axis at $x=1,2,4,8,.....\infty$. Then gravitational force on a mass of $3\text{kg}$ placed at origin is ($G=$ universal gravitational constant)

• A. $4\text{G}$
• B. $\frac{4G}{3}$
• C. $2\text{G}$
• D. $\infty$

Answer 1

The correct option is A $4\text{G}$

By superposition theorem,
Net gravitational force is

$F_g=(\frac{GM{m}_1}{r_1^2}+\frac{GM{m}_2}{r_2^2}+....\infty )$

Since $m_1=m_2=.........\infty =m=1kg$
$F_g=GMm(\frac{1}{r_1^2}+\frac{1}{r_2^2}+....\infty )$
$F_g=G\left(1\right)\left(3\right)(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{4^2}+....\infty )$
$F_g=3G(1+\frac{1}{4}+\frac{1}{16}+....\infty )$
$F_g=3G\left(\frac{1}{1-\frac{1}{4}}\right)=4G$