Question

How far must a particle be on the line joining earth to sun, in order that the gravitational pull on it due to sun is counterbalanced by that due to earth. $($Given orbital radius of earth is ${10}^{8}Km$ and $M_S=3.24\times {10}^5{M}_E)$

• A. $6.4\times {10}^{5}Km$
• B. $1.75\times {10}^{2}Km$
• C. $1.75\times {10}^{9}Km$
• D. $6400Km$

Answer 1

Answer: (B)

$1.75\times {10}^{2}Km$

The distance between sun and earth $d={10}^{8}km={10}^{11}m$
Let the body is at a distance of $xm\Rightarrow$ it is at distance $({10}^{11}-x)m$.
If the net force is zero, $F_s=F_e$
So, $\frac{G{M}_{s}m}{({10}^{11}-x{)}^2}=\frac{G{M}_{e}m}{x^2}$
Solving this we get, $x=1.75\times {10}^{2}km$