A rocket is fired from the earth towards the sun. At what point on its path is the gravitational force on the rocket zero? Mass of the sun is `2 xx 10^(30)kg`, mass of the earth is `6 xx 10^(24) kg`. Neglect the effect of other planets etc. ( orbital radius `= 1.5 xx 10^(11) m )`.

Explanation

Here `M_s = 2 xx 10^30 kg , M_e = 6 xx 10^24 kg , r = 1.50 xx 10^11 m`
Let x be the distance of a point from the earth where gravitational forces on the rocket due to sun and earth become equal and opposite. Then distance of rocket from the sun =(r-x). If m is the mass of rocket then
`(GM_s m)/((r -x)^2) = (GM_e m)/(x^2) " or " ((r -x)^2)/(x^2) = (M_s)/(M_e) "or " (r-x)/(x) = sqrt(M_s/M_e) = sqrt((2 xx 10^30)/(6 xx 10^24)) = (10^3)/(sqrt3) " or " r/x - 1 = (10^3)/(sqrt3)`
or `r/x = 1 + (10^3)/(sqrt3) = (sqrt3 + 10^3)/(sqrt3) "or " x = (rsqrt3)/(sqrt3 + 10^3) = (1.5 xx 10^11 xx sqrt3)/(sqrt3 + 10^3) = (1.5 xx 1.732 xx 10^11)/(1.732 +1000) = 2.59 xx 10^8 m`