### Q.12:- A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 ×10^{30} kg, mass of the earth = 6 × 10^{24} kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10^{11} m).

**Answer:-**Mass of the Sun, *M*_{s} = 2 × 10^{30} kg

Mass of the Earth, *M*_{e} = 6 × 10 ^{24} kg

Orbital radius,* r* = 1.5 × 10^{11} m

Mass of the rocket = *m*

Let

*x*be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:

G*mM*_{s} / (*r* – *x*)^{2} = G*mM*_{e} / *x*^{2}

[ (*r* – *x*) / *x* ]^{2} = *M*_{s} / *M*_{e}

(*r* – *x*) / *x* = [ 2 × 10^{30} / 60 × 10^{24}]^{1/2} = 577.35

1.5 × 10^{11} – *x* = 577.35*x*

578.35*x* = 1.5 × 10^{11}

*x* = 1.5 × 10^{11} / 578.35 = 2.59 × 10^{8} m.