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).

#### Solution 1

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:

`(GmM_s)/(r-x)^2 = GmM_e/x^2`

((r-x)/x)^2 = `M_s/M_e`

`(r-x)/r = ((2xx10^(30))/(60xx10^(24)))^(1/2) = 577.35`

`1.5 xx 10^(11)-x=577.35x`

`x = (1.5xx10^11)/578.35` = `2.59 xx 10^8` m

#### Solution 2

Mass of Sun, M = 2 x 10^{30} kg; Mass of Earth, m = 6 x 10^{24} kg Distance between Sim and Earth, r = 1.5 x 10^{11} m

Let at the point P, the gravitational force on the rocket due to Earth = gravitional force on the rocket due to sun

Let x = distance of the point { from the Earth

Then `(Gm)/x^2 = (GM)/(r-x)^2`

`=>(r-x)^2/x^2 = M/m = (2xx10^(30))/(6xx10)^24 = 10^6/3`

or `(r-x)/x = 10^3/sqrt3 => r/x = 10^3/sqrt3 + 1 =~ 10^3/sqrt3`

or `x = (sqrt3r)/10^3 = (1.732xx1.5xx10^11)/10^3 = 2.6 xx 10^8` m