Question

A rocket is fired from the earth towards the sun. At what distance from the earths centre is the gravitational force on the rocket zero? Mass of the sun = $2\times {10}^{30}kg$, mass of the earth = $6\times {10}^{24}kg$. Neglect the effect of other planets etc. (orbital radius = $1.5\times {10}^{11}m$).

Answer 1

Mass of the Sun, $M_s=2\times {10}^{30}kg$

Mass of the Earth, $M_e=6\times {10}^{24}kg$

Orbital radius, $r=1.5\times {10}^{11}m$

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:

$\frac{Gm{M}_s}{(r-x{)}^2}=\frac{Gm{M}_e}{x^2}$

${\left[\frac{(r-x)}{x}\right]}^2=M_s/M_e$

$(r-x)/x={\left(\frac{2\times {10}^{30}}{60\times {10}^{24}}\right)}^{1/2}=577.35$

$1.5\times {10}^{11}-x=577.35x$

$578.35x=1.5\times {10}^{11}$

$x=1.5\times {10}^{11}/578.35=2.59\times {10}^{8}m$.