Ncert Exercises & Solutions

8.12 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 = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Given,

Mass of the Sun,

$M_{s} = 2 \times 10^{30} kg$

Mass of the Earth,

$M_{e} = 6 \times 10^{24} kg$

Orbital radius,

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

Let the mass of the rocket be m.

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

Then,

$\frac{{GmM}_{s}}{{(r - x)}^{2}} = \frac{{GmM}_{e}}{x^{2}}$
${(\frac{r - x}{x})}^{2} = \frac{M_{s}}{M_{e}}$
$\Rightarrow \frac{r - x}{x} = \sqrt{(\frac{2 \times 10^{30}}{60 \times 10^{24}})} = 577.35$
$\Rightarrow 1.5 \times 10^{11} - x = 577.35 x$
$\Rightarrow 578.35 x = 1.5 \times 10^{11}$
$\Rightarrow x = \frac{1.5 \times 10^{11}}{578.35} = 2.59 \times 10^{8} m .$

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