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

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Solution

Let $d$ be the distance of a point from the earth where gravitational forces on the rocket due to the sun and the earth become equal and opposite.

Then distance of rocket from the sun = $(r - d)$ .

if $m$ is the mass of rocket, then

$\frac{G M_{s} m}{(r - d)^{2}} = \frac{G M_{E} m}{d^{2}} or \frac{(r - d)^{2}}{d^{2}} = \frac{M_{S}}{M_{E}} or \frac{r - d}{d} = \sqrt{\frac{M_{S}}{M_{E}}}$

Given, $M_{S} = 2 \times 10^{30} k g, M_{E} = 6 \times 10^{24} k g$ ,

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

$∴ \frac{r - d}{d} = \sqrt{\frac{2 \times 10^{30}}{6 \times 10^{24}}} = \frac{10^{3}}{\sqrt{3}}$

or $\frac{r}{d} - 1 = \frac{10^{3}}{\sqrt{3}} or \frac{r}{d} = 1 + \frac{10^{3}}{\sqrt{3}} = \frac{\sqrt{3} + 10^{3}}{\sqrt{3}}$

$d = \frac{r \sqrt{3}}{\sqrt{3} + 10^{3}} = \frac{1.5 \times 10^{11} \times \sqrt{3}}{\sqrt{3} + 10^{3}}$

$d = 2.6 \times 10^{8} m$

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A rocket is fired from the earth towards the sun. At what distance from the earth's center is gravitational force on the rocket zero? Mass of the sun =2×1030kg, mass of the earth =6×1024kg. Neglect the effect of other planets etc. (orbital radius=1.5×1011m).