A rocket is fired from the Earth towards the Sun. At what distance from the Earth's center, the gravitational force on the rocket is zero? Mass of the Sun $= 2 \times 10^{30} k g$ and mass of the Earth $= 6 \times 10^{24} k g$ .

2.6 \times 10^{8} m

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3.2 \times 10^{8} m

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3.9 \times 10^{9} m

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2.3 \times 10^{9} m

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Solution

The correct option is A $2.6 \times 10^{8} m$
Let $P$ be a point at a distance $r$ from the Earth's center, where gravitational force due to Sun and Earth are equal and opposite and hence, gravitational force on the rocket is zero.
$\frac{G M_{s} m}{{(x - r)}^{2}} = \frac{G M_{e} m}{r^{2}} \Rightarrow \frac{G M_{s}}{{(x - r)}^{2}} = \frac{G M_{e}}{r^{2}}$
$\frac{2 \times 10^{30}}{{(x - r)}^{2}} = \frac{6 \times 10^{24}}{r^{2}}$
$\frac{{(x - r)}^{2}}{r^{2}} = \frac{2 \times 10^{30}}{6 \times 10^{24}} = \frac{1}{3} \times 10^{6}$
$\Rightarrow \frac{x - r}{r} = \frac{10^{3}}{\sqrt{3}} \Rightarrow r = (\frac{3}{1735}) x$
$\Rightarrow r = \frac{3}{1735} \times 1.5 \times 10^{11} = \frac{3 \times 15 \times 100 \times 10^{8}}{1735}$
$r = 2.594 \times 10^{8} m$
$r = 2.6 \times 10^{8} m$ from Earth.

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