Gravitation

Question : 12 of 25

Marks: +1, -0

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 \times 10^{30} \text{kg}

, mass of the earth

= 6 \times 10^{24} \text{kg}

. Neglect the effect of other planets etc. (orbital radius

= 1.5 \times 10^{11} \text{m}).

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}

\frac{(r-d)^2}{d^2} = \frac{M_S}{M_E}

\frac{r-d}{d} = \sqrt{\frac{M_S}{M_E}}

Given,

M_S = 2 \times 10^{30} \text{kg};

M_E = 6 \times 10^{24} \text{kg};

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

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

= \frac{10^{3}}{\sqrt{3}}

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

\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}}

= 2.6 \times 10^{8} \text{m}

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