question_answer

A rocket is fired from the Earth towards the Sun. At what distance from the Earths centre, the gravitational force on the rocket is zero? Mass of the Sun \[=2\times {{10}^{30}}kg\] and mass of the Earth \[=6\times {{10}^{24}}kg\].

A) \[2.6\times {{10}^{8}}m\]         

B)  \[3.2\times {{10}^{8}}m\]

C)  \[3.9\times {{10}^{9}}m\]

D)  \[2.3\times {{10}^{9}}m\]

Correct Answer: A

Solution :

Let P be a point at a distance r from the Earths centre, where gravitational force due to Sun and Earth are equal and opposite and hence, gravitational force on the rocket is zero.                 \[\frac{G{{M}_{s}}m}{{{(x-r)}^{2}}}=\frac{G{{M}_{e}}m}{{{r}^{2}}}\]                 \[\frac{G{{M}_{s}}}{{{(x-r)}^{2}}}=\frac{G{{M}_{e}}}{{{r}^{2}}}\]                 \[\frac{2\times {{10}^{30}}}{{{(x-r)}^{2}}}=\frac{6\times {{10}^{24}}}{{{r}^{2}}}\]                 \[\frac{{{(x-r)}^{2}}}{r}=\frac{2\times {{10}^{30}}}{6\times {{10}^{24}}}=\frac{1}{3}\times {{10}^{6}}\] \[\Rightarrow \] \[\frac{x-r}{r}=\frac{{{10}^{3}}}{\sqrt{3}}\Rightarrow r=\left( \frac{3}{1735} \right)x\] \[\Rightarrow \] \[r=\frac{3}{1375}\times 1.5\times {{10}^{11}}\]                 \[=\frac{3\times 15\times 100\times {{10}^{8}}}{1735}\]                 \[r=2.594\times {{10}^{8}}\,m\] \[r=2.6\times {{10}^{8}}\,m\] from Earth