Question

A rocket is fired from the earth towards the sun. At what distance from the earth Center 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\[=1.5\times {{10}^{11}}\text{ M}\]

Answer 1

Hint: As in this question the rocket is fired from the earth towards the sun at one point where the gravitational Force is zero. Force on the rocket due to earth as well as by sun should be equal and in the opposite direction so the rocket will be in equilibrium here.

Complete step-by-step answer:
Given Mass of the sun (ms) \[=2\times {{10}^{30}}\text{ kg}\]
Mass of the Earth (Me) \[=6\times {{10}^{24}}\text{ kg}\]
Orbital radius (r) \[=1.5\times {{10}^{11}}\text{ M}\]
Mass of rocket = m
As the Rocket is fired from earth (E) towards sun (s) at point (P) the distance from center of earth
The Force acts on the rocket are
These two gravitational forces are in opposite directions.
As we know gravitational force between two objects which are at distance R from each other.
\[F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{R}^{2}}}\]
G = universal gravitational constant
\[{{m}_{1}}{{m}_{2}}=\] mass of objects
R = Distance between them
By the gravitational Force
Force on Rocket due to Earth = Force on Rocket due to sun
\[\Rightarrow \dfrac{GMeM}{{{x}^{2}}}=\dfrac{GMsm}{{{\left( r-x \right)}^{2}}}\]
By simplification:
\[\Rightarrow \dfrac{6\times {{10}^{24}}}{{{x}^{2}}}=\dfrac{2\times {{10}^{30}}}{{{\left( r-x \right)}^{2}}}\]
\[\Rightarrow {{\left( \dfrac{r-x}{x} \right)}^{2}}=\dfrac{2\times {{10}^{30}}}{6\times {{10}^{24}}}\]
\[\Rightarrow {{\left( \dfrac{r-x}{x} \right)}^{2}}=\dfrac{{{10}^{6}}}{3}\]
\[\Rightarrow \dfrac{r-x}{x}=\dfrac{+{{10}^{3}}}{1.732}\]
\[\Rightarrow 1.732\left( r-x \right)=\pm {{10}^{3}}x\]
So,
\[\Rightarrow 1.732\left( r-x \right)={{10}^{3}}x\]
\[\Rightarrow 1.732r={{10}^{3}}x+1.732x\text{ }..........\text{ 1}\]
And \[1.732\left( r-x \right)=-{{10}^{3}}x\]
\[\Rightarrow 1.732r=-{{10}^{3}}\text{x+1}\text{.732x }..........\text{ 2}\]
Distance will never be Negative so the equation 2 neglected
So by equation 1
\[1.732r=x\left( 1.732+{{10}^{3}} \right)\]
As we know \[1.732<<<{{10}^{3}}\]
So, \[x=\dfrac{1.732\times 1.5\times {{10}^{11}}}{{{10}^{3}}}\text{ }\left[ r=1.5\times {{10}^{11}} \right]\]
\[x=2.6\times {{10}^{8}}\text{M}\]
Therefore, at the distance \[2.6\times {{10}^{8}}\text{M}\] from the earth's center the gravitational Force will be Zero.

Note: In this question at the rocket the force due to earth and sun will be equal and in opposite directions.