Question

A rocket is fired from the earth towards the sun. At what distance from the earth’s 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. $2.3\times {{10}^{9}}m$

Answer 1

Hint: Rocket motion is based on Newton's third law. But we are about to use Newton’s gravitational force formula which describes the force between two particle masses. It is also known as Newton's law of universal gravitation.Let us calculate the gravitational force of both Earth and sun. Then we will equate them to satisfy the zero force condition.

Formula used: $F=\dfrac { GMm }{ { r }^{ 2 } }$

Complete step by step answer:
The law of universal gravitation states that every mass attracts every other mass in the universe with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them. It is given as,

$F=\dfrac { GMm }{ { r }^{ 2 } }$
Where, ‘G’ is the gravitational constant with value of about $6.674\times {{10}^{-11}}N$
      ‘M’ and m are the two masses in kilograms
      ‘r’ is the distance between the masses in meters
Let M1 be the mass of the sun = $2\times {{10}^{30}}kg$
And M2 be the mass of the earth = $6\times {{10}^{24}}kg$

‘r’ is the orbital distance of earth around the sun, which is equal to $1.5\times {{10}^{11}}m$.

We have to assume that ‘x’ is the distance from the earth’s center where the gravitational force on the rocket is zero.

Gravitational force of the earth = $\dfrac { G{ M }_{ 2 }m }{ { x }^{ 2 } }$
Gravitational force of the sun = $\dfrac { G{ M }_{ 1 }m }{ { \left( r-x \right) }^{ 2 } }$

Using Newton’s law and we are equating both the above-mentioned equations,

$\dfrac { G{ M }_{ 1 }m }{ { \left( r-x \right) }^{ 2 } } =\dfrac { G{ M }_{ 2 }m }{ { x }^{ 2 } }$
On cancelling the like terms on both sides and rearranging,
${ \left( \dfrac { r-x }{ x } \right) }^{ 2 }=\dfrac { { M }_{ 1 } }{ { M }_{ 2 } }$
Taking square root on both sides,
${ \left( \dfrac { r-x }{ x } \right) }=\sqrt { \dfrac { 2\times { 10 }^{ 30 } }{ 6\times { 10 }^{ 24 } } }$
By solving the calculations, we get

$1.5\times {{10}^{11}}-x=577.35x$
$1.5\times {{10}^{11}}=577.35x+x$
On taking ‘x’ common in the R.H.S
$1.5\times {{10}^{11}}=\left( 577.35+1 \right)x$
$x=\dfrac { 1.5\times { 10 }^{ 11 } }{ 578.35 }$
$x=2.59\times {{10}^{8}}m$
Therefore, the correct answer for the given question is option (A).

Note: The value is rounded off. At a height of 1000km from the Earth’s surface, the gravitational force on any mass will be about 75% of its value on Earth. Which means, even though a rocket reaches space, the force of gravity will still try to pull it back towards Earth.