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

Let $x$ be the distance from Earth's centre to the null point, so the point is $(R-x)$ from the Sun's centre, where $R$ is the Earth-Sun distance. At this point the two pulls on the rocket (mass $m$) are equal and opposite:

The rocket's mass $m$ cancels because the force law is linear in $m$ on both sides. Cross-multiplying:

Since $0<x<R$, both $x$ and $(R-x)$ are positive, so taking the square root of both sides keeps the positive root:

Expanding and collecting the $x$ terms on one side:

$\sqrt{M_E}\,R = x\left(\sqrt{M_S}+\sqrt{M_E}\right) \implies x=\dfrac{R\sqrt{M_E}}{\sqrt{M_E}+\sqrt{M_S}}$ …

Setting up

Let $x$ be the distance from Earth's centre to the point where the net gravitational force on the rocket is zero, along the line joining Earth and the Sun. The distance from the Sun's centre to that point is then $R-x$, where $R=1.5\times10^{11}$ m is the Earth-Sun distance.

Balancing the two pulls

The gravitational force from Earth on the rocket (mass $m$) is $F_E=\dfrac{GM_Em}{x^2}$, and from the Sun is $F_S=\dfrac{GM_Sm}{(R-x)^2}$. For the net force to be zero, these must be equal:

Notice that both $G$ and the rocket's own mass $m$ cancel completely — this is simply because Newton's law of gravitation makes each force directly proportional to the test mass $m$, so it cancels on both sides of the equation. (This has nothing to do with gravitational and inertial mass being equal — it's just the ordinary algebra of a force that's linear in $m$.)

Solving for $x$

Taking square roots (both $x$ and $R-x$ are positive, since the rocket is between Earth and the Sun):

Equivalently, $x=\dfrac{R\sqrt{M_E}}{\sqrt{M_E}+\sqrt{M_S}}$.

Substituting values …

Step 1: Let $x$ be the distance of the zero-force point from Earth's centre; then $R-x$ is its distance from the Sun's centre, where $R$ is the Earth–Sun distance.

Step 2: Equate the two pulls on the rocket (mass $m$, which cancels since both sides are linear in $m$): $\dfrac{GM_Em}{x^2} = \dfrac{GM_Sm}{(R-x)^2}$.

Step 3: Take square roots: $\dfrac{\sqrt{M_E}}{x} = \dfrac{\sqrt{M_S}}{R-x}$, so $R - x = x\sqrt{M_S/M_E}$, i.e. $R = x\left(1+\sqrt{M_S/M_E}\right)$.

Step 4: Solve for $x$: $x = \dfrac{R\sqrt{M_E}}{\sqrt{M_E}+\sqrt{M_S}}$. …