Animated Solution for Physics - Gravitation: The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where, $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth is

Imagine you are an astronaut blasting off from the surface of the Earth. As your rocket climbs higher and higher into the atmosphere and eventually into the vacuum of space, you feel a profound change. The invisible tether that binds you to the planet—gravity—begins to weaken.

But how exactly does it weaken? It doesn't just fade away randomly; it follows a strict mathematical law dictated by the universe. The acceleration due to gravity at any height $h$ above the Earth's surface is given by the formula:

Here, $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth. Notice how the distance from the center of the Earth is $(R+h)$.

We already know that on the surface of the Earth (where $h=0$), the gravity is simply:

If we divide our first equation by this surface gravity equation, the constants $G$ and $M$ beautifully cancel out, leaving us with a powerful, purely geometric relationship:

This is our master equation. It tells us exactly how gravity scales with height, independent of the Earth's mass!

The problem asks us to find the specific height $h$ where the gravity $g^{\prime }$ is exactly one-ninth of the surface gravity $g$. Mathematically, this means $g^{\prime }=\frac{g}{9}$.

Let's substitute this condition into our master equation:

The $g$ on both sides is just begging to be cancelled out. Let's oblige:

Now, to liberate our variables from the square, we take the square root of both sides. Since distance cannot be negative, we only consider the positive root:

From here, it's a simple matter of cross-multiplication:

Subtracting $R$ from both sides, we arrive at our grand conclusion:

So, you have to travel a distance equal to twice the Earth's radius above the surface to experience a gravity that is one-ninth as strong.

You might recall another formula for gravity at a height: $g^{\prime }\approx g\left(1-\frac{2h}{R}\right)$. Why didn't we use that?

That formula is an approximation derived using the binomial theorem, and it is only valid when $h$ is much, much smaller than $R$ (typically $h<5\%\text{ of }R$). In our case, the gravity dropped significantly (to $1/9$), which implies a very large height. Using the approximation here would lead to a disastrously wrong answer. Always trust the exact master equation when dealing with large distances!