Animated Solution for Physics - Gravitation: A satellite is revolving in a circular orbit at a height $h$ from the Earth's surface (radius of Earth $R,h\ll R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the Earth's gravitational field, is close to (Neglect the effect of atmosphere)

A Satellite in Low Earth Orbit

Imagine you are in a control room, monitoring a satellite peacefully gliding through space. It is in a circular orbit at a height $h$ above the Earth's surface. The problem gives us a crucial piece of information: $h\ll R$. This means the satellite is in what we call a Low Earth Orbit (LEO). Compared to the massive radius of the Earth ($R\approx 6400\text{ km}$), the height $h$ (maybe a few hundred kilometers) is practically negligible.

Because it's in a stable circular orbit, it possesses a specific orbital velocity, $v_o$. This velocity is the exact "sweet spot" speed required so that the satellite's tendency to fly off in a straight line perfectly balances the Earth's gravitational pull pulling it inward.

Breaking Free

Now, the mission changes. We want to send this satellite out into deep space, completely escaping the Earth's gravitational grip. To do this, we need to fire its thrusters and increase its speed. The target speed we need to reach is the escape velocity, $v_e$, from that specific altitude.

The question asks for the minimum increase in its orbital velocity. Mathematically, this is simply the difference between the target escape velocity and its current orbital velocity:

Let's start with the orbital velocity. The formula for a satellite at a distance $r=R+h$ from the Earth's center is:

Since we are given the approximation $h\ll R$, we can safely say $R+h\approx R$. This simplifies our orbital velocity to:

Next, we need the escape velocity from that same orbit. By applying the principle of conservation of mechanical energy (setting total energy at infinity to zero), the escape velocity from a distance $R+h$ is:

Applying the same approximation ($h\ll R$), the escape velocity becomes:

Now, we find the required boost in speed by subtracting the two:

We can factor out the common term $\sqrt{\frac{GM}{R}}$:

To match the options provided, we need to express this in terms of the acceleration due to gravity at the Earth's surface, $g$. We know the standard relation:

Substituting $gR$ for $\frac{GM}{R}$ in our equation, we arrive at the final beautiful result:

This tells us that to escape from a low Earth orbit, a satellite must increase its speed by a factor of $(\sqrt{2}-1)$, which is approximately $0.414$ or $41.4\%$. This is a fundamental concept in orbital mechanics and a favorite for competitive exams!