Animated Solution for Physics - Gravitation: The energy required to take a satellite to a height '$h$' above earth surface (where, radius of earth $=6.4\times 10^3$ km) is $E_1$ and kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1$ and $E_2$ are equal is

Imagine you are standing on the surface of the Earth, looking up at a satellite that needs to be launched into a circular orbit at a specific height $h$. The problem presents us with two distinct energy requirements. First, the energy $E_1$ required to simply lift the satellite from the Earth's surface to that height $h$. Second, the kinetic energy $E_2$ required to keep that satellite moving in a stable circular orbit at that exact height.

Our mission is to find the specific height $h$ where these two energy values are perfectly equal. Let's break this down step by step.

When we lift a satellite, we are doing work against the Earth's gravitational pull. This work done is stored as the change in the satellite's gravitational potential energy.

The gravitational potential energy $U$ of a system is given by the formula:

where $G$ is the universal gravitational constant, $M_e$ is the mass of the Earth, $m$ is the mass of the satellite, and $r$ is the distance from the center of the Earth.

The energy $E_1$ is the difference between the final potential energy at height $h$ and the initial potential energy at the Earth's surface (where $r=R_e$).

Substituting the values, we get:

Simplifying this by handling the negative signs, we arrive at our first master equation:

Now, let's look at the second part of the puzzle. Once the satellite is at height $h$, it needs to travel at a specific orbital velocity $v_o$ to avoid falling back to Earth. This velocity is determined by equating the gravitational force to the required centripetal force.

The orbital velocity $v_o$ is given by:

The kinetic energy $E_2$ required for this circular motion is simply $\frac{1}{2}m{v}_o^2$. Substituting our expression for $v_o$, we get:

The core condition of the problem states that $E_1$ and $E_2$ are equal. Let's set our two equations equal to each other:

Notice how beautifully the physics simplifies the math! The term $G{M}_{e}m$ is common to every single part of the equation. We can safely divide the entire equation by $G{M}_{e}m$, leaving us with a pure geometric relationship:

Let's move the terms containing $h$ to the right side of the equation to group them together:

Taking the common denominator on the right side, we get:

Now, we cross-multiply to solve for $h$:

Subtracting $2R_e$ from both sides, we find a remarkably elegant result:

We have found that the required height is exactly half the radius of the Earth. The problem provides the radius of the Earth as $R_e=6.4\times 10^3\text{ km}$.

Substituting this value into our result:

This matches option (a) perfectly. The beauty of this problem lies in how the complex gravitational constants completely vanish, revealing a simple, fundamental geometric truth about orbits!