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Step-by-Step Solution
Step 1: Express the Potential Energy Required (E₁)
The energy required to move a satellite of mass m from the Earth’s surface (radius R) to an altitude h above the surface is the change in gravitational potential energy:
$E_1 = U_h - U_\text{surface} = \left(-\frac{GMm}{R + h}\right) - \left(-\frac{GMm}{R}\right)$
Rewriting, we get:
$E_1 = GMm \left(\frac{1}{R} - \frac{1}{R + h}\right)\,.$
We can factor out $h$ from the denominator difference to get:
$E_1 = \frac{GMm}{R(R + h)} \times h\,.$
Step 2: Express the Kinetic Energy Required for Circular Orbit (E₂)
For a satellite to orbit at a height h, the required centripetal force is provided by gravity:
$\frac{mv^2}{R+h} = \frac{GMm}{(R+h)^2}\,.$
Solving for $v^2$:
$mv^2 = \frac{GMm}{R + h}\quad \Longrightarrow \quad v^2 = \frac{GM}{R + h}\,.$
The kinetic energy in orbit is therefore:
$E_2 = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{GM}{R+h}\right)
= \frac{GMm}{2(R+h)}\,.$
Step 3: Equate Potential Energy Change (E₁) and Kinetic Energy (E₂)
According to the problem statement, $E_1$ and $E_2$ are equal:
$\frac{GMm}{R(R + h)} \times h = \frac{GMm}{2(R + h)}\,.$
Canceling common terms $GMm/(R+h)$ on both sides, we get:
$\frac{h}{R} = \frac{1}{2}\,.$
Step 4: Solve for the Required Height h
From the above relationship, we find:
$h = \frac{R}{2}\,.$
Since the Earth’s radius R is given as $6.4 \times 10^3\text{ km}$, we have:
$h = \frac{6.4 \times 10^3\text{ km}}{2} = 3.2 \times 10^3\text{ km}\,.$
Final Answer
Thus, the value of h for which the energy needed to lift the satellite equals the kinetic energy required for its circular orbit is:
$\boxed{3.2 \times 10^3\text{ km}}$
