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Step-by-Step Solution
Step 1: Understand the Problem
We have a body released from a height equal to the Earth’s radius, R, above the Earth’s surface.
We want to find its velocity just before it strikes the Earth's surface.
Step 2: Apply Energy Conservation
As the body falls, its gravitational potential energy decreases and is converted into kinetic energy.
We can use the principle of conservation of energy:
$$
\Delta U = -\Delta K.
$$
Step 3: Express Gravitational Potential Energy
The gravitational potential energy of a mass $m$ at a distance $r$ from Earth’s center is
$$
U(r) = -\frac{GMm}{r},
$$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.
Step 4: Calculate Initial and Final Potential Energies
• Initial distance from Earth's center:
$$
r_{\text{initial}} = R + R = 2R.
$$
So,
$$
U_{\text{initial}} = -\frac{GMm}{2R}.
$$
• Final distance from Earth's center just before hitting the surface:
$$
r_{\text{final}} = R.
$$
Hence,
$$
U_{\text{final}} = -\frac{GMm}{R}.
$$
Step 5: Change in Potential Energy
$$
\Delta U = U_{\text{final}} - U_{\text{initial}}
= \left(-\frac{GMm}{R}\right)
- \left(-\frac{GMm}{2R}\right)
= -\frac{GMm}{2R}.
$$
The negative sign shows a decrease in potential energy.
Step 6: Relate $GM$ to $g$
At the Earth’s surface:
$$
g = \frac{GM}{R^2}
\quad\Longrightarrow\quad
GM = g\,R^2.
$$
Substituting this into $\Delta U$:
$$
\Delta U = -\frac{g\,R^2\,m}{2R}
= -\frac{m\,g\,R}{2}.
$$
Since the decrease in potential energy equals the increase in kinetic energy:
$$
K = -\Delta U = \frac{m\,g\,R}{2}.
$$
Step 7: Find the Velocity
Kinetic energy is given by
$$
K = \frac{1}{2} m v^2.
$$
Equating:
$$
\frac{1}{2} m v^2 = \frac{m\,g\,R}{2}.
$$
Simplifying for $v$:
$$
v^2 = g\,R
\quad\Longrightarrow\quad
v = \sqrt{g\,R}.
$$
Step 8: Final Answer
Thus, the velocity of the body just before it strikes the Earth’s surface is
$$
\boxed{\sqrt{g\,R}}.
$$
