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Step-by-Step Solution
Step 1: Identify the physical situation
An object is dropped from a height equal to Earth’s radius $R$ above the Earth’s surface. Thus, the distance from the center of the Earth at the drop point is $2R$. The problem is to determine the velocity with which it strikes the Earth’s surface, ignoring air resistance.
Step 2: Apply energy conservation
We consider the total energy (potential + kinetic) at the starting position and at the final position. Let the mass of the object be $m$, the gravitational constant be $G$, and the mass of the Earth be $M$.
Potential energy of mass $m$ at the surface of Earth (radius $R$) is $U_S = -\frac{GMm}{R}$.
Potential energy of mass $m$ at height $R$ above the Earth’s surface (i.e., at distance $2R$ from Earth's center) is $U_P = - \frac{GMm}{2R}$, since potential energy at a distance $r$ from Earth’s center is $-\frac{GMm}{r}$.
We assume initial kinetic energy is zero since the object is simply “allowed to fall.”
Step 3: Write the energy conservation equation
By conservation of mechanical energy:
$$
\left( \text{Kinetic Energy} + \text{Potential Energy} \right)_{\text{initial}}
=
\left( \text{Kinetic Energy} + \text{Potential Energy} \right)_{\text{final}}
$$
Initially (at height $2R$):
$$
K_i = 0, \quad U_i = -\frac{GMm}{2R}.
$$
Finally (at Earth's surface, radius $R$):
$$
K_f = \frac{1}{2} m v^2, \quad U_f = -\frac{GMm}{R}.
$$
Hence,
$$
0 - \frac{GMm}{2R} = \frac{1}{2} m v^2 - \frac{GMm}{R}.
$$
Step 4: Solve for the final velocity
Rearranging terms, we get:
$$
\frac{1}{2} m v^2 = -\frac{GMm}{2R} + \frac{GMm}{R}.
$$
Factor out $-\frac{GMm}{2R}$:
$$
\frac{1}{2} m v^2 = \frac{GMm}{R} - \frac{GMm}{2R}
= \frac{GMm}{2R}.
$$
Divide both sides by $\tfrac{1}{2}m$:
$$
v^2 = \frac{GM}{R}.
$$
We also know that $g = \frac{GM}{R^2}$ at the Earth’s surface, so:
$$
\frac{GM}{R} = gR.
$$
Therefore,
$$
v = \sqrt{gR}.
$$
Final Answer
The velocity of the object when it reaches the Earth’s surface is $ \sqrt{g R} $.
