An object is dropped from an altitude of one Earth radius above Earth’s surface. If M is the mass of Earth and R is its radius. The speed of the object just before it hits Earth is given by:

Options

A. `sqrt(GM//R)`
B. `sqrt(GM//2R)`
C. `sqrt(2GM//R)`
D. `sqrt(GM//R^2)`

Explanation

To find the speed of an object just before it hits the Earth when dropped from an altitude of one Earth radius above the Earth's surface, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Define the initial and final positions
The object is dropped from a height of \( h = R \) (one Earth radius) above the Earth's surface. Therefore, the initial distance from the center of the Earth is:
\[ d_{\text{initial}} = R + R = 2R \]
where \( R \) is the radius of the Earth.

### Step 2: Calculate initial potential energy
The gravitational potential energy (U) at a distance \( d \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{d} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the object.

The initial potential energy when the object is at a height of \( 2R \) is:
\[ U_{\text{initial}} = -\frac{GMm}{2R} \]

### Step 3: Calculate final potential energy
When the object reaches the surface of the Earth, the distance from the center of the Earth is \( R \). Thus, the final potential energy is:
\[ U_{\text{final}} = -\frac{GMm}{R} \]

### Step 4: Use conservation of energy
According to the conservation of energy, the total mechanical energy remains constant. Therefore, the change in potential energy equals the change in kinetic energy. The change in potential energy (\( \Delta U \)) is:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = -\frac{GMm}{R} - \left(-\frac{GMm}{2R}\right) \]
\[ \Delta U = -\frac{GMm}{R} + \frac{GMm}{2R} = -\frac{GMm}{R} + \frac{GMm}{2R} = -\frac{GMm}{2R} \]

### Step 5: Relate potential energy change to kinetic energy
The change in kinetic energy (\( \Delta K \)) is given by:
\[ \Delta K = K_{\text{final}} - K_{\text{initial}} = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2 \]

Setting the change in potential energy equal to the change in kinetic energy:
\[ -\frac{GMm}{2R} = \frac{1}{2}mv^2 \]

### Step 6: Solve for velocity \( v \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{2R} = \frac{1}{2}v^2 \]
Multiplying both sides by 2 gives:
\[ -\frac{GM}{R} = v^2 \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

### Final Answer
The speed of the object just before it hits the Earth is:
\[ v = \sqrt{\frac{GM}{R}} \]