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 equal to one Earth radius above the 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 equal to one Earth radius (R) above the Earth's surface. Therefore, the initial height (h) from the center of the Earth is:
\[ h_{initial} = R + R = 2R \]
- The final position, just before the object hits the Earth, is at the surface of the Earth:
\[ h_{final} = R \]

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

### Step 3: Calculate initial potential energy
At the initial height (2R):
\[ U_{initial} = -\frac{GMm}{2R} \]

### Step 4: Calculate final potential energy
At the final position (R):
\[ U_{final} = -\frac{GMm}{R} \]

### Step 5: Apply conservation of mechanical energy
The total mechanical energy at the initial position is equal to the total mechanical energy at the final position:
\[ K.E_{initial} + U_{initial} = K.E_{final} + U_{final} \]
Since the object is dropped from rest, the initial kinetic energy \( K.E_{initial} = 0 \):
\[ 0 - \frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{1}{2}mv^2 \]
\[ \frac{GMm}{2R} = \frac{1}{2}mv^2 \]

### Step 7: Cancel mass and solve for \( v^2 \)
Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{2R} = \frac{1}{2}v^2 \]
Multiplying both sides by 2:
\[ \frac{GM}{R} = v^2 \]

### Step 8: Solve for \( v \)
Taking the square root of both sides:
\[ v = \sqrt{\frac{GM}{R}} \]

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