An object is allowed to fall from a height R above the earth, where R is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be

Options

A. `sqrt(gR)`
B. `2sqrt(gR)`
C. `sqrt(2gR)`
D. `sqrt((gR)/(2))`

Answer

To solve the problem of finding the velocity of an object when it strikes the Earth's surface after falling from a height equal to the radius of the Earth (R), we can use the principle of conservation of mechanical energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final States
The object is released from a height R above the Earth's surface. At this point (let's call it point A), it has gravitational potential energy and no kinetic energy (since it starts from rest). When it strikes the Earth's surface (point B), it will have kinetic energy and reduced gravitational potential energy.

### Step 2: Write the Conservation of Energy Equation
According to the conservation of mechanical energy:
\[ \text{Initial Potential Energy (PE)} + \text{Initial Kinetic Energy (KE)} = \text{Final Potential Energy (PE)} + \text{Final Kinetic Energy (KE)} \]
At point A:
- Potential Energy (PE) at A = \(-\frac{GMm}{2R}\) (since the height is R)
- Kinetic Energy (KE) at A = \(0\)

At point B (on the surface of the Earth):
- Potential Energy (PE) at B = \(-\frac{GMm}{R}\)
- Kinetic Energy (KE) at B = \(\frac{1}{2}mv_B^2\)

### Step 3: Set Up the Equation
Setting the initial energy equal to the final energy:
\[ -\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv_B^2 \]

### Step 4: Simplify the Equation
Rearranging the equation gives:
\[ -\frac{GMm}{2R} = -\frac{GMm}{R} + \frac{1}{2}mv_B^2 \]
Adding \(\frac{GMm}{R}\) to both sides:
\[ -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{1}{2}mv_B^2 \]
This simplifies to:
\[ \frac{GMm}{2R} = \frac{1}{2}mv_B^2 \]

### Step 5: Cancel Out the Mass
Dividing through by \(m\) (assuming \(m \neq 0\)):
\[ \frac{GM}{2R} = \frac{1}{2}v_B^2 \]
Multiplying both sides by 2:
\[ \frac{GM}{R} = v_B^2 \]

### Step 6: Solve for \(v_B\)
Taking the square root of both sides gives:
\[ v_B = \sqrt{\frac{GM}{R}} \]

### Step 7: Relate \(GM\) to \(g\)
We know that \(g = \frac{GM}{R^2}\), which allows us to express \(GM\) as:
\[ GM = gR^2 \]
Substituting this back into our equation for \(v_B\):
\[ v_B = \sqrt{\frac{gR^2}{R}} = \sqrt{gR} \]

### Final Answer
Thus, the velocity of the object when it strikes the Earth's surface is:
\[ v_B = \sqrt{gR} \]